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Solution - Absolute value equations

Exact form: t=56,14
t=\frac{5}{6} , \frac{1}{4}
Decimal form: t=0.833,0.25
t=0.833 , 0.25

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
|t2|=|5t+3|
without the absolute value bars:

|x|=|y||t2|=|5t+3|
x=+y(t2)=(5t+3)
x=y(t2)=(5t+3)
+x=y(t2)=(5t+3)
x=y(t2)=(5t+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||t2|=|5t+3|
x=+y , +x=y(t2)=(5t+3)
x=y , x=y(t2)=(5t+3)

2. Solve the two equations for t

9 additional steps

(t-2)=(-5t+3)

Add to both sides:

(t-2)+5t=(-5t+3)+5t

Group like terms:

(t+5t)-2=(-5t+3)+5t

Simplify the arithmetic:

6t-2=(-5t+3)+5t

Group like terms:

6t-2=(-5t+5t)+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=5

Divide both sides by :

(6t)6=56

Simplify the fraction:

t=56

12 additional steps

(t-2)=-(-5t+3)

Expand the parentheses:

(t-2)=5t-3

Subtract from both sides:

(t-2)-5t=(5t-3)-5t

Group like terms:

(t-5t)-2=(5t-3)-5t

Simplify the arithmetic:

-4t-2=(5t-3)-5t

Group like terms:

-4t-2=(5t-5t)-3

Simplify the arithmetic:

4t2=3

Add to both sides:

(-4t-2)+2=-3+2

Simplify the arithmetic:

4t=3+2

Simplify the arithmetic:

4t=1

Divide both sides by :

(-4t)-4=-1-4

Cancel out the negatives:

4t4=-1-4

Simplify the fraction:

t=-1-4

Cancel out the negatives:

t=14

3. List the solutions

t=56,14
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
y=|t2|
y=|5t+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.