Enter an equation or problem
Camera input is not recognized!

Solution - Absolute value equations

Exact form: d=6,411
d=6 , \frac{4}{11}
Decimal form: d=6,0.364
d=6 , 0.364

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
|5d+1|=|6d5|
without the absolute value bars:

|x|=|y||5d+1|=|6d5|
x=+y(5d+1)=(6d5)
x=y(5d+1)=(6d5)
+x=y(5d+1)=(6d5)
x=y(5d+1)=(6d5)

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||5d+1|=|6d5|
x=+y , +x=y(5d+1)=(6d5)
x=y , x=y(5d+1)=(6d5)

2. Solve the two equations for d

10 additional steps

(5d+1)=(6d-5)

Subtract from both sides:

(5d+1)-6d=(6d-5)-6d

Group like terms:

(5d-6d)+1=(6d-5)-6d

Simplify the arithmetic:

-d+1=(6d-5)-6d

Group like terms:

-d+1=(6d-6d)-5

Simplify the arithmetic:

d+1=5

Subtract from both sides:

(-d+1)-1=-5-1

Simplify the arithmetic:

d=51

Simplify the arithmetic:

d=6

Multiply both sides by :

-d·-1=-6·-1

Remove the one(s):

d=-6·-1

Simplify the arithmetic:

d=6

10 additional steps

(5d+1)=-(6d-5)

Expand the parentheses:

(5d+1)=-6d+5

Add to both sides:

(5d+1)+6d=(-6d+5)+6d

Group like terms:

(5d+6d)+1=(-6d+5)+6d

Simplify the arithmetic:

11d+1=(-6d+5)+6d

Group like terms:

11d+1=(-6d+6d)+5

Simplify the arithmetic:

11d+1=5

Subtract from both sides:

(11d+1)-1=5-1

Simplify the arithmetic:

11d=51

Simplify the arithmetic:

11d=4

Divide both sides by :

(11d)11=411

Simplify the fraction:

d=411

3. List the solutions

d=6,411
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
y=|5d+1|
y=|6d5|
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.