Question

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

Answer 1

`y = 70`

Explanation:

`y/5 - 6 = 8`

add `6` to both sides
`y/5 - 6 + 6= 8 + 6`
`y/5 = 14`

multiply `5` to both sides
`y/cancel5 * cancel5 = 14 * 5`

`y = 70`

Answer 2

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

`y=70`.

Explanation:

We start with

`y/5-6=8`

Remember: this equation is like a balanced scale, with `y/5-6` on one side of the scale balancing the `8` on the other side. We can add, subtract, multiply, divide, etc. whatever we want to the equation, just as long as we do the same thing to both sides, keeping the scale balanced.

Step 1: Isolate the `y` term.

We can add 6 to both sides (a.k.a. make the `-6` on the left into a `+6` on the right).

`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 `y` coefficient into 1.

In this form, the equation says that a fifth of `y` is equal to 14. We want to know what number, when divided by 5, equals 14. Well, if one-fifth of `y` is 14, then five of these fifths should equal five 14's.

To solve for `y`, we multiply both sides by 5:

`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` back into the original equation:

`y/5-6stackrel(?"  ")=8`

`color(red)70/5-6stackrel(?"  ")=8`

`14-6stackrel(?"  ")=8`

`"     "8"    "=8`

Since the equation holds, our answer of `y=70` is correct.