The Organic Chemistry Tutor · YouTube · 7:05
This 120-second whiteboard lesson introduces how to graph simple linear inequalities in the xy-plane. It works through three examples: , , and . The key ideas are that fixing x gives a vertical boundary, fixing y gives a horizontal boundary, strict inequalities use dashed lines, inequalities with equality use solid lines, and the solution is a shaded half-plane rather than only the boundary line. This video segment demonstrates how to graph linear inequalities in one and two variables. It begins by concluding the graph of , then introduces to explain that non-strict inequalities use solid boundary lines and 'less than' conditions require shading below the line. The focus then shifts to two-variable inequalities using the example . The presenter identifies the slope and y-intercept from the slope-intercept form, plots points using the rise-over-run method, draws a dashed boundary line for the strict inequality, and shades the region above the line to represent the solution set. This 120-second whiteboard segment works one algebra example: graphing the two-variable linear inequality . The narrator first reads the boundary line’s slope and y-intercept from slope-intercept form, then draws a coordinate plane and plots points on using rise 2 over run 1. Because the inequality includes equality, the boundary is drawn as a solid red line. The solution region is indicated below the line with blue downward arrows. The clip then introduces test points as an alternative way to decide shading, marking and , and substitutes the origin to obtain , showing that the origin side is not part of the solution set. This 65-second whiteboard segment works through graphing the linear inequality . It first reads the boundary line data and , then tests (0,0), obtains the false statement , and rejects that region. Next it tests (4,-2), simplifies to -2 ≤ 5, marks the statement true, and shades the half-plane containing that point. The ending states two equivalent methods: use a test point, or directly shade below the line because the inequality is solved as .
Use the learning inspector for key ideas and moments, or open the reading tabs for the complete notes.
Generated from the video's visuals and explanation; not verbatim speech.
The lesson opens by shifting from one-variable inequalities to the two-dimensional xy-plane. The narrator states that the goal is to graph inequalities in an x-y coordinate system, setting up the idea that the solution will be a region of the plane rather than a single point or interval.
The first example is . The narrator contrasts this with graphing on a number line and asks what happens when both x- and y-axes are present. This frames the central question: how does a condition on only one coordinate appear in two dimensions?
The boundary is identified by replacing the inequality with equality: . The narrator explicitly says this is a vertical line. Visually, a coordinate plane is drawn and a vertical boundary is placed at .
Because the inequality is strict, the boundary is drawn dashed rather than solid. The reasoning given in the video is that x is greater than 3 but not equal to 3, so the line itself is excluded from the solution set.
The solution region is then chosen by direction. Since larger x-values lie to the right, the narrator says to shade toward the right. On screen, blue arrows extend rightward from the dashed vertical line, indicating the half-plane .
The narrator adds a general principle: because an inequality has a range of answers, the appropriate region must be shaded. This reinforces that the graph of an inequality is not just its boundary but an entire solution set.
The second example is . Again the boundary comes from the related equation , which the narrator identifies as a vertical line. The coordinate plane is redrawn with a vertical boundary at .
Here the inequality includes equality, so the boundary is drawn solid instead of dashed. The visual distinction matches the logical one: points with satisfy the inequality and therefore belong to the graph.
Since the condition is “less than,” the solution lies toward smaller x-values, which are to the left. Green arrows are drawn leftward from the solid vertical line, showing the half-plane .
The third example changes the restricted variable from x to y: . The narrator asks how to graph it and where to shade, then explains that fixing y produces a horizontal line rather than a vertical one.
Because the inequality is strict, the horizontal boundary is drawn dashed. This mirrors the first example’s rule: exclusion of the boundary value is represented by a broken line.
Finally, the solution region is above the line, since numbers greater than -4 correspond to larger y-values. Blue upward arrows fill the region above the dashed horizontal line, completing the graph of .
The segment opens by concluding the graph of the inequality . Blue upward arrows are shown shading the region above a dashed red horizontal line at , reinforcing that strict inequalities exclude the boundary line.
The instructor introduces a new example: graphing . A solid red horizontal line is drawn at . The audio explains that because the inequality includes 'or equal to', the boundary line must be solid, indicating that points on the line are part of the solution.
To complete the graph of , blue downward arrows are drawn below the solid red line. The instructor clarifies that the 'less than' condition dictates shading the region below the boundary line, representing all y-values smaller than or equal to 2.
The lesson transitions to two-variable linear inequalities with the example . The instructor writes the general slope-intercept form y = mx + b on the screen to analyze the specific inequality.
By comparing to y = mx + b, the instructor identifies the slope (since there is no explicit number in front of x) and the y-intercept . These values are written on the screen.
The graphing process begins by plotting the y-intercept at the coordinates (0, -2) on the Cartesian plane. A red dot marks this initial point on the vertical axis.
The instructor explains the slope as a rise-over-run ratio of . Red arrows demonstrate moving up one unit and right one unit from the y-intercept to locate the next point on the boundary line, which is plotted at (1, -1). This process is repeated to plot (2, 0).
Because the inequality is strict (greater than, not equal to), the instructor draws a dashed red diagonal line through the plotted points. The dashed style visually confirms that the boundary line itself is not included in the solution set.
Finally, the shading direction is determined. Since the inequality states that y is greater than the expression x - 2, the solution region lies above the boundary line. Blue upward arrows are drawn above the dashed red line to shade the correct area, completing the graph.
The segment opens on a black screen and immediately frames itself as another worked example in graphing linear inequalities.
The inequality written at upper left is . This is a two-variable linear inequality, and its boundary line is the equation .
Because the expression is already in slope-intercept form, the narrator extracts and . Those values describe the boundary line, not yet the whole shaded solution region.
A coordinate plane is drawn. The next task is to graph the boundary line before deciding how to represent the inequality’s solution set.
Starting from the y-intercept , the slope is interpreted as . Moving up 2 and right 1 gives ; repeating the same step gives and then .
Those red points are connected with a solid red line. The solidity matters: since the inequality uses , points satisfying are included in the solution set.
The narrator then determines the shaded side. For an inequality solved for , means the allowed -values are on or below the boundary line, so blue arrows are drawn downward.
An alternative verification method is introduced: choose test points on different sides of the boundary and substitute them into the inequality. The clip marks and as sample points.
The rule stated is that if the substituted statement is true, shade the region containing that test point; if false, do not shade that region. Although the narrator says “equation,” the actual substitution is into the inequality.
Substituting the origin gives , which simplifies to . That statement is false, so the side containing is not the solution side.
The clip ends while the narrator is verbally checking whether 0 is less than negative 3, leaving the visual conclusion clear even though the final spoken sentence is cut short.
The board is working with the inequality . The boundary line is the graph of , with slope and y-intercept ; the board also writes the slope as .
To decide whether the origin belongs to the solution region, substitute into the inequality. This gives , which simplifies to .
A small number line is used to compare the two numbers. Since lies to the right of , it is greater than , so the statement is false.
Because the origin fails the inequality, the region containing is not shaded. The red X beside records that rejection.
The next test point is , shown as a green point below the boundary line. Substitute and into .
The written substitution is . First compute , giving ; then compute , giving .
Now compare the numbers: is less than . Equivalently, is greater than . Either reading expresses the same true order relation.
Since the substituted statement is true, the point lies in the solution set, so the half-plane containing that point should be shaded. The red check mark beside records success.
The clip closes by naming two equivalent strategies for this problem: use a test point, or read the solved inequality directly. Because the inequality is , the solution is the region at or below the boundary line.
This introductory card captures the main setup of the clip: inequalities are being represented in a two-dimensional coordinate system. The video emphasizes that the result is a region of the plane, not merely a line or a point.
When only the x-coordinate is fixed, the boundary in the xy-plane is vertical. The first two examples use this fact: and both begin by identifying vertical boundary lines at and .
When only the y-coordinate is fixed, the boundary is horizontal. The third example uses this fact by turning into the boundary line before deciding where to shade.
If the inequality does not include equality, the boundary line is dashed. The video demonstrates this with and , where the narrator explicitly says the line is dashed because the endpoint value itself is not included.
If the inequality includes equality, the boundary line is solid. The example shows this directly: because the relation is “less than or equal to,” the vertical boundary is drawn as a solid line.
When the inequality restricts x, the solution is a left or right half-plane. Larger x-values are to the right, so is shaded rightward; smaller x-values are to the left, so is shaded leftward.
When the inequality restricts y, the solution is an upper or lower half-plane. In the example , the narrator says the numbers greater than -4 are above the line, so the region above the dashed horizontal boundary is shaded.
The clip repeatedly stresses that an inequality has a range of answers. Therefore the graph is not just the boundary line but the appropriate shaded region representing all ordered pairs that satisfy the inequality.
When graphing linear inequalities, the style of the boundary line depends on the inequality symbol. If the inequality includes equality (≤ or ≥), the boundary line is drawn as a solid line, meaning points on the line are part of the solution. If the inequality is strict (< or >), the boundary line is drawn as a dashed line, meaning points on the line are excluded from the solution.
The direction of shading indicates which side of the boundary line contains the solution set. For inequalities involving only y (like or ), 'less than' means shade below the horizontal line, and 'greater than' means shade above. For two-variable inequalities (like ), if y is greater than the expression, shade above the boundary line; if y is less than the expression, shade below.
Linear inequalities in two variables can often be analyzed using the slope-intercept form, y = mx + b. By rewriting or comparing the inequality to this form, you can easily identify the slope (m) and the y-intercept (b) of the boundary line. For example, in , the slope is 1 and the y-intercept is -2.
To graph an inequality like , first identify the slope and y-intercept from the slope-intercept form. Plot the y-intercept on the coordinate plane. Use the slope (rise over run) to plot additional points on the boundary line. Draw the boundary line as solid or dashed depending on the inequality symbol. Finally, shade the region above or below the line based on whether y is greater than or less than the expression.
The video graphs a linear inequality in two variables. The boundary line is , and the inequality includes all points where the -coordinate is less than or equal to that line’s value.
From , the boundary line has slope and y-intercept . These values are used to plot the line before shading the inequality region.
Starting at , the slope is treated as . This produces the points , , and on the line .
Because the inequality symbol is , equality is allowed. Therefore the boundary line is included in the solution set and is drawn as a solid line rather than a dashed line.
When the inequality is solved for , the relation means the solution lies on or below the boundary line. The video marks this with downward blue arrows.
Another way to determine the correct half-plane is to substitute a test point into the inequality. If the resulting statement is true, shade the region containing that point; if false, do not shade it.
Substituting the origin gives , which simplifies to . This is false, so the side of the boundary containing is not part of the solution set.
The inequality is graphed by first drawing its boundary line . Since the inequality symbol is , points on the line are included, so the boundary is solid rather than dashed.
From , the coefficient of is the slope , and the constant term is the y-intercept . The board also writes the slope as .
Choose a point not on the boundary line, substitute its coordinates into the inequality, and check whether the resulting numerical statement is true. A true statement means the region containing that point is shaded; a false statement means it is not.
Substituting the origin gives , hence . On a number line, is to the right of , so and the inequality is false. Therefore the region containing is not shaded.
Substituting gives , then , then . Since is less than , the statement is true, so the half-plane containing is the solution region.
Explore conditions, steps and evidence. Supplementary explanations are labeled separately from content shown in the video.
The inequality is written at the upper left of the screen.
The narrator says, "So let's say that x is greater than three."
x
Horizontal coordinate in the xy-plane; in this example it is the variable constrained by the inequality.
Real numbers
The inequality is written at the upper left of the screen.
The narrator says, "Now let's say if y is greater than negative four."
y
Vertical coordinate in the xy-plane; in this example it is the variable constrained by the inequality.
Real numbers
The narrator explains that because x is greater than 3 but not equal to it, a dashed line is used and shading goes to the right.
Strict inequality stating that the horizontal coordinate must be larger than 3.
All points (x,y) with
The narrator says, "It's less than or equal to negative two," and explains that the boundary is a solid vertical line shaded to the left.
Non-strict inequality stating that the horizontal coordinate is at most -2.
All points (x,y) with
The narrator says, "Now let's say if y is greater than negative four," and explains that the boundary is a dashed horizontal line shaded above.
Strict inequality stating that the vertical coordinate must be larger than -4.
All points (x,y) with
The variable y appears in the inequalities , , and .
y
vertical coordinate variable in the Cartesian plane
real numbers
The variable x appears in the inequality and in the slope-intercept form y = mx + b.
x
horizontal coordinate variable in the Cartesian plane
real numbers
The symbol m appears in y = mx + b and is assigned the value 1 for the example .
The speaker says, 'The number in front of x is the slope.'
m
slope of a line in slope-intercept form
real numbers
The symbol b appears in y = mx + b and is assigned the value -2 for the example .
The speaker says, 'the constant you see on the outside that's the y intercept which is negative two'.
b
y-intercept of a line in slope-intercept form
real numbers
The narrator says “Let’s say y is less than or equal to 2x minus 3.”
The inequality written on the board is .
y
Vertical coordinate variable in the linear inequality and graph.
Real numbers; plotted as the vertical axis coordinate.
The narrator says “Let’s say y is less than or equal to 2x minus 3.”
The inequality written on the board is .
x
Horizontal coordinate variable in the linear inequality and graph.
Real numbers; plotted as the horizontal axis coordinate.
The narrator says “the slope is 2.”
The board shows .
m
Slope of the boundary line .
Real number; here .
The narrator states, "In this lesson, we're going to talk about graphing inequalities in an x y coordinate system."
The video introduces the task of representing inequalities as regions in a two-dimensional Cartesian coordinate system rather than only on a one-dimensional number line.
The setting is an x-y coordinate system.
The narrator says, "X equals three is basically a vertical line."
A vertical boundary is drawn at on the coordinate plane.
When only the x-coordinate is fixed at a constant value, the corresponding graph in the xy-plane is a vertical line.
c is a constant real number.
The narrator says, "So whenever y is equal to a number, it's going to be a horizontal line as opposed to a vertical line."
A horizontal boundary is drawn at on the coordinate plane.
When only the y-coordinate is fixed at a constant value, the corresponding graph in the xy-plane is a horizontal line.
c is a constant real number.
For , the narrator says, "Now x being greater than three but not equal to it, instead of drawing a solid line, we're gonna draw a dashed line."
For , the narrator says, "And because it's less than and equal to, we're gonna draw a solid line as opposed to a dashed line."
The first example uses a red dashed vertical line; the second uses a red solid vertical line; the third uses a red dashed horizontal line.
A strict inequality such as > or < is graphed with a dashed boundary line, while a non-strict inequality such as or is graphed with a solid boundary line.
Use dashed for strict inequalities.
Use solid for inequalities that include equality.
For , the narrator says, "Because x is greater than three, we need to shade towards the right, towards the large x values."
For , the narrator says, "And less than is towards the left side... you have to shade everywhere to the left of this line."
Blue arrows point right from the dashed line ; green arrows point left from the solid line .
For inequalities that restrict x, shade to the right when x is greater than the boundary value and to the left when x is less than the boundary value.
Boundary is a vertical line .
Right shading corresponds to or .
Left shading corresponds to or .
The narrator says, "The numbers that are greater than negative four are above this line. So we need to shade in this region."
Blue arrows point upward from the dashed horizontal line .
For inequalities that restrict y, shade above the boundary when y is greater than the boundary value and below the boundary when y is less than the boundary value.
Boundary is a horizontal line .
Upward shading corresponds to or .
Downward shading corresponds to or .
The narrator says, "Because it's an inequality, there's a range of answers, so you have to shade the appropriate region."
Each example ends with a whole half-plane indicated by arrows rather than a single curve or point.
Because an inequality allows many possible values rather than exactly one boundary value, its graph is a shaded region of the coordinate plane.
The relation is an inequality rather than an equation.
The speaker states, 'because it's less than or equal to, it's a solid line as opposed to a dashed line.'
A solid red horizontal line is drawn at .
When graphing an inequality that includes equality (such as ≤ or ≥), the boundary line is drawn as a solid line to indicate that points on the line are part of the solution set.
The inequality must contain an 'or equal to' component (≤ or ≥).
The speaker states, 'Now, it's greater than but not equal to, so we're going to draw a dashed line through these points.'
A dashed red diagonal line is drawn through the plotted points.
When graphing a strict inequality (such as < or >), the boundary line is drawn as a dashed line to indicate that points on the line are not part of the solution set.
The inequality must be strict (< or >) without an 'or equal to' component.
The equation y = mx + b is written on the screen.
The speaker explains, 'So this equation is in slope intercept form. The number in front of x is the slope... and the constant you see on the outside that's the y intercept'.
The slope-intercept form of a linear equation is y = mx + b, where m represents the slope of the line and b represents the y-intercept.
The equation must be a linear function of x.
The speaker says, 'If the slope is one, that means that rise over run is one over one... up one is the rise, one to the right is the run.'
Red arrows show moving up one unit and right one unit from the y-intercept to find the next point.
Slope can be interpreted as the ratio of vertical change (rise) to horizontal change (run). A slope of 1 means moving up 1 unit for every 1 unit moved to the right.
The slope is given as a fraction or integer.
The narrator introduces “one more example” and states “Let’s say y is less than or equal to 2x minus 3.”
The board writes , then identifies and .
The clip works with a two-variable linear inequality. The boundary relation is obtained by replacing with equality, giving the line . The inequality itself describes all ordered pairs for which the -value is no greater than .
The inequality is in two variables and .
The boundary line is .
The relation includes points on the boundary line.
The narrator explicitly states, "X equals three is basically a vertical line."
In the xy-plane, the equation graphs as a vertical line.
c is a constant.
For any constant c.
The narrator explicitly states, "So whenever y is equal to a number, it's going to be a horizontal line as opposed to a vertical line."
In the xy-plane, the equation graphs as a horizontal line.
c is a constant.
For any constant c.
The narrator says that because x is greater than 3 but not equal to it, a dashed line is drawn instead of a solid line.
If the inequality does not include equality, the boundary line is drawn dashed.
The inequality is strict, such as > or <.
For the examples shown in the clip.
The narrator says that because the relation is less than and equal to, a solid line is drawn as opposed to a dashed line.
If the inequality includes equality, the boundary line is drawn solid.
The inequality is non-strict, such as or .
For the examples shown in the clip.
The narrator says, "Because it's an inequality, there's a range of answers, so you have to shade the appropriate region."
An inequality has a range of answers, so its graph is represented by shading a region.
The relation being graphed is an inequality.
For the examples shown in the clip.
The narrator states “the slope is 2, the y-intercept is negative 3.”
The board writes and under .
For the inequality , the associated boundary line has slope and y-intercept .
The inequality is written in the form .
Applies to the specific boundary line shown in this example.
The narrator says “because it’s less than or equal to, it’s going to be a solid line as opposed to a dashed line.”
The boundary line is drawn solid.
If a linear inequality includes equality, such as , its boundary line is drawn as a solid line rather than a dashed line.
The inequality relation includes equality.
The graph is of a linear inequality in two variables.
Stated for this example and contrasted generally with a dashed line.
The narrator says “because it’s less than the function, we need to shade in the region below it.”
Blue arrows point downward from the boundary line.
The spoken wording “less than the function” is imprecise relative to the displayed inequality.
For , the solution region is the set of points on or below the line .
The inequality is solved for .
The comparison is .
Applies to the specific inequality graphed in the clip.
The narrator says “If the equation is true, then you should shade in that region. If it’s not true, do not shade it.”
The narrator says “equation,” but the demonstrated substitution is into the inequality .
After substituting a test point into the inequality, shade the region containing the test point if the resulting statement is true; do not shade it if the statement is false.
The test point is used to determine which side of the boundary line satisfies the inequality.
General method stated verbally in the clip.
The board shows , then , with a red X beside it.
"So this is not true, which means that we should not shade in this region."
Substituting into gives , which is false; therefore the region containing is not shaded.
The inequality is .
The test point is .
For the specific point (0,0).
The board shows -2 ≤ , then -2 ≤ 8-3, then -2 ≤ 5, with a red check mark.
"Both ways you read it, it's a true statement, which means that we should shade in the region that contains that point."
Substituting into gives , which is true; therefore the region containing is shaded.
The inequality is .
The test point is .
For the specific point (4,-2).
"So now you have two ways in determining which is the appropriate region to shade. You can use a test point, or you can use the fact that the equation says y is less than or equal to 2x minus 3, which means that you should shade below the line."
For this inequality, the correct half-plane can be found either by testing a point or by reading the solved form and shading below the boundary line.
The inequality is written in solved form with isolated.
The boundary line separates the plane into two candidate regions.
For linear inequalities of the displayed form.
The narrator introduces , identifies as a vertical line, chooses a dashed boundary because equality is excluded, and shades to the right.
A coordinate plane is drawn, then a red dashed vertical line at , then blue arrows pointing right.
Start with the inequality to be graphed.
Given in the video.
Identify the boundary line by replacing the inequality with equality.
The narrator states that x equals 3 is a vertical line.
Draw the boundary as dashed because the inequality is strict.
The narrator says that since x is greater than 3 but not equal to it, a dashed line is used.
Shade the half-plane containing larger x-values.
The narrator says to shade toward the right, toward the large x values.
The graph of is the region to the right of the dashed vertical line .
The narrator introduces , identifies the boundary as a vertical line, chooses a solid line because equality is included, and shades to the left.
A coordinate plane is drawn, then a red solid vertical line at , then green arrows pointing left.
Start with the inequality to be graphed.
Given in the video.
Identify the boundary line by replacing the inequality with equality.
The narrator says that whenever x is equal to a number, it is a vertical line.
Draw the boundary as solid because the inequality includes equality.
The narrator says that because it is less than and equal to, a solid line is drawn.
Shade the half-plane containing smaller x-values.
The narrator says less than is toward the left side and to shade everywhere to the left of the line.
The graph of is the region to the left of the solid vertical line .
The narrator introduces , identifies the boundary as a horizontal line, chooses a dashed line because equality is excluded, and shades above.
A coordinate plane is drawn, then a red dashed horizontal line at , then blue arrows pointing upward.
Start with the inequality to be graphed.
Given in the video.
Identify the boundary line by replacing the inequality with equality.
The narrator says that whenever y is equal to a number, it is a horizontal line.
Draw the boundary as dashed because the inequality is strict.
The narrator says that because it is greater than but not equal to, it is a dashed line.
Shade the half-plane containing larger y-values.
The narrator says the numbers greater than negative four are above this line.
The graph of is the region above the dashed horizontal line .
The speaker explains the steps to graph : drawing a horizontal line at 2, making it solid because of the 'or equal to', and shading below because it is 'less than'.
A solid red horizontal line is drawn at , followed by blue downward arrows indicating the shaded region.
Identify the boundary line as a horizontal line at .
The inequality only involves the variable y, meaning x can take any real value while y is constrained.
Draw the boundary line as a solid line.
The inequality is 'less than or equal to' (≤), which includes the boundary line in the solution set.
Shade the region below the boundary line.
The inequality is 'less than' (≤), meaning all y-values must be smaller than or equal to 2.
The graph of is a solid horizontal line at with the region below it shaded.
The speaker explains how to graph by identifying the slope and y-intercept, plotting points, drawing a dashed line, and shading above.
The equation y = mx + b is written, and are identified, points are plotted, a dashed line is drawn, and blue upward arrows shade the region above.
Recognize the inequality is in slope-intercept form.
The expression matches the structure y > mx + b.
Identify the slope and y-intercept.
The coefficient of x is 1 (implicit), and the constant term is -2.
Plot the y-intercept at (0, -2).
The y-intercept b is the point where the line crosses the y-axis ().
Use the slope (rise/run = ) to plot subsequent points like (1, -1) and (2, 0).
A slope of 1 means for every 1 unit moved right, the line moves 1 unit up.
Draw a dashed line through the plotted points.
The inequality is strict (>), so the boundary line is not included in the solution set.
Shade the region above the dashed line.
The inequality is y > ..., meaning the solution consists of points with y-values greater than the line's y-values for a given x.
The graph of is a dashed line with a slope of 1 and y-intercept of -2, with the region above the line shaded.
The narrator starts at negative 3, explains rise 2 and run 1, and names the points , , and .
Red points are plotted successively on the coordinate plane and then connected by a solid red line.
Start from the y-intercept of the boundary line.
The video identifies from and says to start with negative 3.
Express the slope as a vertical change of 2 for every horizontal change of 1.
The narrator states that because the slope is 2, the rise is 2 and the run is 1.
Move up 2 units and right 1 unit from the intercept to get the next point.
This follows directly from the stated rise/run interpretation of slope.
Repeat the same rise and run to obtain the next point.
The narrator says going up 2 over 1 again takes us to .
Repeat once more to obtain the next point.
The narrator continues the pattern to and says “and so forth.”
The plotted red points lie on the boundary line , and connecting them gives the solid boundary used for the inequality graph.
The narrator says to replace with 0 and with 0, then asks whether 0 is less than negative 3.
The board writes and simplifies to .
The final spoken conclusion is cut off after the question, but the displayed inequality is false.
Choose the origin as a test point.
The narrator explicitly selects the origin, , as the first test point.
Substitute and into the inequality.
This is direct substitution into .
Simplify the right-hand side.
Since , the inequality reduces to .
The origin does not satisfy , because is false; therefore the side containing is not the solution side.
The displayed chain is , then .
The narrator explains that 0 is to the right of -3 and therefore greater, so the statement is not true.
Substitute and into .
Direct substitution of the test point coordinates into the inequality.
Evaluate to get .
Arithmetic simplification.
On a number line, 0 lies to the right of -3, so 0 is greater than -3.
Order property of real numbers on the usual number line.
Since , the statement cannot be true.
Negation of the proposed inequality from the previous comparison.
The origin fails the inequality, so the region containing is not part of the shaded solution set.
The displayed chain is -2 ≤ , then -2 ≤ 8-3, then -2 ≤ 5.
"Four times two is eight, eight minus three is five... negative two is less than five, and if we read it the other way, five is greater than negative two."
The audio phrase "Four times two is eight" reverses the written order , but the arithmetic result shown on the board is consistent.
Substitute and into .
Direct substitution of the test point coordinates into the inequality.
Evaluate to get .
Multiplication.
Evaluate to get .
Subtraction.
Negative two is less than five; equivalently, five is greater than negative two.
Order comparison of real numbers.
The point satisfies the inequality, so the region containing is shaded.
is written on screen.
The narrator explains the boundary line, dashed style, and rightward shading.
A red dashed vertical line at is drawn, followed by blue arrows pointing right.
Graph the inequality in the xy-coordinate system.
The inequality is .
The coordinate system has x- and y-axes.
Represent all points satisfying as a graph.
Find the boundary line by setting x equal to 3.
The narrator states that x equals 3 is a vertical line.
Use a dashed boundary because the inequality is strict.
The narrator says that since x is greater than 3 but not equal to it, a dashed line is drawn.
Shade the side with larger x-values.
The narrator says to shade toward the right, toward the large x values.
The solution is the half-plane to the right of the dashed vertical line .
The visual arrows point away from the boundary into the region of larger x-values, matching .
is written on screen.
The narrator explains that the boundary is vertical, solid, and shaded to the left.
A red solid vertical line at is drawn, followed by green arrows pointing left.
Graph the inequality in the xy-coordinate system.
The inequality is .
The coordinate system has x- and y-axes.
Represent all points satisfying as a graph.
Find the boundary line by setting x equal to -2.
The narrator says that whenever x is equal to a number, it is a vertical line.
Use a solid boundary because equality is included.
The narrator says that because it is less than and equal to, a solid line is drawn.
Shade the side with smaller x-values.
The narrator says less than is toward the left side and to shade everywhere to the left of the line.
The solution is the half-plane to the left of the solid vertical line .
The visual arrows point into the region of smaller x-values, matching .
is written on screen.
The narrator explains that the boundary is horizontal, dashed, and shaded above.
A red dashed horizontal line at is drawn, followed by blue arrows pointing upward.
Graph the inequality in the xy-coordinate system.
The inequality is .
The coordinate system has x- and y-axes.
Represent all points satisfying as a graph.
Find the boundary line by setting y equal to -4.
The narrator says that whenever y is equal to a number, it is a horizontal line.
Use a dashed boundary because the inequality is strict.
The narrator says that because it is greater than but not equal to, it is a dashed line.
Shade the side with larger y-values.
The narrator says the numbers greater than negative four are above this line.
The solution is the half-plane above the dashed horizontal line .
The visual arrows point upward into the region of larger y-values, matching .
The speaker introduces the example: 'Let's say if y is less than or equal to two.'
The inequality is written on the screen.
Graph the linear inequality .
The inequality is .
Produce the correct graphical representation of the solution set on the Cartesian plane.
Draw a horizontal line at .
Since the inequality only restricts y, the boundary is a horizontal line.
Make the line solid.
The 'or equal to' condition means the boundary is included.
Shade the area below the line.
The 'less than' condition means y-values must be smaller than 2.
A solid horizontal line at with the region below it shaded.
Visual confirmation of the solid line and downward shading matches the inequality conditions.
The speaker introduces the example: 'Let's say that y is greater than x minus two.'
The inequality is written on the screen.
Graph the linear inequality .
The inequality is .
Produce the correct graphical representation of the solution set on the Cartesian plane.
Identify the slope and y-intercept from the slope-intercept form.
Comparing to y > mx + b gives and .
Plot the y-intercept at (0, -2).
This is the starting point for drawing the boundary line.
Use the slope of 1 to plot additional points, such as (1, -1) and (2, 0).
Rise over run is , meaning move up 1 and right 1 from the previous point.
Draw a dashed line through the points.
The strict inequality '>' means the boundary line is excluded.
Shade the region above the line.
The inequality y > ... indicates that solution points have higher y-values than the boundary line.
A dashed line passing through (0, -2) with a slope of 1, with the region above it shaded.
Visual confirmation of the dashed line and upward shading matches the inequality conditions.
The narrator says “Let’s try one more example” and then graphs step by step.
The board shows , , , , test points and , and the substitution .
A coordinate plane is drawn, red points are plotted on the line, a solid red boundary line is drawn, blue arrows indicate shading below, and green test points are marked.
The clip ends during the test-point explanation, so the full verbal conclusion for the second test point is not heard.
Graph the solution set of the inequality in the coordinate plane.
The inequality is .
From the inequality, the boundary line is .
The slope is and the y-intercept is .
Two test points are chosen: and .
Determine the boundary line, decide whether it is solid or dashed, and identify which region to shade.
Write the given inequality.
This is the example introduced verbally and visually at the start of the clip.
Read the slope and y-intercept from slope-intercept form.
The narrator explicitly identifies the slope as 2 and the y-intercept as negative 3.
Plot points on the boundary line using the y-intercept and slope .
Starting at , each step uses rise 2 and run 1, as stated by the narrator.
Draw the boundary line through the plotted points as a solid line.
Because the inequality is , equality is included, so the boundary is not dashed.
Indicate the solution region below the boundary line.
The narrator says that since is less than or equal to the boundary expression, the shading goes below the line; blue arrows show this visually.
Test the origin by substitution.
The narrator chooses as a test point and substitutes it into the inequality.
Mark a second test point on the other side of the boundary line.
The narrator selects and explains that a true substitution means shade that region, while a false one means do not shade it.
The graph consists of the solid line with the region on or below that line shaded as the solution set.
The origin test gives , which is false, confirming that the side containing is not shaded. The visual shading arrows point downward from the boundary line, matching the rule for .
The worked inequality is .
A solid red boundary line, blue downward arrows, and test points (0,0) and (4,-2) are shown.
The narrator determines which region to shade using both test points and the inequality direction.
Determine the solution region for the linear inequality and identify which side of the boundary line should be shaded.
Inequality: .
Boundary line slope: .
y-intercept: .
Candidate test points shown: and .
Identify the correct half-plane solution region for .
Draw the boundary line by replacing with equality.
The boundary of a linear inequality in two variables is the corresponding equation.
Use a solid line because the inequality includes equality.
The symbol means points on the line satisfy the inequality.
Test the origin by substituting .
Test point method.
Since is to the right of on a number line, .
Number-line order of real numbers.
Test the point by substituting and .
Test point method.
Negative two is less than five, so the inequality holds.
Order comparison of real numbers.
Because satisfies the inequality, shade its half-plane.
A true test point identifies the solution region containing that point.
Alternatively, read directly from the solved inequality that is less than or equal to the line value.
For inequalities solved for , corresponds to the region at or below the boundary line.
The solution is the closed half-plane on or below the line ; the origin is not included, while is included.
The two checks agree: gives the false statement , and gives the true statement ; the direct rule from also says to shade below the line.
A horizontal line is drawn first, then a vertical line crossing it, forming an x-y coordinate system.
horizontal axis
vertical axis
origin intersection
A blank black screen becomes a coordinate plane.
The axes remain perpendicular and centered on the screen.
This establishes the two-dimensional setting needed to graph an inequality in x and y.
A red dashed vertical line is drawn at , with arrowheads indicating extension up and down.
red dashed vertical line
arrowheads
The boundary appears to the right of the y-axis at the marked x-value 3.
The line stays vertical and does not move horizontally once placed.
The dashed style visually encodes that points on are not included in .
Multiple blue horizontal arrows are drawn starting from the dashed line and pointing to the right.
blue arrows
dashed vertical boundary
Arrows accumulate across several heights, all directed rightward from the boundary.
All arrows point in the same direction, away from the boundary toward larger x-values.
The arrows indicate the solution half-plane to the right of .
A red solid vertical line is drawn at , with arrowheads indicating extension up and down.
red solid vertical line
arrowheads
The boundary appears to the left of the y-axis at the marked x-value -2.
The line remains vertical and continuous.
The solid style visually encodes that points on are included in .
Multiple green horizontal arrows are drawn starting from the solid line and pointing to the left.
green arrows
solid vertical boundary
Arrows accumulate across several heights, all directed leftward from the boundary.
All arrows point in the same direction, away from the boundary toward smaller x-values.
The arrows indicate the solution half-plane to the left of .
A red dashed horizontal line is drawn at , with arrowheads indicating extension left and right.
red dashed horizontal line
arrowheads
The boundary appears below the x-axis at the marked y-value -4.
The line remains horizontal and dashed.
The dashed style visually encodes that points on are not included in .
Multiple blue vertical arrows are drawn starting from the dashed line and pointing upward.
blue arrows
dashed horizontal boundary
Arrows accumulate across several horizontal positions, all directed upward from the boundary.
All arrows point in the same direction, away from the boundary toward larger y-values.
The arrows indicate the solution half-plane above .
Blue arrows are drawn pointing downwards from the solid red line at .
The speaker says, 'because it's less than, we need to shade below it rather than above it.'
solid red horizontal line at
blue downward arrows
Blue arrows appear sequentially below the red line to indicate the shaded region.
The red boundary line remains solid and horizontal.
The downward arrows visually represent the 'less than' condition, showing that all points with y-values smaller than 2 are part of the solution.
Blue arrows are drawn pointing upwards from the dashed red diagonal line.
The speaker says, 'Notice that y is greater than the function. So we're going to shade above the line rather than below it.'
dashed red diagonal line
blue upward arrows
Blue arrows appear sequentially above the red dashed line to indicate the shaded region.
The red boundary line remains dashed and diagonal.
The upward arrows visually represent the 'greater than' condition, showing that all points with y-values greater than the line's y-value for a given x are part of the solution.
The screen begins black, then white handwriting appears at upper left: , followed by and .
The narrator introduces the example and identifies slope 2 and y-intercept negative 3.
Black background
White handwritten inequality
White handwritten
White handwritten
The inequality is written first.
The slope value is added below it.
The y-intercept value is added below the slope.
The example remains the same inequality throughout this interval.
All writing is in white on a black background.
The visual setup establishes the algebraic form before graphing, making the slope and intercept explicit for the next steps.
A horizontal line and then a vertical line are drawn to form axes, followed by tick marks on both axes.
The narrator says “Let’s go ahead and graph this function.”
Horizontal axis
Vertical axis
Tick marks
A horizontal axis is drawn.
A vertical axis is added to create a cross-shaped coordinate system.
Tick marks are placed along both axes.
The inequality and extracted values remain visible at upper left.
The drawing color stays white for the axes and ticks.
The coordinate plane provides the framework for plotting the boundary line of the inequality.
A red point appears at , then additional red points appear at , , and as red arrows illustrate upward and rightward movement.
The narrator explains starting at negative 3 and using rise 2, run 1 to reach the listed points.
Red point at
Red point at
Red point at
Red point at
Red directional arrows
The first red point is placed at the y-intercept.
Red arrows show moving up and right.
Successive red points are added along the same linear pattern.
The underlying line being represented is .
The slope pattern remains rise 2 over run 1 for each step.
The animation visually converts the algebraic slope into geometric movement on the coordinate plane.
The narrator emphasizes that because it is an inequality, there is a range of answers and the appropriate region must be shaded.
One might think the graph of an inequality is just the boundary line itself.
The boundary line only separates included from excluded points; the actual solution is a whole shaded region.
The narrator contrasts dashed and solid lines based on whether equality is included.
One might draw the same kind of boundary line regardless of whether equality is allowed.
Strict inequalities use dashed boundaries, while inequalities that include equality use solid boundaries.
The speaker explicitly contrasts solid and dashed lines based on the presence of 'or equal to'.
Students might draw a dashed line for a non-strict inequality (≤ or ≥) or a solid line for a strict inequality (< or >).
The boundary line must be solid if the inequality includes equality (≤ or ≥), and dashed if it is strict (< or >).
The narrator explicitly contrasts “solid line” with “dashed line” based on “less than or equal to.”
A learner may draw the boundary line as dashed even when the inequality includes equality.
For or , the boundary line is included in the solution set and should be drawn solid; dashed lines are for strict inequalities such as or .
The narrator says “If the equation is true…” while demonstrating substitution into the inequality .
The wording is audible, but it is unclear whether this is a slip of speech or informal usage.
The test-point step is described using the word “equation,” which could confuse learners into thinking they substitute into instead of the inequality.
The actual substitution shown is into the inequality ; the truth value of the inequality, not the boundary equation, determines shading.
The narrator says to shade below because the inequality is “less than the function.”
Blue arrows point downward from the boundary line.
A learner may shade above the line even though the inequality compares to a smaller-or-equal expression.
When the inequality is solved for and has or , the solution region is below the boundary line; here that is shown by downward blue arrows.
"So this is not true, which means that we should not shade in this region."
A red X is placed next to .
A test point that makes the inequality false might be mistaken as evidence to shade its region anyway.
If the substituted statement is false, the region containing that test point is not the solution region.
"negative two is less than five, and if we read it the other way, five is greater than negative two. Both ways you read it, it's a true statement."
Learners may think changing the reading direction of an inequality changes its truth value.
The statements and express the same order relation and are both true.
After introducing graphing in an x-y coordinate system, the narrator immediately explains that is a vertical line.
Understanding the xy-coordinate setting is needed before recognizing that fixing x produces a vertical boundary.
In the third example, the narrator moves from the coordinate-system setup to the statement that y equal to a number gives a horizontal line.
Recognizing horizontal boundaries depends on the same coordinate-plane framework introduced at the start.
In each example, the narrator first decides dashed versus solid and only then explains which side to shade.
The boundary type is determined first, then the solution region is chosen by direction of shading.
The first two examples both use vertical boundaries determined by x-values.
The rule for shading left or right applies specifically when the boundary is a vertical line coming from an x-inequality.
The third example uses a horizontal boundary determined by a y-value and then shades above it.
The rule for shading above or below applies specifically when the boundary is a horizontal line coming from a y-inequality.
The narrator explicitly contrasts x = constant giving a vertical line with y = constant giving a horizontal line.
Fixing x yields a vertical boundary, while fixing y yields a horizontal boundary.
The speaker uses the slope-intercept form to determine the slope and y-intercept for graphing.
The slope-intercept form is applied to extract the necessary parameters (slope and y-intercept) to graph the boundary line of the inequality.
The speaker uses the rise over run interpretation to find subsequent points on the line.
The rise over run method is applied to plot additional points on the boundary line after the y-intercept is established.
The board moves from to and .
The narrator identifies slope and y-intercept directly from the written inequality.
The method of reading slope and intercept is applied to the specific inequality to prepare for graphing its boundary line.
The narrator links “less than or equal to” to a solid line.
The boundary line is drawn solid after the inequality symbol has been established.
The inequality symbol in the example determines the rule that the boundary line must be solid rather than dashed.
After the solid red boundary is drawn, blue arrows are added below it.
The narrator first explains the solid line, then explains shading below.
The boundary line must be drawn before the solution region can be indicated relative to it.
The narrator says “Another way in which you can determine where to shade is by using test points.”
Green test points are added after the blue shading arrows already indicate the region.
The clip presents two ways to determine the shaded region: directly from the inequality direction, and alternatively by testing points on either side of the boundary.
The narrator explains that because x is greater than 3 but not equal to it, a dashed line is used.
The narrator explains that because the inequality is less than or equal to, a solid line is used.
The narrator states that x equal to a number gives a vertical line and y equal to a number gives a horizontal line.
The narrator says to shade right for larger x-values and left for smaller x-values.
The narrator says the numbers greater than negative four are above the line, so that region is shaded.
The narrator says an inequality has a range of answers, so the appropriate region must be shaded.
The speaker demonstrates how to graph .
The speaker asks, 'Now what if we have an equation with two variables?' and proceeds to graph .
The speaker explains the difference between solid and dashed lines based on the inequality symbol.
The speaker explains shading below for 'less than' and above for 'greater than'.
The board shows and under .
The narrator states the slope is 2 and the y-intercept is negative 3.
The narrator says “because it’s less than or equal to, it’s going to be a solid line as opposed to a dashed line.”
The boundary line is drawn solid.
Covered · Audio introduces graphing inequalities in an x-y coordinate system.
Covered · First worked example graphs with a dashed vertical boundary and rightward shading.
Covered · Brief transition between examples with no new mathematical content beyond moving to the next problem.
Covered · Second worked example graphs with a solid vertical boundary and leftward shading.
Covered · Brief transition between examples with no new mathematical content beyond moving to the next problem.
Covered · Third worked example graphs with a dashed horizontal boundary and upward shading.
Covered · Concludes the previous example , establishing the context of graphing inequalities.
Covered · Introduces and graphs the inequality , explaining solid boundary lines and shading below.
Covered · Introduces and graphs the two-variable inequality , detailing slope-intercept form, point plotting, dashed boundary lines, and shading above.
Covered · Black screen before the first spoken introduction; it leads directly into the example setup.
Covered · The inequality is introduced and rewritten informationally as , .
Covered · Coordinate axes and tick marks are drawn in preparation for graphing the boundary line.
Covered · Points on the boundary line are plotted using the y-intercept and slope .
Covered · The boundary line is drawn solid because the inequality includes equality.
Covered · Blue arrows indicate that the solution region is below the boundary line.
Covered · The narrator introduces test points and marks and on opposite sides of the line.
Covered · The origin is substituted into the inequality, producing , which is false.
Covered · Final fraction of a second contains no new mathematical content beyond the completed substitution display.
Covered · The clip opens with the inequality, boundary-line data, and the failed test at the origin.
Covered · Attention shifts to the second test point (4,-2); the earlier algebra panel is cleared before the new substitution is written.
Covered · The substitution for (4,-2) is written step by step and identified as true.
Covered · The narrator summarizes two methods for choosing the shaded region and circles the ≤ symbol.
Reviewed English material v1: card "Graphing inequalities in the xy-plane" at 0:00 — This introductory card captures the main setup of the clip: inequalities are being represented in a two-dimensional coordinate system. The video emphasizes that the result is a region of the plane, not merely a line or a point.
YouTube5:27Another application of Inequalities.
Inequalities
YouTube8:00Another application of Inequalities.
Inequalities