Mario's Math Tutoring · YouTube · 7:13
This introductory algebra whiteboard segment teaches how to solve absolute value equations by first interpreting absolute value as distance. The instructor reviews |-2|=2 and |2|=2 on a number line, then solves 3|x-4|+ by isolating |x-4|=4. He explains the equation geometrically as numbers four units from 4, giving 0 and 8, and algebraically by splitting into or . He checks both solutions and introduces |2x-3|=-2 as a trick example because an absolute value cannot equal a negative number. This 150-second Algebra 1 whiteboard segment teaches how to solve absolute value equations. It first shows that |2x-3| = -2 has no solution because an absolute value cannot equal a negative number. It then solves || = 3 by splitting into or , obtaining or , and checking both by substitution. After a short promotional interruption, the lesson returns to more complicated equations: -1|x-4|+ and 5|x-10|-5=20. In both cases, the instructor isolates the absolute value expression first by undoing outside operations, then splits into two linear equations, yielding or and or respectively. The board also displays a general absolute deviation template, |X - TARGET VALUE| = DEVIATION, and the clip ends as a word problem is introduced but not yet solved. This whiteboard algebra segment teaches how to model a tolerance problem with an absolute value equation. The instructor introduces the formula |x - TARGET VALUE| = DEVIATION, applies it to a basketball hoop set at 10 ft with an allowed deviation of .25 in, converts 10 ft to 120 in, writes |h - 120| = .25, splits it into h - 120 = .25 or h - 120 = -.25, and solves to obtain and . He then interprets these as the maximum and minimum allowable heights.
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 on a whiteboard titled 6.5 SOLVING ABSOLUTE VALUE EQUATIONS. Several practice equations are already written, including |2x-3|=-2, ||=3, -|x-4|+, 5|x-10|-5=20, and the worked example 3|x-4|+.
The instructor reviews the meaning of the vertical bars. On the right side of the board, ABSOLUTE VALUE is illustrated with |-2|=2 and |2|=2, together with a number line labeled DISTANCE FROM ZERO. The key idea is that absolute value measures how far a number is from zero, so both -2 and 2 are two units away.
The first solving strategy is to isolate the absolute value expression. Starting from 3|x-4|+, the instructor works from the outside in: subtract 2 from both sides to obtain 3|x-4|=12, then divide both sides by 3 to obtain |x-4|=4.
The isolated equation |x-4|=4 is then interpreted geometrically. Since subtraction gives a difference and difference can be understood as distance, |x-4| means the distance between x and 4. A number line centered at 4 shows that x can be 4 units to the left or 4 units to the right, giving or .
The instructor next presents the algebraic method. From |x-4|=4, split the equation into two cases: or . Adding 4 to both sides in each case gives or . This matches the geometric number-line result.
The reason for considering both cases is that the expression inside absolute value bars could originally have been negative; taking absolute value turns that distance positive. The solutions are checked directly: |0-4|=|-4|=4 and |8-4|=|4|=4, so both satisfy the isolated equation.
The segment closes by introducing the board example |2x-3|=-2 as a trick question. Because an absolute value represents distance and cannot equal a negative number, this equation has no real solution; the instructor begins this explanation just before the clip ends.
The clip opens on a whiteboard titled 6.5 Solving Absolute Value Equations. Problem ① is |2x-3| = -2. The instructor points out that an absolute value cannot produce a negative result, so this equation has no solution, and he writes NO SOLN. under it.
He moves to problem ②, || = 3. Because the absolute value expression is already isolated and the right side is positive, he rewrites it as two separate linear equations: or .
Solving each case gives from the first equation and from the second. He combines them with “or,” stating the solution set as or .
To verify, he substitutes both answers back into the original equation. For , |-2+5|=|3|=3. For , |-8+5|=|-3|=3. Both checks confirm the solutions.
From about 45 to 70 seconds, the math board is replaced by a promotional segment for an Algebra 1 video course. No new mathematical content is introduced in this interval.
Returning to the board, the instructor restates the core idea: split an isolated absolute value equation into a positive case and a negative case. He then turns to problem ③, -1|x-4|+.
Before splitting, he isolates the absolute value expression by working from the outside in. First he subtracts 2 from both sides to get -1|x-4|=-9. Then he multiplies both sides by -1 to obtain |x-4|=9.
Now he splits into or . Adding 4 to both sides in each case yields or . He notes that these can again be checked by substitution.
For problem ④, 5|x-10|-5=20, he repeats the same strategy. Add 5 to both sides to get 5|x-10|=25, then divide by 5 to isolate |x-10|=5.
He splits this into or . Adding 10 to both sides gives or , completing the fourth worked example.
At the end, he introduces one more problem, a word problem. The board also shows a general template for absolute deviation: |X - TARGET VALUE| = DEVIATION, but the word problem itself is not solved within this clip.
The segment opens on a whiteboard lesson titled 6.5 SOLVING ABSOLUTE VALUE EQUATIONS. At the lower left, a new word problem asks for an absolute value equation modeling a basketball hoop installed at 10 ft with an absolute deviation of .25 inches.
The instructor first points to the general tool at the lower right: ABSOLUTE DEVIATION: |x - TARGET VALUE| = DEVIATION. He explains that x is the actual variable, TARGET VALUE is what you are aiming for, and DEVIATION is how far off the actual value is from the target.
He emphasizes that deviation can go in either direction: the hoop may be slightly higher or slightly lower than target, but not by more than .25 inches. This motivates the use of absolute value rather than a one-sided inequality in the setup.
To apply the formula, he chooses h for the actual height. Because the deviation is given in inches, he converts the target height from 10 ft to 120 in. With matching units, he writes the model |h - 120| = .25.
Next he solves the absolute value equation by splitting it into two linear equations: h - 120 = .25 or h - 120 = -.25. Adding 120 to both sides of each equation gives and .
He then interprets the results in context. The larger value, 120.25 in, corresponds to 10 ft in, the maximum allowable height. The smaller value, 119.75 in, is the minimum allowable height. He closes by restating the reusable formula |x - TARGET VALUE| = DEVIATION.
The vertical bars in expressions such as |-2| and |2| denote absolute value. Absolute value measures distance from zero on the number line, so both -2 and 2 have absolute value 2. This distance interpretation explains why absolute value outputs are nonnegative.
When solving an absolute value equation, undo the operations outside the bars first. For 3|x-4|+, subtract 2 from both sides and then divide by 3 so that the absolute value expression stands alone.
The expression |x-4| represents the distance between x and 4. The equation |x-4|=4 asks for numbers that are exactly 4 units away from 4. On a number line, those points are 4 units left of 4 and 4 units right of 4.
After isolation, convert |x-4|=4 into two linear equations: one where the inside equals the positive right-hand side and one where it equals the negative right-hand side. This produces both possible solutions.
From , add 4 to get . From , add 4 to get . Both values check in the isolated equation because |8-4|=4 and |0-4|=4.
The board summarizes the structure as |x - TARGET VALUE| = EQUATION. Mathematically, this means the distance between x and a target value equals a specified amount. In the worked example, the target value is 4 and the distance is 4.
An absolute value cannot equal a negative number because it represents distance. The equation |2x-3|=-2 is therefore impossible over the real numbers. The video identifies it as a trick example and begins this explanation before ending.
If an equation has the form |expression| = c with , then there is no real solution. The video demonstrates this with |2x-3| = -2 and labels it NO SOLN.
When the absolute value expression is isolated and the right-hand side is nonnegative, rewrite the equation as two cases: or . This is used for ||=3.
Split into or , then subtract 5 in each case to obtain or .
Substitute each candidate back into the original equation. For ||=3, gives |3|=3 and gives |-3|=3, so both are valid.
If extra operations surround the absolute value, undo them first using inverse operations. Only after the absolute value expression stands alone should you split into two cases.
Subtract 2 to get -1|x-4|=-9, multiply by -1 to get |x-4|=9, then split into or . Adding 4 gives or .
Add 5 to get 5|x-10|=25, divide by 5 to get |x-10|=5, then split into or . Adding 10 gives or .
The board gives a general model for word problems involving deviation from a target value: the absolute value of the difference between X and the target equals the deviation.
The lesson introduces a reusable modeling formula for tolerance problems: the absolute value of the difference between the actual variable and the target value equals the allowed deviation. The instructor describes deviation as how far off you are from what you are aiming for.
For a hoop target of 10 ft with allowed deviation .25 in, the instructor first converts 10 ft to 120 in so the units match. Using h for actual height, the situation is modeled by |h - 120| = .25.
Because the deviation is stated in inches, the target height cannot stay as 10 ft inside the same equation. Converting to 120 inches keeps all quantities dimensionally consistent before solving.
To solve |h - 120| = .25, rewrite it as two linear equations joined by or: h - 120 = .25 or h - 120 = -.25. This captures the idea that the actual height may be above or below the target by the same amount.
Adding 120 to both sides of each linear equation gives the two boundary solutions and . These are the extreme heights allowed by the .25-inch deviation from the 120-inch target.
The instructor explains that 120.25 inches is 10 ft in, the maximum allowable height, while 119.75 inches is the minimum allowable height. For a story problem, he stresses writing the conclusion with units.
Explore conditions, steps and evidence. Supplementary explanations are labeled separately from content shown in the video.
The instructor says that the vertical bars mean absolute value.
Whiteboard examples use notation such as |2x-3|, ||, |x-4|, and |x-10|.
| |
Absolute value; the distance of the enclosed quantity from zero.
Real-valued expressions inside the bars.
The worked equation is 3|x-4|+, and the algebraic split uses or .
x
Unknown real number whose distance from 4 is being solved for in the example.
Real numbers.
The board states |x - TARGET VALUE| = EQUATION.
In the example |x-4|=4, the instructor interprets 4 as the reference point on the number line.
TARGET VALUE
Reference point in an absolute value expression; |x-a| represents the distance between x and a.
Real number reference point.
The board writes |x - TARGET VALUE| = EQUATION.
The instructor solves |x-4|=4 by splitting into positive and negative scenarios.
The word EQUATION on the board is used informally to mean the right-hand constant or target distance, not a full equation.
EQUATION
Right-hand side value in an isolated absolute value equation; interpreted as the required distance from the target value.
Nonnegative real number for solvable cases.
A number line is drawn with marks at -2, 0, and 2 under DISTANCE FROM ZERO.
The board shows |-2|=2 and |2|=2.
-2, 0, 2
Sample points used to show that both -2 and 2 are two units from 0.
Number line positions.
The algebraic work ends with and circled.
The number line for |x-4|=4 marks 0 and 8 as four units left and right of 4.
,
Solutions to the example equation 3|x-4|+ after isolating |x-4|=4.
Real numbers.
Throughout the worked examples, vertical bars denote absolute value, as in |2x-3|, ||, |x-4|, and |x-10|.
||
Absolute value of the enclosed expression.
Real-valued algebraic expressions.
The unknown variable x appears in all four worked equations and their solution steps.
x
Unknown real number to be solved for in each equation.
Real numbers.
The instructor says to split the absolute value equation into two equations and uses “or” between them.
Written solution lines use “or” between the two resulting linear equations or answers.
Logical disjunction connecting the positive-case and negative-case equations or solutions.
Used between equations and between solution values.
The instructor says, "I'll just use h for height."
He writes h in the equation |h - 120| = .25 and in the two resulting linear equations.
h
actual height of the basketball hoop, measured in inches after unit conversion
real numbers; in this problem it represents a physical height in inches
The instructor says the target height is 10 feet and converts it to 120 inches.
The number 120 appears inside |h - 120| = .25 and in h - 120 = .25 or h - 120 = -.25.
120
target height of the basketball hoop expressed in inches
constant length in inches
The instructor says the deviation is .25 and explains that the hoop should not be off by more than .25 inches.
The number .25 appears on the right side of |h - 120| = .25 and as +.25 and -.25 in the split equations.
.25
allowed absolute deviation from the target height, in inches
positive constant length in inches
The instructor says absolute value tells you the distance from zero.
The board lists ABSOLUTE VALUE: |-2|=2 and |2|=2.
A number line labeled DISTANCE FROM ZERO shows -2, 0, and 2 with equal distances marked.
The vertical bars denote absolute value. For a real number, its absolute value is its distance from zero on the number line. Because distance is nonnegative, both |-2| and |2| equal 2, and taking an absolute value gives a positive result in these examples.
Applies to real numbers or real-valued expressions inside absolute value bars.
The value represented is a distance, so it is nonnegative.
The instructor says the first thing is to isolate or get the absolute value quantity by itself, working from the outside in.
The board transforms 3|x-4|+ by subtracting 2 and dividing by 3 to obtain |x-4|=4.
To solve an absolute value equation, first undo operations outside the absolute value bars. In the example 3|x-4|+, subtract 2 from both sides to get 3|x-4|=12, then divide both sides by 3 to isolate |x-4|=4.
Use inverse operations on both sides of the equation.
The goal is to leave the absolute value expression alone on one side.
The instructor explains that minus means difference and difference means distance in geometry, so the distance between some number and 4 is 4 units.
The isolated equation is |x-4|=4.
A number line centered at 4 marks solutions 4 units to the left and right.
An expression such as |x-4| represents the distance between x and 4 on the number line. The equation |x-4|=4 asks for numbers whose distance from 4 is exactly 4 units.
x is treated as a real number on a number line.
The right-hand side is interpreted as a distance.
The instructor says to split the equation into two equations, one with the positive scenario and one with the negative scenario.
The board writes or , then solves to and .
Once the absolute value expression is isolated, set the inside expression equal to both the positive and negative versions of the right-hand side. For |x-4|=4, this gives or . Solving each equation gives the two possible solutions.
The absolute value expression must be isolated first.
Both the positive and negative cases must be considered because the inside expression could have been negative before applying absolute value.
The board writes * ABSOLUTE DEVIATION: |x - TARGET VALUE| = EQUATION.
The instructor connects the example |x-4|=4 to distance from a target value.
The board label EQUATION is informal; mathematically it functions as the right-hand distance value.
The board summarizes the structure of an absolute value equation as |x - TARGET VALUE| = EQUATION. In words, this means the distance between x and a target value equals a specified amount. In the worked example, the target value is 4 and the specified distance is 4.
Used as a template for interpreting absolute value equations.
The right-hand side should be nonnegative if real solutions are expected.
“A negative value when you take the absolute value. So this one is actually no solution.”
Problem ① on the board is |2x-3| = -2, and the instructor writes “NO SOLN.” beneath it.
The video states that if an equation requires an absolute value expression to equal a negative number, then there is no solution. This is demonstrated directly on problem ①, where |2x-3| = -2 is labeled NO SOLN.
The right-hand side is negative.
The left-hand side is an absolute value expression.
“So this one, what we're going to do is we're going to split it up into two equations: equals 3 or equals negative 3.”
Under problem ②, the board shows or .
When an absolute value expression is already isolated and equals a nonnegative constant, the instructor solves it by rewriting it as two separate linear equations: one with the positive constant and one with its opposite.
The absolute value expression is isolated on one side.
The constant on the other side is nonnegative.
“Now the nice thing about algebra is you can check your answer. You put it back in, negative 2 plus 5 is 3, absolute value of 3 is 3. Negative 8 plus 5 is negative 3, absolute value of negative 3 still gives us positive 3.”
After solving, the instructor verifies each candidate by substituting it back into the original absolute value equation and evaluating the absolute value.
Use after obtaining candidate solutions from the two-case split.
“Again, our goal is to get that part in absolute value by itself. So I'm going to work from the outside in by doing the opposite operation.”
For problem ③, the board shows subtracting 2 and then multiplying by -1 before splitting.
For equations with extra operations around the absolute value, the instructor first undoes those outer operations step by step until the absolute value expression stands alone, then splits into two equations.
The absolute value expression is not initially isolated.
Outer operations must be undone using inverse operations.
On the right side of the whiteboard, a starred note reads ABSOLUTE DEVIATION: |X - TARGET VALUE| = DEVIATION.
The word problem itself is only introduced at the end of the clip; no full setup or solution is shown within this segment.
The board includes a general modeling formula for absolute deviation problems: the absolute value of the difference between a variable quantity and a target value equals the allowed deviation.
Used for word problems involving deviation from a target value.
The instructor introduces "this absolute deviation equation" and explains x is the variable, the target value is what you are aiming for, and the difference is the deviation.
The board displays ABSOLUTE DEVIATION: |x - TARGET VALUE| = DEVIATION.
The video presents a modeling formula for situations involving tolerance around a target. The absolute value of the difference between the actual variable and the target value equals the allowed deviation. In words, deviation means how far off the actual value is from the target.
The variable, target value, and deviation must be expressed in compatible units.
The deviation is used here as a nonnegative allowed distance from the target value.
The instructor reads the story problem about setting up a basketball hoop at a height of 10 feet with an absolute deviation of .25 inches.
The board text reads: YOU WANT TO SET UP A BASKETBALL HOOP AT A HEIGHT OF 10 FT WITH AN ABSOLUTE DEVIATION OF .25 INCHES. WRITE AN ABSOLUTE VALUE EQUATION TO MODEL THIS SITUATION AND SOLVE.
The method is to translate a tolerance statement into an absolute value equation. The actual quantity is compared with the target quantity, and the allowed error becomes the right-hand side of the absolute value equation.
The problem gives a target value and an absolute deviation.
The requested model is an absolute value equation.
The instructor says that when we take the absolute value, it always gives us a positive result.
The board shows |-2|=2 and |2|=2.
Near the end, the instructor begins explaining that |2x-3|=-2 is a trick question because absolute value always has to be positive.
The spoken statement uses positive result; the standard mathematical statement is nonnegative result, since |0|=0.
For the examples shown, taking an absolute value gives a nonnegative distance; the instructor phrases this as always giving a positive result.
The expression inside the absolute value bars is a real number or real-valued expression.
For the displayed examples |-2| and |2|, and for the general rule being introduced in the lesson.
The board lists example ① |2x-3|=-2.
The instructor calls it a trick question and begins saying that absolute value always has to be positive, so you can never have...
The clip ends before the instructor completes the sentence or explicitly writes no solution.
The conclusion is strongly indicated but not fully stated within the provided segment.
The equation |2x-3|=-2 is presented as a trick question because an absolute value cannot equal a negative number; therefore it has no real solution.
The equation is over the real numbers.
The absolute value expression is already isolated on one side.
For real-valued absolute value expressions, no real x satisfies an equation of the form |expression| = negative number.
“A negative value when you take the absolute value. So this one is actually no solution.”
Problem ① is |2x-3| = -2 and is marked NO SOLN.
If an equation has the form |expression| = c with , then the equation has no real solution.
c is negative.
The left side is an absolute value expression.
For real-valued expressions and constants.
“we're going to split it up into two equations: equals 3 or equals negative 3.”
The board rewrites ||=3 as or .
For a nonnegative constant c, an isolated equation |u| = c is solved by the two cases or .
The absolute value expression is isolated.
c >= 0.
For real-valued u and c.
The instructor says, "deviation means how far off are you," then explains the hoop could be a little higher or a little lower but not by more than .25 inches.
The board labels the right side of |x - TARGET VALUE| = DEVIATION as DEVIATION.
In this lesson, deviation means the amount by which the actual value differs from the target value, whether the actual value is above or below the target.
There is a target value.
There is an actual variable value.
The comparison is made through absolute value.
for the variable and target value in the modeled situation
The isolated equation |x-4|=4 is written on the board.
A number line is drawn with 4 marked and arrows or spacing indicating 4 units left and right.
The instructor says the answer could be four units to the right or four units to the left, giving 0 or 8.
Start with the example equation written on the board.
Given equation from the whiteboard.
After isolating the absolute value expression, interpret it as a distance equation.
Subtract 2 from both sides and divide by 3, as shown in the preceding algebra.
Read |x-4| as the distance from x to 4 on the number line.
The instructor explains that minus means difference and difference means distance in geometry.
Move 4 units left from 4 to reach 0, or 4 units right from 4 to reach 8.
Number line interpretation of a fixed distance from a target value.
The geometric method gives the solutions and .
The board writes or , then and circled.
The instructor says he would use the algebra method and split it into two equations.
Begin with the isolated absolute value equation.
Result of isolating the absolute value expression in 3|x-4|+.
Split the absolute value equation into a positive case and a negative case.
If an absolute value equals 4, the inside expression may be +4 or -4.
Solve the first linear equation by adding 4 to both sides.
Addition property of equality applied to .
Solve the second linear equation by adding 4 to both sides.
Addition property of equality applied to .
The algebraic method also gives or , matching the geometric number-line solution.
The instructor checks , whose absolute value is 4, and , whose absolute value is still 4.
Substitute into |x-4| and evaluate the absolute value.
Direct substitution and the definition of absolute value as distance from zero.
Substitute into |x-4| and evaluate the absolute value.
Direct substitution and the definition of absolute value as distance from zero.
Both and satisfy |x-4|=4.
The board shows ||=3, then or , then or .
The instructor narrates splitting into two equations and solving each separately.
Start from the given absolute value equation.
Given problem ② on the board.
Rewrite the absolute value equation as two linear equations.
Apply the two-case method for an isolated absolute value equal to a positive number.
Solve the first linear equation.
Subtract 5 from both sides of .
Solve the second linear equation.
Subtract 5 from both sides of .
Combine the two results as the solution set.
The original absolute value equation is satisfied by either case.
The solutions to ||=3 are or .
The board shows -1|x-4|+, then -1|x-4|=-9, then |x-4|=9, then or , then or .
The instructor explains working from the outside in, subtracting 2, multiplying by -1, then splitting into two equations.
Start from the given equation.
Given problem ③ on the board.
Remove the added constant from the left side.
Subtract 2 from both sides.
Make the coefficient of the absolute value positive.
Multiply both sides by -1.
Split the isolated absolute value equation into two linear cases.
Apply the two-case method with right-hand side 9.
Solve the first linear equation.
Add 4 to both sides of .
Solve the second linear equation.
Add 4 to both sides of .
State the final solution set.
Either case satisfies the original equation.
The solutions to -1|x-4|+ are or .
The board shows 5|x-10|-5=20, then 5|x-10|=25, then |x-10|=5, then or , then or .
The instructor says to work from the outside in, divide by 5, then split into two equations.
Start from the given equation.
Given problem ④ on the board.
Move the constant term to the right side.
Add 5 to both sides.
Isolate the absolute value expression completely.
Divide both sides by 5.
Split into two linear equations.
Apply the two-case method with right-hand side 5.
Solve the first linear equation.
Add 10 to both sides of .
Solve the second linear equation.
Add 10 to both sides of .
State the final solution set.
Either case satisfies the original equation.
The solutions to 5|x-10|-5=20 are or .
The instructor writes and narrates the setup |h - 120| = .25, then solves h - 120 = .25 or h - 120 = -.25 by adding 120 to both sides.
The board shows |h - 120| = .25, then h - 120 = .25 OR h - 120 = -.25, then +, then and .
Use the absolute deviation formula with h as the actual height, 120 as the target height in inches, and .25 as the allowed deviation in inches.
Modeling step from the formula |x - TARGET VALUE| = DEVIATION after converting 10 ft to 120 in.
Remove the absolute value by allowing the inside expression to be either +.25 or -.25.
Definition of absolute value as distance from zero; equivalently, |A|=b gives or when b is nonnegative.
Add 120 to both sides of the first linear equation.
Addition property of equality.
Simplify the first equation to obtain the larger possible height.
Arithmetic simplification.
Add 120 to both sides of the second linear equation.
Addition property of equality.
Simplify the second equation to obtain the smaller possible height.
Arithmetic simplification.
The boundary heights satisfying the equation are inches and inches.
The instructor says the maximum height could be 10 feet and one quarter of an inch, and the lowest it could be would be 119.75 inches.
audio；The spoken interpretation mixes feet-and-inches wording for the upper bound with inches-only wording for the lower bound.
This is the height .25 inches above the 120-inch target.
It comes from solving h-120=.25.
Rewrite 120.25 inches as 10 feet plus one quarter inch.
Since 120 inches = 10 feet, the extra .25 inch is one quarter of an inch.
This is the height .25 inches below the 120-inch target.
It comes from solving h-120=-.25.
The solution values represent the extreme allowable heights: 10 ft in above target and 119.75 in below target.
The board shows 3|x-4|+ and the full solution process.
The instructor says, let's look at an example here and I'll show you how it works.
A number line around 4 is used to show solutions 0 and 8.
Solve the absolute value equation 3|x-4|+.
Equation: 3|x-4|+.
Unknown: real number x.
Goal: find all x satisfying the equation.
Isolate the absolute value expression, interpret it as a distance equation, and solve for x by both geometric and algebraic methods.
Subtract 2 from both sides to begin removing operations outside the absolute value bars.
Inverse operation for the +2 term; the instructor says to work from the outside in.
Simplify both sides after subtracting 2.
Arithmetic simplification shown on the board.
Divide both sides by 3 to isolate the absolute value expression.
Inverse operation for multiplication by 3.
The equation is now in isolated absolute value form.
Simplification of the previous division step.
Split the isolated absolute value equation into two linear equations.
The inside expression can equal either the positive or negative version of the right-hand side.
Add 4 to both sides in each linear equation.
Addition property of equality.
or .
The instructor checks both values: |0-4|=|-4|=4 and |8-4|=|4|=4, so both satisfy |x-4|=4.
The board lists ① |2x-3|=-2.
The instructor calls it a trick question and begins explaining that absolute value always has to be positive.
The explanation is cut off before the instructor explicitly states no solution.
No algebraic work is shown for this example within the clip.
Consider the equation |2x-3|=-2 listed as example ① on the board.
Equation: |2x-3|=-2.
The absolute value expression is already isolated.
The right-hand side is negative.
Recognize whether the equation can have a real solution based on the meaning of absolute value.
The equation asks for an absolute value to equal -2.
Given equation from the board.
An absolute value represents distance and cannot equal a negative number.
Definition of absolute value as distance from zero; the instructor states absolute value always has to be positive.
Within this clip, the instructor identifies it as a trick question and indicates that it cannot be satisfied because the right-hand side is negative; the explicit final phrase no solution is not completed before the clip ends.
The check is conceptual: a nonnegative absolute value cannot equal -2.
Problem ① on the board is |2x-3| = -2, and the instructor writes NO SOLN.
“So this one is actually no solution.”
Solve |2x-3| = -2.
Equation: |2x-3| = -2.
Determine whether real solutions exist.
Inspect the equation.
This is the given problem.
Conclude there is no solution.
An absolute value expression cannot equal a negative number.
No solution.
The conclusion follows directly from the nonnegativity of absolute value.
Problem ② is ||=3, rewritten as or , with solutions or circled.
The instructor solves both equations and then checks both answers by substitution.
Solve || = 3.
Equation: || = 3.
Find all real x satisfying the equation.
Split the absolute value equation into two cases.
Use the two-case method for |u|=c with .
Solve the positive case.
Subtract 5 from both sides of .
Solve the negative case.
Subtract 5 from both sides of .
State the solution set.
Both cases are valid.
Check the first solution.
Substitute into the original equation.
Check the second solution.
Substitute into the original equation.
or .
Substitution gives in both cases.
Problem ③ is -1|x-4|+, transformed on the board to |x-4|=9 and then to or .
The instructor explicitly says to work from the outside in and then split into two equations.
Solve -1|x-4| + .
Equation: -1|x-4| + .
Find all real solutions.
Subtract 2 from both sides.
Undo the addition outside the absolute value term.
Multiply both sides by -1.
Isolate the absolute value expression with a positive coefficient.
Split into two linear equations.
Apply the two-case method to |x-4|=9.
Solve the first case.
Add 4 to both sides of .
Solve the second case.
Add 4 to both sides of .
State the final answer.
Both cases satisfy the isolated equation.
or .
The instructor says the answers can be plugged back in and checked.
Problem ④ is 5|x-10|-5=20, transformed on the board to |x-10|=5 and then to or .
The instructor says to divide by 5, then split into two equations, then add 10 to both sides in each case.
Solve 5|x-10| - .
Equation: 5|x-10| - .
Find all real solutions.
Add 5 to both sides.
Undo the subtraction outside the absolute value term.
Divide both sides by 5.
Isolate the absolute value expression.
Split into two linear equations.
Apply the two-case method to |x-10|=5.
Solve the first case.
Add 10 to both sides of .
Solve the second case.
Add 10 to both sides of .
State the final answer.
Both cases satisfy the isolated equation.
or .
The instructor indicates the result is complete after solving both cases.
The instructor reads and solves the basketball hoop story problem aloud.
The board contains the full worked example from the prompt through the answers and .
You want to set up a basketball hoop at a height of 10 ft with an absolute deviation of .25 inches. Write an absolute value equation to model this situation and solve.
Target height = 10 ft
Allowed absolute deviation = .25 in
Variable chosen for actual height = h
Write an absolute value equation modeling the situation and solve for the boundary heights.
Convert the target height to inches so it matches the deviation unit.
Unit consistency before writing the equation.
Set the absolute difference between actual height and target height equal to the allowed deviation.
Apply the absolute deviation formula |x - TARGET VALUE| = DEVIATION.
Split the absolute value equation into two linear equations.
If |A|=b with , then or .
Solve the first linear equation by adding 120 to both sides.
Addition property of equality and arithmetic.
Solve the second linear equation by adding 120 to both sides.
Addition property of equality and arithmetic.
or , interpreted as the maximum and minimum allowable heights.
The instructor checks the meaning verbally by describing 120.25 in as 10 ft in and 119.75 in as the lower boundary.
The upper-right board area shows ABSOLUTE VALUE: |-2|=2, |2|=2, and DISTANCE FROM ZERO with a number line marked -2, 0, 2.
The instructor points out that -2 and 2 are each two units from zero.
Number line with marks at -2, 0, and 2.
Labels |-2|=2 and |2|=2.
Heading DISTANCE FROM ZERO.
The instructor gestures between the absolute value examples and the number line.
Equal distances from 0 to -2 and from 0 to 2 are visually emphasized.
Zero remains the reference point.
The distance interpretation applies to both negative and positive inputs.
The diagram shows that absolute value measures distance from zero, so both -2 and 2 have absolute value 2.
A second number line is drawn below the worked example with 4 marked and solutions placed four units away.
The instructor says the number could be four units to the right or four units to the left of 4.
Some small handwritten labels on the number line are difficult to read, but the spoken explanation identifies the key values as 4, 0, and 8.
Number line centered conceptually at 4.
Marks or labels for 0 and 8.
Distance indication of 4 units from 4.
The reference point changes from 0 in the definition example to 4 in the equation example.
Two solution locations are identified symmetrically around 4.
The equation still expresses a distance condition.
The distance value remains 4 units.
The visual model translates |x-4|=4 into the statement that x is 4 units away from 4, giving or .
Arrows are drawn from |x-4|=4 down to the two equations and .
The resulting solutions and are circled.
Isolated equation |x-4|=4.
Two downward branches.
Linear equations and .
Circled solutions and .
One absolute value equation is transformed into two separate linear equations.
Each branch is solved independently.
Both branches come from the same isolated equation.
The positive and negative cases are treated symmetrically.
The branching diagram shows the standard method of splitting an absolute value equation into two cases.
The whiteboard is organized with lesson title 6.5 Solving Absolute Value Equations at top left, four numbered solve problems in the middle-left area, reference notes on the right, and a word-problem prompt at bottom left.
Title box: 6.5 Solving Absolute Value Equations
Problems ①–④
Right-side reference notes
Bottom-left word problem prompt
The instructor writes solutions under each numbered problem in sequence.
During 45–70 seconds, the scene cuts away to a promotional overlay and then returns to the same board.
The main board content remains the same set of absolute value equations and reference notes.
The right-side notes stay visible whenever the board is shown.
The visual layout separates worked examples from conceptual reference material, making the solution procedure easy to track problem by problem.
From about 45 to 70 seconds, the board view is replaced by the instructor speaking against a plain wall with a red promotional card reading Algebra 1 Video Course, Step by Step Lessons, 13 Chapters 87 Lessons!
Instructor on plain background
Red course advertisement card
The mathematical board disappears during the promo segment.
The board returns afterward with the previously written solutions still present.
No new mathematical content is introduced during the promo segment.
This interval is non-mathematical promotion inserted between the second example and the later examples.
A whiteboard fills the background with the title 6.5 SOLVING ABSOLUTE VALUE EQUATIONS, several earlier numbered examples, the new story problem at lower left, and the ABSOLUTE DEVIATION formula at lower right.
The instructor points first to the story problem, then to the formula, then writes the worked solution beneath the story problem.
Title 6.5 SOLVING ABSOLUTE VALUE EQUATIONS
Earlier solved examples 1-4
Story problem 5
ABSOLUTE DEVIATION formula
Handwritten solution lines for h
The instructor shifts attention from the written story problem to the general formula.
He then writes |h - 120| = .25 below the story problem.
He adds the split equations h - 120 = .25 OR h - 120 = -.25.
He writes + under the two equations.
He circles the final answers and .
The board remains a static whiteboard lecture setting throughout the clip.
The general formula |x - TARGET VALUE| = DEVIATION stays visible while the specific example is worked.
The visual layout connects a general absolute-deviation model to a concrete word problem by placing the formula beside the example and building the solution directly underneath the prompt.
While explaining deviation, the instructor gestures upward and downward with his hands near chest level.
Instructor’s hands
Spoken explanation of deviation
His hands move to indicate one direction above the target and another direction below it.
The target itself is not redrawn during this gesture sequence.
The gestures visually reinforce that absolute deviation allows error in either direction, higher or lower than the target value.
Example ① on the board is |2x-3|=-2.
The instructor says this is a trick question because absolute value always has to be positive and begins, you can never have...
The clip ends before the instructor finishes the explicit conclusion.
A student may try to solve |2x-3|=-2 as though the right-hand side were an ordinary allowable distance.
Since absolute value represents distance and is nonnegative, an isolated absolute value expression cannot equal -2. The video presents this as a trick question and indicates there is no real solution, although the final words are cut off.
The instructor says to consider both cases because whatever was originally inside could have been negative and absolute value makes it positive.
The board splits |x-4|=4 into or .
A student may solve only and miss the second solution.
The video emphasizes splitting into both a positive scenario and a negative scenario, producing both and .
“A negative value when you take the absolute value. So this one is actually no solution.”
Problem ① is explicitly marked NO SOLN.
One might try to split an equation like |2x-3| = -2 into two cases as if it had ordinary solutions.
The video shows that if the absolute value expression is set equal to a negative number, the correct conclusion is no solution.
“Again, our goal is to get that part in absolute value by itself. So I'm going to work from the outside in by doing the opposite operation.”
One might immediately split an equation such as -1|x-4|+ or 5|x-10|-5=20 into two cases before isolating the absolute value.
The instructor emphasizes undoing the outside operations first so the absolute value expression stands alone, and only then splitting into two equations.
The instructor explicitly converts 10 feet to 120 inches and says he did so because the deviation was in inches and he wanted the units to be the same.
One might write an equation using 10 directly together with .25 without converting units.
The target height must be rewritten as 120 inches so that both terms in |h - 120| = .25 are measured in inches.
The instructor says the hoop could be a little higher or a little lower, but not by more than .25 inches.
One might think deviation refers only to being too high.
Here deviation means distance from the target in either direction, which is why the absolute value equation splits into plus and minus cases.
The instructor first defines absolute value as distance, then says the first step in solving is to isolate the absolute value quantity.
Understanding absolute value as distance is used before applying the isolation method to an equation.
The board first transforms 3|x-4|+ into |x-4|=4, then splits |x-4|=4 into two equations.
The two-equation split is applied only after the absolute value expression has been isolated.
The specific equation |x-4|=4 is connected to the board template |x - TARGET VALUE| = EQUATION.
The specific distance interpretation |x-4|=4 is generalized into the template |x - TARGET VALUE| = EQUATION.
The number line gives and .
The algebraic split also gives and .
The geometric number-line method and the algebraic two-case method produce the same solutions for |x-4|=4.
The instructor uses the rule that absolute value has to be positive to explain why |2x-3|=-2 is a trick question.
The concluding sentence is cut off before the explicit no-solution statement.
The nonnegativity property of absolute value is applied to identify |2x-3|=-2 as impossible over the real numbers.
The instructor first rules out problem ① because the right side is negative, then introduces the split method for problem ② with a positive right side.
The clip contrasts the no-solution case |u| = negative number with the solvable case |u| = positive number handled by splitting into two equations.
“our goal is to get that part in absolute value by itself” followed by splitting into two equations.
For problems ③ and ④, isolating the absolute value expression is required before the two-case split can be applied correctly.
After solving by splitting, the instructor says the nice thing about algebra is you can check your answer and substitutes both solutions back in.
The checking method is applied to the solutions obtained from the two-case split.
The right-side note |X - TARGET VALUE| = DEVIATION appears alongside the worked absolute value equations.
The clip does not fully solve a word problem using this template; it only displays the formula and mentions a word problem at the end.
The deviation template is an application form of an absolute value equation that would later be solved using the same isolation and splitting ideas.
The instructor introduces the absolute deviation formula and then applies it to the basketball problem using h, 120, and .25.
The board moves from |x - TARGET VALUE| = DEVIATION to |h - 120| = .25.
The general absolute deviation formula is instantiated in the basketball example by replacing x with h, TARGET VALUE with 120, and DEVIATION with .25.
The worked example |h - 120| = .25 is immediately rewritten as h - 120 = .25 OR h - 120 = -.25.
The method for splitting an absolute value equation is used to solve the specific basketball equation.
The instructor converts 10 ft to 120 in before writing the equation.
Unit conversion must happen before the story problem can be modeled correctly with matching units.
The instructor explains the vertical bars mean absolute value and distance from zero.
The instructor says the first thing is to isolate the absolute value quantity by itself.
The instructor says to split into positive and negative scenarios because the inside could originally have been negative.
A number line around 4 shows solutions 4 units left and right.
The board lists |2x-3|=-2 as a trick example.
The instructor begins explaining that absolute value always has to be positive.
The explicit final conclusion is not fully audible in the clip.
“So this one is actually no solution.”
|2x-3| = -2 marked NO SOLN.
“we're going to split it up into two equations: equals 3 or equals negative 3.”
“You put it back in, negative 2 plus 5 is 3, absolute value of 3 is 3...”
“our goal is to get that part in absolute value by itself... work from the outside in”
Starred board note: ABSOLUTE DEVIATION: |X - TARGET VALUE| = DEVIATION.
Only the template is visible; the actual word problem solution is not completed in this clip.
The board shows ABSOLUTE DEVIATION: |x - TARGET VALUE| = DEVIATION.
The instructor writes |h - 120| = .25 for the basketball hoop problem.
Covered · The entire 150-second segment is covered: introduction to absolute value as distance, worked solution of 3|x-4|+ by geometric and algebraic methods, checking solutions, general template on the board, and the beginning of the trick example |2x-3|=-2. No additional mathematical events occur outside these items.
Covered · Problem ① is identified as having no solution because the absolute value equals -2.
Covered · Problem ② is solved by splitting into two linear equations and then checking both answers.
Covered · Non-mathematical promotional interruption; no new algebra content is presented.
Covered · The lesson returns to the board and solves problem ③ by isolating the absolute value first, then splitting.
Covered · Problem ④ is solved by adding 5, dividing by 5, splitting into two cases, and solving each linear equation.
Covered · The instructor introduces a final word problem and the board shows the absolute deviation template, but the word problem is not solved within this clip.
Covered · The instructor reads the basketball hoop word problem written on the board.
Covered · He introduces the absolute deviation formula and explains deviation as distance from target in either direction.
Covered · He converts 10 ft to 120 in and writes the model |h - 120| = .25.
Covered · He splits the absolute value equation and solves to get and .
Covered · He interprets the answers verbally as maximum and minimum allowable heights and closes the segment.
Reviewed English material v1: card "Absolute value means distance from zero" at 0:10 — The vertical bars in expressions such as |-2| and |2| denote absolute value. Absolute value measures distance from zero on the number line, so both -2 and 2 have absolute value 2. This distance interpretation explains why absolute value outputs are nonnegative.
Reviewed English material v1: card "Split |u| = c into two linear equations" at 2:40 — When the absolute value expression is isolated and the right-hand side is nonnegative, rewrite the equation as two cases: or . This is used for ||=3.
YouTube12:48Another explanation of Absolute value.
Absolute value
YouTube4:49Another explanation of Absolute value.
Absolute value
YouTube9:14Application of Absolute value, complementing the current explanation.
Absolute value