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Solution - Absolute value equations

Exact form: t=12,-16
t=\frac{1}{2} , -\frac{1}{6}
Decimal form: t=0.5,0.167
t=0.5 , -0.167

Other Ways to Solve

Absolute value equations

Step-by-step explanation

1. Rewrite the equation without absolute value bars

Use the rules:
|x|=|y|x=±y and |x|=|y|±x=y
to write all four options of the equation
|8t2|=|2t+3|
without the absolute value bars:

|x|=|y||8t2|=|2t+3|
x=+y(8t2)=(2t+3)
x=y(8t2)=(2t+3)
+x=y(8t2)=(2t+3)
x=y(8t2)=(2t+3)

When simplified, equations x=+y and +x=y are the same and equations x=y and x=y are the same, so we end up with only 2 equations:

|x|=|y||8t2|=|2t+3|
x=+y , +x=y(8t2)=(2t+3)
x=y , x=y(8t2)=(2t+3)

2. Solve the two equations for t

11 additional steps

(8t-2)=(-2t+3)

Add to both sides:

(8t-2)+2t=(-2t+3)+2t

Group like terms:

(8t+2t)-2=(-2t+3)+2t

Simplify the arithmetic:

10t-2=(-2t+3)+2t

Group like terms:

10t-2=(-2t+2t)+3

Simplify the arithmetic:

10t2=3

Add to both sides:

(10t-2)+2=3+2

Simplify the arithmetic:

10t=3+2

Simplify the arithmetic:

10t=5

Divide both sides by :

(10t)10=510

Simplify the fraction:

t=510

Find the greatest common factor of the numerator and denominator:

t=(1·5)(2·5)

Factor out and cancel the greatest common factor:

t=12

10 additional steps

(8t-2)=-(-2t+3)

Expand the parentheses:

(8t-2)=2t-3

Subtract from both sides:

(8t-2)-2t=(2t-3)-2t

Group like terms:

(8t-2t)-2=(2t-3)-2t

Simplify the arithmetic:

6t-2=(2t-3)-2t

Group like terms:

6t-2=(2t-2t)-3

Simplify the arithmetic:

6t2=3

Add to both sides:

(6t-2)+2=-3+2

Simplify the arithmetic:

6t=3+2

Simplify the arithmetic:

6t=1

Divide both sides by :

(6t)6=-16

Simplify the fraction:

t=-16

3. List the solutions

t=12,-16
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
y=|8t2|
y=|2t+3|
The equation is true where the two lines cross.

Why learn this

We encounter absolute values almost daily. For example: If you walk 3 miles to school, do you also walk minus 3 miles when you go back home? The answer is no because distances use absolute value. The absolute value of the distance between home and school is 3 miles, there or back.
In short, absolute values help us deal with concepts like distance, ranges of possible values, and deviation from a set value.