Rob Englot · YouTube · 8:58
This 150-second clip introduces a lesson on solving linear equations with rational coefficients. The presenter writes the section title, explains that rational coefficients may appear as fractions, decimals, or related forms, and then sets up a fraction example with two equations: a) and b) . The clip ends during the setup stage, with the stated method of finding a common denominator and multiplying through, but no solution is carried out. This video demonstrates the algebraic method for solving a system of linear equations with rational coefficients. The instructor explains and shows how to find a common denominator for each equation, multiply every term by that denominator to eliminate the fractions, and then reduce the resulting expressions to obtain a simpler system with integer coefficients. A brief correction highlights the importance of watching for sign errors during the multiplication step. The final transformed system is prepared for solving via substitution or elimination. This 150-second whiteboard lesson solves a two-equation linear system that began with rational coefficients. The page already shows the original equations and , together with their cleared-denominator forms and the reduced integer system and . The instructor chooses elimination because the y-coefficients are opposites, adds the equations to get , and finds . He then back-substitutes into the first original equation, simplifies , multiplies by 3 to clear denominators, and solves -2y = 4 to obtain . The ordered-pair solution (2, -2) is written, and a check in the second original equation is begun but not completed within the clip. This 88-second whiteboard clip is the verification stage of a handwritten algebra example titled Section 1.6 Solving Linear Equation with Rational #'s / Fraction Example. The earlier solution work is already on the page, and the instructor now checks the pair in the equation . He substitutes the values, multiplies to get and , explains why a common denominator of 20 is needed, rewrites the terms as and by multiplying by forms of 1, subtracts to obtain , reduces it to , and concludes the solution is correct. The last twenty seconds contain no new math, only a page adjustment and a preview of the next tutorial.
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 clip opens on a blank sheet of graph paper, then the instructor begins speaking and writes the heading "Section 1.6 Solving linear equation with Rational #s" across the top.
With the title in place, the speaker explains that the lesson concerns linear equations whose coefficients are rational numbers, and he names several ways those coefficients can appear: fractions, decimals, bracketed expressions, fractions with binomial tops, and possibly roots.
He then narrows the focus to the first case, saying this tutorial will use a fraction example, and writes the subheading "Fraction Example" below the title.
Under that heading he writes the first equation, labeled a): .
He immediately adds a second equation, labeled b): .
After both equations are on the page, the instructor states the first goal: decide how to eliminate the denominators. He emphasizes that each equation will be treated independently at this stage.
Turning to equation a), he points out the denominators 2 and 3 and says a common denominator must be found before clearing fractions.
He writes the procedural note "Find Common Denominator" and then "x Multiply through," setting up the next algebraic step.
The clip ends before he actually computes the common denominator or rewrites either equation without fractions.
The video begins by outlining the strategy for solving a system of linear equations that contain rational coefficients. The core method is to find a common denominator for each individual equation and multiply every term through by that value to eliminate the fractions.
Applying this to the first equation, the denominators are 2, 3, and 3. The least common multiple is 6. The instructor multiplies each term—()x, -()y, and —by 6. This results in the intermediate expression ().
For the second equation, the denominators are 4, 5, and 10. The instructor notes that 10 is not a multiple of 4, so the least common multiple must be 20. Every term in the equation is multiplied by 20, yielding ().
With the fractions cleared, the next step is to reduce the equations by performing the division in each term. For the first equation, 6 divided by 2 is 3, 12 divided by 3 is 4, and 42 divided by 3 is 14, simplifying the equation to .
Turning to the second equation, 60 divided by 4 gives 15x. The instructor briefly catches and corrects a sign error, ensuring the middle term remains positive. Twenty divided by 5 is 4, giving +4y, and 220 divided by 10 is 22. The simplified equation is .
The original system with rational coefficients has now been successfully transformed into a system with integer coefficients: and . This new system is much easier to solve using standard methods like substitution or elimination.
The page opens with the full setup already visible: the title identifies Section 1.6 on solving linear equations with rational numbers, and two original fractional equations are shown above their cleared-denominator versions. Beneath them is the reduced integer system and . The instructor frames the next move as choosing the elimination method, because the system has already been prepared for it.
The key observation is that the coefficients of are opposites: in the first equation and in the second. That means the equations can be added directly, and the -terms will cancel. The instructor notes that this convenience is not guaranteed in every problem; sometimes extra manipulation is needed first, but here the setup is already ideal.
Adding the two equations term by term gives , since and , while . Dividing by yields . This is the first completed unknown, and the rest of the solution will be built from it.
To find , the instructor switches from elimination to back-substitution. He explicitly uses one of the original equations rather than the reduced integer system, writing . Substituting turns the two-variable problem into a one-variable equation in .
The product simplifies to , so the equation becomes . At this stage the remaining obstacle is the fractions. The instructor therefore applies the same strategy used earlier on the page: clear denominators by multiplying every term by a common denominator.
Because the denominators present are and , the common denominator is . Multiplying through by transforms into . This step preserves equivalence while removing fractions, making the equation easier to isolate.
Now the equation is purely linear in . Subtracting from both sides gives , and dividing by gives . Together with the earlier result , the system’s solution is the ordered pair .
The instructor then states the final procedural step: verification. Since was found by substituting into the first original equation, he checks the pair in the second original equation instead. He begins writing , signaling that the solution should satisfy the unused equation as well. The clip ends before that arithmetic check is completed.
The page is already filled with earlier work from a lesson on solving linear equations with rational coefficients. At this moment the instructor is not solving from scratch; he is checking a candidate solution by substituting into the displayed equation and aiming to recover the right-hand side .
He evaluates the two products one at a time. The first term becomes , and the second becomes . This is the ordinary rule for multiplying a fraction by an integer: keep the denominator and multiply the numerator.
Now the subtraction cannot be done directly because the denominators are different. The instructor therefore pauses the arithmetic to explain that a common denominator is required before combining and .
He chooses 20 as the common denominator and justifies the rewrite carefully: multiply the first fraction by and the second by . Since each of those factors equals 1, the values do not change; only the form of the fractions changes so they can be subtracted.
With the equivalent fractions and in place, the subtraction is straightforward. The numerators combine to give , and then the fraction is reduced by canceling the common factor 2, yielding .
That final simplified value matches the right-hand side of the original equation, so the instructor states that the check is true and the solution is correct. The mathematical point of this segment is that verification succeeds when substitution and simplification produce an identity.
After the proof is complete, there is no further algebra. He slides the paper upward to keep the lower writing in view and closes by saying this was the first fraction example and that the next tutorial will use fractions with binomial tops.
The video introduces a lesson on solving linear equations with rational coefficients, meaning the numbers multiplying the variables are rational numbers. The spoken introduction and the handwritten title both identify this as Section 1.6.
The speaker says rational coefficients may show up as fractions, decimals, expressions with brackets, fractions with binomial tops, and even roots. Only the fraction case is actually written in this clip.
The presenter writes two example equations under the heading "Fraction Example": a) and b) . These are the concrete problems used to begin the lesson.
For equations with fractional coefficients, the instructor states that the first goal is to eliminate the denominators. He writes the plan "Find Common Denominator" and then "x Multiply through."
When introducing the denominator-clearing step, the speaker explicitly says each equation is treated independently. In this clip, he begins with equation a) and its denominators 2 and 3.
To solve a linear system with rational coefficients, find a common denominator for each equation. Multiply every term in that equation by the common denominator to eliminate the fractions, then reduce the resulting integer-coefficient equation.
When determining the common denominator for an equation with multiple fractions, find the least common multiple of all the denominators. For example, for denominators 4, 5, and 10, the LCM is 20, not 10, because 10 is not a multiple of 4.
When multiplying through an equation to clear fractions, carefully check the sign of each term. A common mistake is accidentally changing a plus sign to a minus sign or vice versa during the arithmetic.
After multiplying by the common denominator, simplify each fraction by dividing the numerator by the denominator. For instance, ()x reduces to 3x, and () reduces to 14.
In the reduced system and , the -coefficients are and . Because they are exact opposites, adding the equations eliminates immediately. The instructor emphasizes that this direct setup is convenient but not universal; other systems may require prior manipulation.
Term-by-term addition produces , hence . Dividing by yields . This is the first solved variable and the basis for the next step.
After obtaining , the instructor substitutes into the first original fractional equation, , giving . This reduces the problem to one equation in one unknown.
The substituted equation simplifies to . To remove fractions, every term is multiplied by , the common denominator of and . The equivalent equation becomes .
From , subtract from both sides to get , then divide by to obtain . Combined with , the solution pair is .
The instructor treats checking as the last step and begins substituting into the second original equation, . The visible start of the check is , though the clip ends before the verification is completed.
The clip’s main task is verification, not solving. The instructor takes the already-found values and , inserts them into , and checks whether the two sides become the same number.
Each product is simplified separately before any addition or subtraction is attempted. Multiplying by 2 gives , and multiplying by gives .
The instructor explicitly stops before subtracting because and do not share a denominator. A common denominator is required so the numerators can be combined directly.
To reach denominator 20, he multiplies by and . The key idea stated aloud is that these factors equal 1, so the fractions change form but not value.
Once both terms are written over 20, the subtraction is performed on the numerators alone. This produces , which is then reduced to .
Because the simplified left-hand side equals the right-hand side, the instructor declares the check true and the solution correct. The final mathematical takeaway is that a valid solution must make the original equation a true statement.
Explore conditions, steps and evidence. Supplementary explanations are labeled separately from content shown in the video.
Equation a) is written as "", so x appears as the variable multiplied by .
Equation b) is written as "", so x appears again as the variable multiplied by .
x
Unknown variable appearing in both linear equations.
Video does not state a domain for x.
Equation a) contains "- ".
Equation b) contains "+ ".
y
Second unknown variable appearing in both linear equations.
Video does not state a domain for y.
The first equation is labeled "a)" on the page.
Speaker says, "That's going to be equation A."
a)
Label identifying the first displayed equation.
Label only; not a mathematical variable.
The second equation is labeled "b)" on the page.
Speaker says, "and then B will be ..."
b)
Label identifying the second displayed equation.
Label only; not a mathematical variable.
x appears in equation a) as the variable multiplied by and later by .
x
unknown variable in the first equation of the system
real number (assumed from context)
y appears in equation a) as the variable multiplied by -2/3 and later by -12/3.
y
unknown variable in the first equation of the system
real number (assumed from context)
x appears in equation b) as the variable multiplied by and later by .
x
unknown variable in the second equation of the system
real number (assumed from context)
y appears in equation b) as the variable multiplied by and later by .
y
unknown variable in the second equation of the system
real number (assumed from context)
x appears in the system and , then in the substitution step .
x
unknown variable in the two-variable linear system; solved value is .
real numbers
y appears in , , and later in and -2y = 4.
y
unknown variable in the two-variable linear system; solved value is .
real numbers
The page shows the reduced integer system and under 'Reduce the equations'.
first reduced equation obtained from the original fractional equation a).
linear equation in two variables
The page shows the reduced integer system and under 'Reduce the equations'.
second reduced equation obtained from the original fractional equation b).
linear equation in two variables
Speaker says the tutorial deals with "solving equations, linear equations that have rational coefficients or rational numbers in them."
Handwritten title reads "Section 1.6 Solving linear equation with Rational #s".
This clip introduces a lesson on solving linear equations whose coefficients are rational numbers. The speaker explicitly frames the section as working with rational coefficients, and the written heading states the same topic.
The equations under discussion are linear.
The coefficients are rational numbers.
Speaker says, "So you could have fractions, decimals, and other types, maybe even roots and stuff like that."
Speaker then says the examples will look at fractions, brackets, fractions with binomial tops, and decimals.
The mention of roots is spoken as a possible type but no root example is shown in this clip.
The speaker identifies several ways rational coefficients can appear in examples: fractions, decimals, expressions involving brackets, and fractions with binomial numerators. This is a verbal overview of the kinds of coefficient forms the section will address.
These are forms mentioned by the speaker, not all demonstrated within this clip.
Speaker says the first goal is to decide how to eliminate the denominators in each equation.
Speaker says, "We've got to find a common denominator so that we can eliminate the denominator there between two and three."
Handwritten note begins "Find Common Denominator" and then "x Multiply through".
The clip stops before the multiplication step is carried out.
For an equation with fractional coefficients, the speaker’s stated first step is to find a common denominator and then multiply the whole equation through by it. In equation a), the denominators mentioned are 2 and 3, so the method is introduced before any algebraic simplification is performed.
Applies to equations containing fractional coefficients.
Each equation is treated independently in the setup described here.
The speaker explains to find the common denominator for each equation and multiply through in each expression.
The written note says 'Find Common Denominator x Through'.
To solve a linear system with rational coefficients, first find a common denominator for the fractions in each equation. Then multiply every term in that equation by the common denominator to eliminate the fractions, and finally reduce the resulting integer-coefficient equation.
The system consists of linear equations with rational coefficients.
Each equation is processed separately using its own common denominator.
The speaker states, 'Then we reduce the equation.'
The written note says 'Reduce The equations'.
After multiplying through by the common denominator, simplify each fraction by dividing the numerator by the denominator to obtain an equivalent equation with integer coefficients.
Fractions have been cleared by multiplying every term by a common denominator.
The instructor says elimination is the easiest method here because there is a negative 4y and a positive 4y, so one can just add the equations.
The displayed system is and , whose y-coefficients are opposites.
When two linear equations in the same variables have opposite coefficients for one variable, adding the equations eliminates that variable directly. In this clip the y-terms are -4y and +4y, so addition removes y and leaves an equation only in x.
Two equations involve the same unknowns.
One pair of corresponding coefficients are exact opposites.
Equations are aligned term-by-term before addition.
After finding , the instructor says they would solve for y given that and chooses one of the original equations.
He writes , then simplifies to .
Once one variable is found, substitute its value into one of the original equations and solve the resulting one-variable equation for the remaining unknown. Here is substituted into the first original fractional equation rather than the reduced integer system.
At least one variable has already been solved.
The chosen equation still contains the remaining unknown.
Substitution uses the original equation or any equivalent equation from the system.
The instructor says to eliminate the denominator by using a common denominator, which here is 3, and multiply everything by 3.
From he writes .
If an equation contains fractions, multiply every term by a common denominator to obtain an equivalent equation without fractions. In this clip the denominators are 1 and 3, so multiplying by 3 converts into .
All terms are multiplied by the same nonzero number.
The multiplier is a common denominator of the fractional terms.
The instructor says the last step is to reassure ourselves by a check and will use the other equation.
He begins writing ... for verification.
The clip ends before the verification arithmetic is completed on screen.
After obtaining an ordered pair, verify it by substituting into an equation not used for back-substitution. This confirms that the pair satisfies the whole system, not just one equation. Here the solution (2,-2) is checked in the second original equation.
A candidate ordered pair has been found.
At least one original equation remains available for substitution checking.
The speaker says is negative two and that this will give eleven tenths.
The page shows the verification setup .
The clip demonstrates checking a previously obtained solution by replacing the variables in the second equation with the values already found on the page, namely and . The purpose is not to solve again but to verify that the ordered pair satisfies the equation.
The candidate solution values must already be known.
The equation being checked must contain those variables.
He says using our fraction skills, this is six over four times one, which is four, and this would be negative two over five.
The written intermediate terms are and .
The instructor evaluates each product by multiplying the numerator by the integer while keeping the denominator unchanged: and . This is the first arithmetic step after substitution.
The integer is treated as having denominator 1.
Only the numerator changes when multiplying directly by the integer.
He says we have to do a common denominator here to subtract these two, and the common denominator here would be twenty.
He rewrites and with denominator 20.
Before subtracting and , the instructor converts both fractions to equivalent fractions with the same denominator, choosing 20. This makes the numerators directly comparable so the subtraction can be performed in one step.
The fractions must be rewritten as equivalent fractions.
The chosen denominator must be a common multiple of the original denominators.
Speaker says, "each of them are going to be treated independently."
The clip does not show whether the two equations are later combined into a system or solved separately.
When clearing denominators in this example, each equation is handled independently rather than together.
There are multiple displayed equations.
The current step is eliminating denominators.
For each equation in the displayed set, the denominator-clearing step is applied separately.
The instructor states that because there is a negative 4y and a positive 4y, one can just add the equations.
Adding and yields .
If two linear equations contain opposite coefficients for the same variable, adding the equations eliminates that variable.
The equations are written in aligned form.
The corresponding coefficients of one variable are negatives of each other.
For a pair of linear equations in the same variables.
The page records 'Solution (2, -2)' after deriving and .
The instructor says, 'So our solution is equal to 2 negative 2.'
The simultaneous solution of the displayed system is and .
The system is and , equivalently derived from the original fractional equations.
For the specific two-equation system shown in the clip.
The page shows and the intermediate work leads to the same value.
The speaker says this is going to give us eleven tenths and later says we are correct there.
For the displayed equation , the values and make the equation true.
The equation being checked is .
The candidate values are and .
For these specific values of and .
He says if you're multiplying by five over five, that's one, you're not changing the value, and you're not changing the value by multiplying by four over four.
Multiplying a fraction by or leaves its value unchanged because each factor equals 1.
The factor used is a fraction with equal nonzero numerator and denominator.
For the fractions shown in this worked example.
Equation a) is written as "".
Speaker identifies the denominators in equation a) as 2 and 3 and says a common denominator is needed.
Written plan says "Find Common Denominator" and "x Multiply through".
The actual common denominator value is not written before the clip ends.
No multiplied equation is shown.
The first displayed equation is written on the page.
Directly observed from the handwritten formula.
The speaker points out the denominators involved in equation a).
Audio states the need to eliminate the denominator "between two and three," and the written equation also shows on the right-hand side.
The chosen first step is to find a common denominator for the fractional terms.
Explicitly written on the page and stated in the audio.
After finding the common denominator, the next planned step is to multiply the entire equation through by it.
Explicitly written on the page and stated in the audio.
The clip establishes the method for equation a): identify the denominators, find a common denominator, and multiply through. The execution is not reached within this segment.
The speaker identifies the common denominator for equation a) as 6 and multiplies through.
The written steps show , then , and finally .
Start with the original first equation.
Given in the video.
Multiply every term in the equation by the common denominator 6.
Multiplication property of equality.
Perform the multiplication in the numerators.
Arithmetic multiplication.
Reduce each fraction by dividing the numerator by the denominator.
Simplifying fractions.
The first equation simplifies to .
The speaker identifies the common denominator for equation b) as 20 and multiplies through.
The written steps show , then , and finally .
Start with the original second equation.
Given in the video.
Multiply every term in the equation by the common denominator 20.
Multiplication property of equality.
Perform the multiplication in the numerators.
Arithmetic multiplication.
Reduce each fraction by dividing the numerator by the denominator.
Simplifying fractions.
The second equation simplifies to .
The instructor adds the two displayed equations and writes , then .
He says, 'We just add the equations. So I have 18x ... equals 36, so x is equal to 2.'
Start with the first reduced equation.
Given equation on the page.
Start with the second reduced equation.
Given equation on the page.
Add the equations term by term.
Elimination by addition when y-coefficients are opposites.
The y-terms cancel and the x-terms combine.
Arithmetic simplification.
Divide both sides by 18.
Solving a one-variable linear equation.
The elimination step gives .
The instructor substitutes into , obtains , multiplies by 3 to get , then writes -2y = 4 and .
He narrates each algebraic step, including multiplying everything by 3 and dividing by -2.
Substitute into the first original equation.
Back-substitution of the known value.
Compute ()(2)=1.
Fraction multiplication and simplification.
Multiply every term by 3 to clear denominators.
Equivalent transformation using a common denominator.
Subtract 3 from both sides.
Isolating the y-term.
Divide both sides by -2.
Solving a one-variable linear equation.
The remaining unknown is , so the ordered pair is (2,-2).
The full visible chain is , then , then , then .
The narration follows the same order: substitute, multiply, find common denominator, subtract, reduce, conclude correctness.
Start from the second equation and replace by 2 and by .
Substitution of the candidate solution into the equation being checked.
Evaluate each product separately.
Multiplying a fraction by an integer multiplies the numerator; the sign of carries into the second term.
Rewrite both fractions with denominator 20.
Use equivalent fractions obtained by multiplying by and , which do not change value.
Subtract the numerators while keeping the common denominator.
Rule for subtracting fractions with the same denominator.
Reduce the result to lowest terms.
Divide numerator and denominator by their common factor 2.
Compare the simplified left-hand side with the right-hand side of the original equation.
A true numerical equality confirms the candidate pair satisfies the equation.
The ordered pair checks in the displayed equation, so the verification succeeds.
Speaker says, "So in the first tutorial I'm going to go through a fraction example."
The page shows "Fraction Example" followed by two labeled equations.
Equation a) is "" and equation b) is "".
The clip does not solve either equation.
It is not shown whether these are meant as a system or as separate practice equations, although the speaker says they are treated independently for denominator elimination.
Solve linear equations with rational coefficients, beginning by eliminating denominators.
Equation a):
Equation b):
Set up the first step for removing fractional denominators from each equation.
The presenter labels the equations before discussing the method.
Observed directly from the handwriting and audio naming.
The speaker focuses first on equation a) and notes the denominators 2 and 3.
Audio explicitly discusses equation a) and the denominators between 2 and 3.
The written plan states that the next task is to find a common denominator.
Written on the page beneath the equations.
The written plan states that after finding the common denominator, one should multiply through.
Written on the page and echoed in the audio.
No final solution is given in this clip; only the setup step is completed.
The clip ends before any cleared-denominator equation or solution is produced, so verification is not available within this segment.
The full system and all transformation steps are written on the paper.
The speaker narrates the process of finding common denominators, multiplying through, and reducing the equations.
Solve the system of linear equations: a) , b) .
Equation a):
Equation b):
Transform the system into one with integer coefficients to prepare for solving.
Identify the least common multiple of the denominators 2, 3, and 3.
Finding a common denominator to clear fractions.
Multiply equation a) by 6 and simplify.
Multiplication property of equality and fraction reduction.
Identify the least common multiple of the denominators 4, 5, and 10.
Finding a common denominator to clear fractions.
Multiply equation b) by 20 and simplify.
Multiplication property of equality and fraction reduction.
The transformed system is and .
The speaker notes that this new system can now be solved using substitution or elimination.
The page presents the original fractional system and , then the reduced system and .
The instructor solves the system by elimination, substitutes into the first original equation, finds , and begins a check in the second equation.
The final verification line is not completed within the clip.
Solve the system given by the original equations and , using the reduced integer system shown on the page.
Original equation a): .
Original equation b): .
Reduced system shown: and .
Find the ordered pair (x,y) satisfying the system and begin verifying it.
Use the reduced integer system because the y-coefficients are opposites.
Preparation for elimination.
Add the two equations to eliminate y.
Opposite coefficients cancel under addition.
Solve for x.
Divide by 18.
Substitute into the first original equation.
Back-substitution to find y.
Simplify the substituted term.
()(2)=1.
Multiply through by 3 to remove fractions.
Common denominator clearing.
Subtract 3 from both sides.
Isolate the y-term.
Divide by -2.
Solve for y.
Begin checking the solution in the second original equation.
Verification step stated by the instructor.
Solution (2,-2)
The instructor starts the check by substituting (2,-2) into the second original equation, but the clip ends before the arithmetic is finished.
The page heading includes Fraction Example, and the worked line is .
The instructor narrates the whole check and ends by saying we are correct there.
Verify that the previously found solution , satisfies the second equation of the displayed system.
Equation to check: .
Candidate values: , .
Show whether substituting the candidate values makes the equation true.
Substitute the known values into the left-hand side.
Direct substitution for verification.
Compute the two products.
Fraction-times-integer multiplication.
Convert to a common denominator of 20.
Equivalent fractions created by multiplying by and .
Subtract the numerators.
Same-denominator fraction subtraction.
Reduce the fraction.
Cancel the common factor 2.
Yes; the left-hand side simplifies to , matching the right-hand side, so is verified.
The final comparison is the true statement .
A hand writes the heading across the top of graph paper.
The visible text becomes "Section 1.6 Solving linear equation with Rational #s".
The exact spacing and capitalization vary in the handwriting, but the wording is legible.
graph paper
hand
pen
heading text
Blank page becomes titled with the section name.
The camera remains fixed overhead.
No equations are present yet.
The visual establishes the lesson topic before any example is introduced.
The hand writes "Fraction Example" and then adds two labeled equations underneath.
The equations become visible as a) and b) .
heading "Fraction Example"
equation a)
equation b)
pen
paper
A new subheading is added.
Two equations are written one after the other.
The page changes from title-only to a worked example setup.
The equations remain in fractional form throughout this interval.
The visual introduces the concrete example that the rest of the clip analyzes.
The hand writes a short method note below the equations.
The visible note reads "Find Common Denominator" and then "x Multiply through".
The symbol before "Multiply through" is a multiplication sign written as "x".
method note
multiplication sign
paper
A procedural reminder is added beneath the example.
No algebraic transformation of the equations is shown yet.
The visual records the intended next step: clear fractions by multiplying through by a common denominator.
A hand writes mathematical equations and steps on graph paper throughout the video.
Hand
Pen
Graph paper
Mathematical equations
The hand writes the initial system of equations.
The hand writes the multiplication steps for equation a).
The hand writes the simplified equation a).
The hand writes the multiplication steps for equation b).
The hand writes the simplified equation b).
The camera angle remains fixed on the graph paper.
The lighting remains consistent.
The visual progression shows the step-by-step algebraic manipulation required to clear fractions from a linear system.
A handwritten sheet titled 'Section 1.6 Solving linear equation with Rational #' shows two original fractional equations at the top, their cleared-denominator forms, and below them the reduced system and .
Title line
Original equations a) and b)
Intermediate cleared-denominator equations
Reduced system and
No new mathematics is introduced at the very start; the page already contains the setup for the worked example.
The original fractional equations remain visible throughout.
The reduced integer system remains visible throughout.
The visual organization separates the problem statement from the solution work, making the elimination setup explicit before any new steps are written.
The pen writes 'Solve using elimination', then 'add the equation', draws a line under the system, and writes followed by .
System of two equations
Written instruction 'Solve using elimination'
Sum line
Result
Result
The y-terms disappear visually after addition.
A new one-variable equation appears below the system.
x is isolated in the next line.
The original two equations stay above the new work.
The operation being demonstrated is addition of equations.
The writing sequence shows the causal structure of elimination: align equations, add them, cancel opposite terms, then solve the remaining equation.
The pen writes 'Solve for y given ', then substitutes into , simplifies to , multiplies by 3 to get , then writes -2y = 4 and .
Instruction 'Solve for y given '
Substituted equation
Simplified equation
Cleared-denominator equation
Final y-value
x is replaced by 2.
Fractions are removed by multiplying by 3.
The equation is transformed stepwise until y is isolated.
The chosen equation is the first original equation.
Each new line is an equivalent transformation of the previous one.
The visual progression demonstrates how a known variable value is inserted into an original fractional equation and then reduced to a simple linear equation in one unknown.
The pen writes 'Solution (2, -2)' and then begins the check by writing ... on the lower left.
The check expression is incomplete by the end of the clip.
Ordered pair (2,-2)
Check expression using the second original equation
The solved values are packaged as an ordered pair.
A new verification line is started in the unused equation.
The solution pair remains (2,-2).
The verification uses the second original equation.
The page transitions from solving to validating, indicating that finding a candidate pair is not the final mathematical step until substitution confirms it.
A single sheet of graph paper fills the frame, with the title Section 1.6 Solving Linear Equation with Rational #'s and Fraction Example at the top, a solved system in the middle, and the verification work added at the bottom left.
Graph-paper worksheet
Handwritten title Section 1.6 Solving Linear Equation with Rational #'s
Handwritten subtitle Fraction Example
Previously written system-solving work
Bottom-left verification arithmetic
New lines are added sequentially under the existing work as the instructor writes the check.
Near the end the sheet is slid upward so more of the lower writing remains in view.
The camera remains fixed overhead.
The earlier solved work above stays visible while the new verification is written below it.
The visual layout separates prior solution work from the current verification step, making the clip function as a check of an already obtained answer rather than a fresh solve.
The hand writes , then , then the converted fractions and , then , and finally .
Pen tip
Successive handwritten fractions
Equality signs linking the steps
Each new line depends on the previous one: products first, then common-denominator rewrite, then subtraction, then reduction.
The target right-hand side is present from the start of the check.
The substituted values and do not change during the computation.
The writing order visually encodes the algebraic procedure: substitute, simplify products, rename fractions equivalently, subtract, reduce, compare.
Speaker mentions "maybe even roots and stuff like that" while listing possible coefficient forms.
Only fractional-coefficient equations are actually written in this clip.
One might infer from the spoken list that roots, decimals, and bracketed forms are all being solved in this segment.
In this clip, only a fraction example is written and discussed; roots and decimals are mentioned verbally as other possible cases but are not demonstrated.
The speaker says, 'Oops, this is a plus sign here, right? Plus sign, plus sign, make sure we don't make those mistakes.'
The speaker corrects a written minus sign to a plus sign in the intermediate step for equation b).
When multiplying terms by a common denominator, it is easy to accidentally change a plus sign to a minus sign or vice versa.
Carefully check the sign of each term before and after multiplying through to ensure the equation remains equivalent.
The instructor says it won't always be set up like that and sometimes you'll have to do some manipulation before.
Learners may think any linear system can be solved by simply adding the equations as written.
Direct addition works here only because the y-coefficients are already opposites; otherwise the equations must first be manipulated.
The instructor calls the check the last step to reassure ourselves that the answer is correct.
Students may stop after obtaining (2,-2) and assume no further validation is needed.
The video explicitly models a final check by substituting the pair into the other original equation.
The instructor explicitly warns that multiplying by or is multiplying by one, so you are not changing the value.
Students may think that changing to or to alters the number.
The video stresses that these are equivalent fractions produced by multiplying by forms of 1, so the values remain the same while the denominators become compatible for subtraction.
He says we have to do a common denominator here to subtract these two.
Learners may attempt to subtract and without first rewriting them.
The clip shows that the subtraction is postponed until both fractions share denominator 20, after which only the numerators are subtracted.
The speaker first states the general topic and then says, "So in the first tutorial I'm going to go through a fraction example."
The written title is followed by the written "Fraction Example" and two equations.
The general topic of solving linear equations with rational coefficients is instantiated by the displayed fraction example.
After showing the equations, the speaker explains the first goal is to eliminate denominators.
The written note "Find Common Denominator" and "x Multiply through" appears directly under the example.
The example motivates the specific method of clearing fractional coefficients by finding a common denominator and multiplying through.
The speaker lists fractions among the possible rational-number forms.
The written example uses fractional coefficients.
The displayed example is the fraction case within the broader set of rational-coefficient forms mentioned by the speaker.
The speaker explains the general method and then applies it to the specific equations.
The general method for clearing fractions is applied to derive the simplified first equation.
The speaker explains the general method and then applies it to the specific equations.
The general method for clearing fractions is applied to derive the simplified second equation.
The speaker states, 'Then we reduce the equation.'
The derivation of the simplified equations depends on the method of reducing fractions after multiplying through.
After obtaining by elimination, the instructor says then we would solve for y given that we know .
The next written step substitutes into the first original equation.
Elimination produces one variable value, and that result is then applied through back-substitution to determine the other variable.
After substitution gives , the instructor multiplies by 3 to obtain .
He explains eliminating the denominator by using the common denominator 3.
Back-substitution creates a fractional one-variable equation, and clearing denominators is the method used to simplify it.
The instructor says the last step is to reassure ourselves by a check and will use the other equation.
He begins substituting (2,-2) into the second original equation.
Once the ordered pair is found by substitution, the next procedural step is verification in the remaining original equation.
The page shows the original fractional equations above and the reduced integer system below, with arrows indicating transformation.
The clip does not re-derive the reduction; it is already written on the page at the start.
The elimination method is applied after the original rational system has already been converted to an integer system by clearing denominators.
The verification begins with substituted values and only then proceeds to fraction multiplication.
The arithmetic simplification can begin only after the candidate values have been inserted into the equation.
The speaker says a common denominator is needed to subtract these two, then performs the subtraction after rewriting.
Same-denominator subtraction is applied only after the fractions have been converted to denominator 20.
Opening narration states the subject of the tutorial.
Title text is visible on the page.
Speaker explains the first goal is to eliminate denominators.
Written note says "Find Common Denominator" and "x Multiply through".
Both equations are written clearly on the page.
Speaker says each equation is treated independently.
The speaker explains the process of finding a common denominator and multiplying through.
The speaker explicitly finds the common denominators 6 and 20 for the two equations.
The speaker corrects a sign error and warns about making such mistakes.
The instructor explains that because there is a negative 4y and a positive 4y, one can just add the equations.
The instructor says they could choose one of those original equations and then substitutes into the first one.
The instructor says the common denominator would be 3 and to multiply everything by 3.
The instructor says the last step is to reassure ourselves by a check and will use the other equation.
The instructor says we know that we are correct there after the arithmetic matches.
The page shows substitution into .
Covered · Brief opening pause before speech begins; no mathematical content beyond the start of the tutorial setup.
Covered · Spoken introduction and handwritten section title establish the topic of solving linear equations with rational coefficients.
Covered · Speaker lists possible rational-coefficient forms such as fractions, decimals, brackets, binomial tops, and roots.
Covered · The presenter writes "Fraction Example" and the two equations a) and b).
Covered · Speaker explains the first goal is to eliminate denominators, treats each equation independently, and writes the plan to find a common denominator and multiply through.
Covered · The entire clip covers the method and application of clearing fractions from a linear system.
Covered · The page already displays the original fractional system, the cleared-denominator forms, and the reduced integer system; the instructor introduces elimination as the preferred method.
Covered · The instructor explains that the opposite y-coefficients allow direct addition, writes the elimination steps, and derives .
Covered · Using , the instructor substitutes into the first original equation, clears denominators by multiplying by 3, and solves to get .
Covered · The solution (2,-2) is written, and the instructor begins a verification in the second original equation; the clip ends before the check is completed.
Covered · The clip opens with the substituted verification line and the statement that will give .
Covered · The instructor computes and from the substituted products.
Covered · He explains the need for a common denominator and rewrites the fractions using and .
Covered · The subtraction, reduction to , and conclusion that the solution is correct are shown.
Covered · No new mathematics is introduced; the presenter slides the page up and announces that the next tutorial will use fractions with binomial tops.
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