# How do I solve for #y# in #y/5-6=8#?

##### 2 Answers

#### Explanation:

add

multiply

Isolate the term with the variable. Introduce a multiplication (or division) to turn the variable coefficient into 1. Double-check the answer.

#### Explanation:

We start with

#y/5-6=8#

Remember: this equation is like a balanced scale, with

**Step 1:** Isolate the

We can add 6 to both sides (a.k.a. make the

#y/5-6 color(red)(" "+6)=8color(red)(" "+6)#

#y/5cancel(-6)cancel(+6)=8+6#

#y/5color(white)(" "-6+6)=14#

**Step 2:** Turn the

In this form, the equation says that a fifth of *one*-fifth of *five* of these fifths should equal *five* 14's.

To solve for

#y/5color(red)(xx 5)" "=14 color(red)(xx 5)#

Multiplication by 5 and division by 5 are opposites; they cancel each other out.

#y/cancel(5)xxcancel(5)" "=14 xx 5#

#" "y" "=70#

**Step 3:** Double-check!

So apparently, when we divide 70 by 5, and then subtract 6, we should get 8. Is this true?

Substitute 70 for

#y/5-6stackrel(?" ")=8#

#color(red)70/5-6stackrel(?" ")=8#

#14-6stackrel(?" ")=8#

#" "8" "=8#

Since the equation holds, our answer of