Solve the following equations. ( A ) 2 ∣ x ∣ = 10 2\left|x\right|=10

x = { 5 , − 5 } x=\left\lbrace5,-5\right\rbrace x = { 5 , − 5 }

x = { 20 , − 20 } x=\left\lbrace20,-20\right\rbrace x = { 20 , − 20 }

x = { 12 , − 12 } x=\left\lbrace12,-12\right\rbrace x = { 12 , − 12 }

x = { 10 , − 10 } x=\left\lbrace10,-10\right\rbrace x = { 10 , − 10 }

Solve the following equations. ( B ) ∣ 3 x − 5 ∣ − 6 = 2 \left|3x-5\right|-6=2

x = { 13 3 , − 1 } x=\left\lbrace\frac{13}{3},-1\right\rbrace x = { 3 13 , − 1 }

x = { − 13 3 , 1 } x=\left\lbrace-\frac{13}{3},1\right\rbrace x = { − 3 13 , 1 }

x = { 3 , − 1 3 } x=\left\lbrace3,-\frac13\right\rbrace x = { 3 , − 3 1 }

Solve the following equations. ( C ) 0.4 ∣ 3 y + 2 ∣ = 3.6 0.4\left|3y+2\right|=3.6

y = { 7 3 , − 11 3 } y=\left\lbrace\frac73,-\frac{11}{3}\right\rbrace y = { 3 7 , − 3 11 }

y = { 14 , − 38 3 } y=\left\lbrace14,-\frac{38}{3}\right\rbrace y = { 14 , − 3 38 }

y = { − 7 3 , 11 3 } y=\left\lbrace-\frac73,\frac{11}{3}\right\rbrace y = { − 3 7 , 3 11 }

y = { − 14 , 38 3 } y=\left\lbrace-14,\frac{38}{3}\right\rbrace y = { − 14 , 3 38 }

Solve the following equations with absolute value. ( A ) 3 ∣ 4 x + 1 ∣ + 7 = 2 3\left|4x+1\right|+7=2

x = { 1 2 , − 1 } x=\left\lbrace\frac12,-1\right\rbrace x = { 2 1 , − 1 }

x = { − 1 6 , 2 3 } x=\left\lbrace-\frac16,\frac23\right\rbrace x = { − 6 1 , 3 2 }

x = { 1 6 , − 2 3 } x=\left\lbrace\frac16,-\frac23\right\rbrace x = { 6 1 , − 3 2 }

Solve the following equations with absolute value. ( B ) 5 ∣ 6 x − 9 ∣ + 3 = 3 5\left|6x-9\right|+3=3

x = − 3 2 x=-\frac32 x = − 2 3

Solve the following equations with 2 absolute values. ( A ) ∣ 3 x + 4 ∣ = ∣ − 2 x + 7 ∣ \left|3x+4\right|=\left|-2x+7\right|

x = { 3 5 , − 11 } x=\left\lbrace\frac35,-11\right\rbrace x = { 5 3 , − 11 }

x = { 3 , − 11 } x=\left\lbrace3,-11\right\rbrace x = { 3 , − 11 }

x = { 3 5 , 3 } x=\left\lbrace\frac35,3\right\rbrace x = { 5 3 , 3 }

x = { − 3 5 , 3 } x=\left\lbrace-\frac35,3\right\rbrace x = { − 5 3 , 3 }

Solve the following equations with 2 absolute values. ( B ) ∣ 2 x − 3 4 ∣ = ∣ x + 1 2 ∣ \left|\frac{2x-3}{4}\right|=\left|x+\frac12\right|

x = { − 5 2 , 1 6 } x=\left\lbrace-\frac52,\frac16\right\rbrace x = { − 2 5 , 6 1 }

x = { − 5 2 , 5 6 } x=\left\lbrace-\frac52,\frac56\right\rbrace

x = { 5 2 , − 1 6 } x=\left\lbrace\frac52,-\frac16\right\rbrace

x = { 5 2 , − 5 6 } x=\left\lbrace\frac52,-\frac56\right\rbrace x = { 2 5 , − 6 5 }

2. Linear Equations and Inequalities

Absolute Value Equations

2. Linear Equations and Inequalities

Absolute Value Equations: Videos & Practice Problems

Video Lessons

Topic summary

Solving absolute value equations involves isolating the absolute value expression and rewriting $|x| = a$ as two equations: $x = a$ and $x = - a$ when $a$ is positive. If $a = 0$ , solve $x = 0$ . No solutions exist if $a$ is negative. For equations with two absolute values, set the inner expressions equal and also equal with opposite signs. This approach enhances understanding of polynomial terms, coefficients, and expressions in standard form.

concept

Solving Absolute Value Equations

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Solving Absolute Value Equations Video Summary

The absolute value of a number represents its distance from zero on the number line, regardless of direction. For instance, the absolute value of 2 is 2, and the absolute value of -2 is also 2, since both are two units away from zero. This fundamental concept helps in solving absolute value equations, which may initially seem complex but can be approached systematically.

When solving an equation involving absolute value, such as $|x| = 2$, the goal is to find all values of $x$ whose distance from zero is 2. This means $x$ can be either 2 or -2, since both satisfy the equation. Substituting these values back into the original equation confirms they are valid solutions. Thus, the solution set is $\{2, -2\}$.

This example illustrates a general rule for absolute value equations: if $|x| = a$ where $a$ is a positive number, then $x = a$ or $x = -a$. This rule extends beyond simple variables to expressions inside the absolute value. For example, in the equation $|x + 1| = 2$, the expression inside the absolute value, $x + 1$, can equal 2 or -2. Solving these two linear equations separately, $x + 1 = 2$ and $x + 1 = -2$, yields $x = 1$ and $x = -3$ respectively. Both values satisfy the original equation, so the solution set is $\{1, -3\}$.

It is important to isolate the absolute value expression on one side of the equation before applying this rule. For example, to solve $|x + 1| + 3 = 5$, first subtract 3 from both sides to isolate the absolute value: $|x + 1| = 2$. Then apply the rule to find the solutions as before.

In summary, solving absolute value equations involves isolating the absolute value expression and then rewriting the equation $|expression| = a$ as two separate equations: $expression = a$ and $expression = -a$. Solving these linear equations provides all possible solutions. This method is effective when $a$ is a positive number, and further considerations are needed when $a$ is zero or negative.

concept

Solving Absolute Value Equations Special Cases

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Solving Absolute Value Equations Special Cases Video Summary

When solving equations involving the absolute value expression, such as |x| = a, the value of a plays a crucial role in determining the number and type of solutions. The absolute value represents the distance from zero, which is always non-negative. Therefore, if a is positive, the equation |x| = a can be rewritten as two separate equations: x = a and x = -a. Solving these gives two solutions because the distance from zero can be achieved in two directions on the number line.

When a = 0, the equation |x| = 0 means the distance from zero is zero, which only happens if the expression inside the absolute value is exactly zero. This simplifies to a single equation, x = 0, yielding one unique solution.

However, if a is negative, such as in |x| = -2, the equation has no solution. Since distance cannot be negative, the absolute value expression cannot equal a negative number. This results in a contradiction, meaning no values of x satisfy the equation.

In summary, solving absolute value equations depends on the sign of the constant on the right side. For positive values, split into two linear equations; for zero, solve the single equation where the inside equals zero; and for negative values, recognize that no solution exists. This understanding is essential for correctly handling absolute value equations in algebra.

Problem

Solve the following equations.
(A) $2 ∣ x ∣ = 10$

$x=\left\lbrace20,-20\right\rbrace$

$x=\left\lbrace12,-12\right\rbrace$

$x=\left\lbrace10,-10\right\rbrace$

$x=\left\lbrace5,-5\right\rbrace$

Problem

Solve the following equations.
(B) $∣ 3 x - 5 ∣ - 6 = 2$

$x=\left\lbrace\frac{13}{3},-1\right\rbrace$

$x=\left\lbrace-\frac{13}{3},1\right\rbrace$

$x=\left\lbrace3,-\frac13\right\rbrace$

No solution

Problem

Solve the following equations.
(C) $0.4 ∣ 3 y + 2 ∣ = 3.6$

$y=\left\lbrace14,-\frac{38}{3}\right\rbrace$

$y=\left\lbrace\frac73,-\frac{11}{3}\right\rbrace$

$y=\left\lbrace-\frac73,\frac{11}{3}\right\rbrace$

$y=\left\lbrace-14,\frac{38}{3}\right\rbrace$

Problem

Solve the following equations with absolute value.
(A) $3 ∣ 4 x + 1 ∣ + 7 = 2$

$x=\left\lbrace\frac12,-1\right\rbrace$

$x=\left\lbrace-\frac16,\frac23\right\rbrace$

$x=\left\lbrace\frac16,-\frac23\right\rbrace$

No solution

Problem

Solve the following equations with absolute value.
(B) $5 ∣ 6 x - 9 ∣ + 3 = 3$

$x=-\frac32$

$x=0$

$x=\frac32$

$x=3$

concept

Solving Equations Involving 2 Absolute Values

Video duration:

Solving Equations Involving 2 Absolute Values Video Summary

Solving equations involving absolute values requires understanding how to handle the expressions inside the absolute value symbols. When dealing with a single absolute value equation, the goal is to isolate the absolute value expression on one side, resulting in a form like |x + 1| = 2. This allows us to rewrite the equation as two separate linear equations by setting the expression inside equal to both the positive and negative values of the number outside the absolute value. For example, x + 1 = 2 and x + 1 = -2 lead to solutions x = 1 and x = -3, respectively. These solutions are then combined into a solution set, denoted by curly brackets: {1, -3}.

When an equation contains two absolute values set equal to each other, such as |x + 1| = |2x - 4|, a similar approach applies. The key is to recognize that if the absolute values of two expressions are equal, then either the expressions themselves are equal or one is the negative of the other. This leads to two separate equations without absolute values: x + 1 = 2x - 4 and x + 1 = -(2x - 4). The second equation requires distributing the negative sign across the entire expression inside the absolute value, resulting in x + 1 = -2x + 4.

Solving these linear equations involves isolating the variable by performing inverse operations. For the first equation, adding 4 to both sides and then subtracting x yields 5 = x. For the second, distributing the negative sign, subtracting 1 from both sides, and adding 2x to both sides simplifies to 3x = 3, which further reduces to x = 1 after dividing both sides by 3. The solution set for the original equation is therefore {1, 5}.

This method of solving equations with two absolute values relies on the principle that |A| = |B| implies either A = B or A = -B. By converting the absolute value equation into two linear equations, one can apply standard algebraic techniques to find all possible solutions. This approach is essential for solving more complex absolute value equations and understanding the behavior of absolute value functions in algebra.

Problem

Solve the following equations with 2 absolute values.
(A) $∣ 3 x + 4 ∣ = ∣ - 2 x + 7 ∣$

$x=\left\lbrace\frac35,-11\right\rbrace$

$x=\left\lbrace3,-11\right\rbrace$

$x=\left\lbrace\frac35,3\right\rbrace$

$x=\left\lbrace-\frac35,3\right\rbrace$

Problem

Solve the following equations with 2 absolute values.
(B) $∣ \frac{2 x - 3}{4} ∣ = ∣ x + \frac{1}{2} ∣$

$x = {- \frac{5}{2}, \frac{5}{6}}$

$x = {\frac{5}{2}, - \frac{1}{6}}$

$x=\left\lbrace-\frac52,\frac16\right\rbrace$

$x=\left\lbrace\frac52,-\frac56\right\rbrace$

example

Solving Equations Involving 2 Absolute Values Example 1

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Solving Equations Involving 2 Absolute Values Example 1 Video Summary

To solve the equation involving absolute values, such as |3x + 4| = |-8|, the first step is to simplify any absolute values of constants. Since the absolute value of a number represents its distance from zero on the number line, |-8| equals 8. This simplifies the equation to |3x + 4| = 8. When an absolute value expression equals a positive number, it can be split into two separate linear equations: one where the expression inside the absolute value equals the positive number, and another where it equals the negative of that number. This gives the two equations 3x + 4 = 8 and 3x + 4 = -8.

Solving the first equation, subtract 4 from both sides to isolate the term with the variable:
\$3x + 4 - 4 = 8 - 4$
which simplifies to
$3x = 4$.
Next, divide both sides by 3 to solve for x:
$x = \frac{4}{3}\(.

For the second equation, similarly subtract 4 from both sides:
\)3x + 4 - 4 = -8 - 4$
which simplifies to
$3x = -12$.
Dividing both sides by 3 gives:
$x = \frac{-12}{3} = -4$.

Thus, the solutions to the original absolute value equation are x = \frac{4}{3} and x = -4. These values satisfy the condition that the expression inside the absolute value is either equal to the positive or negative value of the constant on the other side of the equation. Understanding how to break down absolute value equations into two linear cases is essential for solving such problems efficiently.

example

Solving Equations Involving 2 Absolute Values Example 2

Video duration:

Solving Equations Involving 2 Absolute Values Example 2 Video Summary

To solve the equation involving absolute values, |2x - 4| = |-2x + 3|, we use the property that if two absolute values are equal, then either the expressions inside are equal or they are negatives of each other. This leads to two separate equations to solve.

First, set the expressions inside the absolute values equal:

\$2x - 4 = -2x + 3\(

To isolate the variable, add 4 to both sides:

\)2x = -2x + 7\(

Next, add \)2x\( to both sides to combine like terms:

\)4x = 7\(

Divide both sides by 4 to solve for \)x\(:

\)x = \frac{7}{4}\(

Second, set the first expression equal to the negative of the second expression:

\)2x - 4 = -(-2x + 3)\(

Distribute the negative sign on the right side:

\)2x - 4 = 2x - 3\(

Add 4 to both sides:

\)2x = 2x + 1\(

Subtract \)2x\( from both sides:

\)0 = 1\(

This is a contradiction, indicating no solution exists for this case.

Therefore, the only solution to the original absolute value equation is:

\)x = \frac{7}{4}$

This method highlights how absolute value equations can be broken down into two linear equations, and by carefully solving each, we determine the valid solutions. Understanding how to manipulate and solve these equations is essential for mastering absolute value problems in algebra.

example

Solving Equations Involving 2 Absolute Values Example 3

Video duration:

Solving Equations Involving 2 Absolute Values Example 3 Video Summary

To solve the equation involving two absolute values on the same side, such as |x - 3| - |3x - 9| = 0, the key step is to isolate one absolute value on each side. By adding |3x - 9| to both sides, the equation becomes |x - 3| = |3x - 9|, which is a standard form for solving absolute value equations.

When two absolute values are equal, the expressions inside can either be equal or negatives of each other. This leads to two separate equations:
1. x - 3 = 3x - 9
2. x - 3 = -(3x - 9)

Solving the first equation, x - 3 = 3x - 9, involves isolating the variable terms and constants. Adding 9 to both sides gives x + 6 = 3x. Subtracting x from both sides results in 6 = 2x. Dividing both sides by 2 yields x = 3.

For the second equation, x - 3 = -3x + 9, distribute the negative sign on the right side. Adding 3 to both sides gives x = -3x + 12. Adding 3x to both sides results in 4x = 12. Dividing both sides by 4 yields x = 3.

Both equations lead to the same solution, x = 3. Therefore, the original equation has a single solution: {3}. This approach highlights how absolute value equations can be transformed and solved by considering both positive and negative scenarios of the expressions inside the absolute values.

Here’s what students ask on this topic:

To solve an absolute value equation where the absolute value is isolated, such as $| x | = a$ with $a$ positive, you rewrite it as two separate equations: $x = a$ and $x = - a$ . This is because absolute value represents distance from zero, so the expression inside the absolute value can be either positive or negative $a$ . After rewriting, solve each equation separately to find the solution set. For example, if $|x|=3$ , then $x=3$ or $x=-3$ . Always ensure the absolute value expression is isolated on one side before applying this method.

When an absolute value equation is set equal to zero, such as $| x | = 0$ , the only solution is when the expression inside the absolute value equals zero. This is because the distance from zero is zero only if the value itself is zero. So, you solve the equation inside the absolute value: $x = 0$ . For example, if $|x+1|=0$ , then $x+1=0$ and solving gives $x=-1$ . This results in a single solution.

Absolute value represents the distance from zero, which is always non-negative. Therefore, an equation like $| x | = - a$ where $a$ is positive, has no solution because distance cannot be negative. No value of $x$ will satisfy the equation since absolute value cannot produce a negative result. For example, $|x| = -2$ is impossible, so the solution set is empty. This is an important rule to remember when solving absolute value equations.

When you have an equation with two absolute values equal, such as $| f (x) | = | g (x) |$ , you rewrite it as two separate equations without absolute values: $f (x) = g (x)$ and $f (x) = - g (x)$ . Then solve each equation separately. This works because the absolute values are equal if the expressions inside are either equal or negatives of each other. For example, if $|x+1| = |2x-4|$ , solve $x+1=2x-4$ and $x+1=-(2x-4)$ to find all solutions.

When the absolute value expression is not isolated, first isolate it by performing inverse operations. For example, in $|x+1| + 3 = 5$ , subtract 3 from both sides to get $|x+1| = 2$ . Once isolated, rewrite the equation as two separate equations without absolute values: $x+1=2$ and $x+1=-2$ . Solve each to find $x=1$ and $x=-3$ . Always isolate the absolute value before applying the rule to rewrite the equation as two linear equations.

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Absolute Value Equations: Videos & Practice Problems

Solving Absolute Value Equations

Solving Absolute Value Equations Video Summary

Solving Absolute Value Equations Special Cases

Solving Absolute Value Equations Special Cases Video Summary

Solve the following equations.
(A) $2 ∣ x ∣ = 10$

Solve the following equations.
(B) $∣ 3 x - 5 ∣ - 6 = 2$

Solve the following equations.
(C) $0.4 ∣ 3 y + 2 ∣ = 3.6$

Solve the following equations with absolute value.
(A) $3 ∣ 4 x + 1 ∣ + 7 = 2$

Solve the following equations with absolute value.
(B) $5 ∣ 6 x - 9 ∣ + 3 = 3$

Solving Equations Involving 2 Absolute Values

Solving Equations Involving 2 Absolute Values Video Summary

Solve the following equations with 2 absolute values.
(A) $∣ 3 x + 4 ∣ = ∣ - 2 x + 7 ∣$

Solve the following equations with 2 absolute values.
(B) $∣ \frac{2 x - 3}{4} ∣ = ∣ x + \frac{1}{2} ∣$

Solving Equations Involving 2 Absolute Values Example 1

Solving Equations Involving 2 Absolute Values Example 1 Video Summary

Solving Equations Involving 2 Absolute Values Example 2

Solving Equations Involving 2 Absolute Values Example 2 Video Summary

Solving Equations Involving 2 Absolute Values Example 3

Solving Equations Involving 2 Absolute Values Example 3 Video Summary

Here’s what students ask on this topic:

How do you solve an absolute value equation when the absolute value is isolated?

What is the solution to an absolute value equation when the absolute value equals zero?

Why do absolute value equations have no solution when set equal to a negative number?

How do you solve an equation with two absolute values set equal to each other?

What steps should you follow to solve an absolute value equation that is not initially isolated?

Your Intermediate Algebra tutors

Absolute Value Equations: Videos & Practice Problems

Solving Absolute Value Equations

Solving Absolute Value Equations Video Summary

Solving Absolute Value Equations Special Cases

Solving Absolute Value Equations Special Cases Video Summary

Solve the following equations.(A) 2∣x∣=102\left|x\right|=10

Solve the following equations.(B) ∣3x−5∣−6=2\left|3x-5\right|-6=2

Solve the following equations.(C) 0.4∣3y+2∣=3.60.4\left|3y+2\right|=3.6

Solve the following equations with absolute value.(A) 3∣4x+1∣+7=23\left|4x+1\right|+7=2

Solve the following equations with absolute value.(B) 5∣6x−9∣+3=35\left|6x-9\right|+3=3

Solving Equations Involving 2 Absolute Values

Solving Equations Involving 2 Absolute Values Video Summary

Solve the following equations with 2 absolute values.(A) ∣3x+4∣=∣−2x+7∣\left|3x+4\right|=\left|-2x+7\right|

Solve the following equations with 2 absolute values.(B) ∣2x−34∣=∣x+12∣\left|\frac{2x-3}{4}\right|=\left|x+\frac12\right|

Solving Equations Involving 2 Absolute Values Example 1

Solving Equations Involving 2 Absolute Values Example 1 Video Summary

Solving Equations Involving 2 Absolute Values Example 2

Solving Equations Involving 2 Absolute Values Example 2 Video Summary

Solving Equations Involving 2 Absolute Values Example 3

Solving Equations Involving 2 Absolute Values Example 3 Video Summary

Here’s what students ask on this topic:

How do you solve an absolute value equation when the absolute value is isolated?

How do you solve an absolute value equation when the absolute value is isolated?

What is the solution to an absolute value equation when the absolute value equals zero?

What is the solution to an absolute value equation when the absolute value equals zero?

Why do absolute value equations have no solution when set equal to a negative number?

Why do absolute value equations have no solution when set equal to a negative number?

How do you solve an equation with two absolute values set equal to each other?

How do you solve an equation with two absolute values set equal to each other?

What steps should you follow to solve an absolute value equation that is not initially isolated?

What steps should you follow to solve an absolute value equation that is not initially isolated?

Your Intermediate Algebra tutors

Solve the following equations.
(A) $2 ∣ x ∣ = 10$

Solve the following equations.
(B) $∣ 3 x - 5 ∣ - 6 = 2$

Solve the following equations.
(C) $0.4 ∣ 3 y + 2 ∣ = 3.6$

Solve the following equations with absolute value.
(A) $3 ∣ 4 x + 1 ∣ + 7 = 2$

Solve the following equations with absolute value.
(B) $5 ∣ 6 x - 9 ∣ + 3 = 3$

Solve the following equations with 2 absolute values.
(A) $∣ 3 x + 4 ∣ = ∣ - 2 x + 7 ∣$

Solve the following equations with 2 absolute values.
(B) $∣ \frac{2 x - 3}{4} ∣ = ∣ x + \frac{1}{2} ∣$