Question

How do you solve `sqrt(7r+2)+3=7` and check your solution?

Answer 1

See a solution process below:

Explanation:

First, subtract `color(red)(3)` from each side of the equation to isolate the radical while keeping the equation balanced:

`sqrt(7r + 2) + 3 - color(red)(3) = 7 - color(red)(3)`

`sqrt(7r + 2) + 0 = 4`

`sqrt(7r + 2) = 4`

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

`(sqrt(7r + 2))^2 = 4^2`

`7r + 2 = 16`

Then, subtract `color(red)(2)` from each side of the equation to isolate the `r` term while keeping the equation balanced:

`7r + 2 - color(red)(2) = 16 - color(red)(2)`

`7r + 0 = 14`

`7r = 14`

Now, divide each side of the equation by `color(red)(7)` to solve for `r` while keeping the equation balanced:

`(7r)/color(red)(7) = 14/color(red)(7)`

`(color(red)(cancel(color(black)(7)))r)/cancel(color(red)(7)) = 2`

`r = 2`

To validate the solution substitute `color(red)(2)` for `color(red)(r)` in the original equation and calculate the result to ensure both sides of the equation are equal (remember, the square root of a number produces a positive and negative result):

`+-sqrt(7color(red)(r) + 2) + 3 = 7` becomes:

`+-sqrt((7 * color(red)(2)) + 2) + 3 = 7`

`+-sqrt(14 + 2) + 3 = 7`

`+-sqrt(16) + 3 = 7`

`-4 + 3 = 7` and `4 + 3 = 7`

`-1 != 7` and `7 = 7`

The negative result of the radical is an extraneous solution.

The positive result shows the solution is correct.