Question

How do you solve `sqrt(6x-5)+10=3`?

Answer 1

See the entire solution process below:

Explanation:

First, subtract `color(red)(10)` from each side of the equation to isolate the square root term while keeping the equation balanced:

`sqrt(6x - 5) + 10 - color(red)(10) = 3 - color(red)(10)`

`sqrt(6x - 5) + 0 = -7`

`sqrt(6x - 5) = -7`

Next, square both sides to eliminate the radical while keeping the equation balanced:

`(sqrt(6x - 5))^2 = (-7)^2`

`6x - 5 = 49`

Then, add `color(red)(5)` to each side of the equation to isolate the `x` term while keeping the equation balanced:

`6x - 5 + color(red)(5) = 49 + color(red)(5)`

`6x - 0 = 54`

`6x = 54`

Now, divide each side of the equation by `color(red)(6)` to solve for `x` while keeping the equation balanced:

`(6x)/color(red)(6) = 54/color(red)(6)`

`(color(red)(cancel(color(black)(6)))x)/cancel(color(red)(6)) = 9`

`x = 9`

Answer 2

no solution.

Explanation:

`color(blue)"Isolate the square root"` by subtracting 10 from both sides.

`sqrt(6x-5)cancel(+10)cancel(-10)=3-10`

`rArrsqrt(6x-5)=-7`

`color(blue)"square both sides"`

`(sqrt(6x-5))^2=(-7)^2`

`rArr6x-5=49`

add 5 to both sides.

`6xcancel(-5)cancel(+5)=49+5`

`rArr6x=54`

divide both sides by 6

`(cancel(6) x)/cancel(6)=54/6`

`rArrx=9`

`color(blue)"As a check"`

Substitute this value into the left side of the equation and if equal to the right side then it is the solution.

`"left side "=sqrt((6xx9)-5)+10`

`=color(white)(left side)=sqrt49+10`

`=color(white)("left side)=7+10`

`color(white)(xxxxxxxx)=17!=3`

`rArr" there is no solution"`