Question

How do you solve `sqrt((5y)/6)-10=4` and check your solution?

Answer 1

See a solution process below:

Explanation:

First, add `color(red)(10)` to each side of the equation to isolate the radical while keeping the equation balanced:

`sqrt((5y)/6) - 10 + color(red)(10) = 4 + color(red)(10)`

`sqrt((5y)/6) - 0 = 14`

`sqrt((5y)/6) = 14`

Next, square each side of the equation to eliminate the radical while keeping the equation balanced:

`(sqrt((5y)/6))^2 = 14^2`

`(5y)/6 = 196`

Now, multiply each side of the equation by `color(red)(6)/color(blue)(5)` to solve for `y` while keeping the equation balanced:

`color(red)(6)/color(blue)(5) xx (5y)/6 = color(red)(6)/color(blue)(5) xx 196`

`cancel(color(red)(6))/cancel(color(blue)(5)) xx (color(blue)(cancel(color(black)(5)))y)/color(red)(cancel(color(black)(6))) = 1176/5`

`y = 1176/5`

To check the solution we need to substitute `color(red)(1176/5)` for `color(red)(y)`, calculate each side of the equation and ensure the two results are equal:

`sqrt((5color(red)(y))/6) - 10 = 4` becomes:

`sqrt((5 xx color(red)(1176/5))/6) - 10 = 4`

`sqrt((color(red)(cancel(color(black)(5))) xx color(red)(1176/color(black)(cancel(color(red)(5)))))/6) - 10 = 4`

`sqrt(1176/6) - 10 = 4`

`+-sqrt(196) - 10 = 4`

`+-14 - 10 = 4`

`4 = 4` or `-24 = 4`

The `-14` result of the square root is extraneous.

Therefore, `4 = 4` and the solution is shown to be correct.

Answer 2

`y = 235.2`

Explanation:

`sqrt((5y)/6) - 10 = 4`

`sqrt((5y)/6) = 4 + 10`

`sqrt((5y)/6) = 14`

`(5y)/6 = 14^2`

`(5y)/6 = 196`

`5y = 196 xx 6`

`5y = 1176`

`y = 1176 ÷ 5`

`color(blue)(y = 235.2`

We can now substitute `y` for `235.2` to prove our answer.

`sqrt((5y)/6) - 10 = 4`

`sqrt((5 xx 235.2)/6) - 10 = 4`

`sqrt(1176/6) - 10 = 4`

`sqrt(196) - 10 = 4`

`14 - 10 = 4`