Question

How do you solve `2= \frac { 10+ r } { 7}`?

Answer 1

See a solution process below:

Explanation:

First, multiply each side of the equation by `color(red)(7)` to eliminate the fraction while keeping the equation balanced:

`color(red)(7) xx 2 = color(red)(7) xx (10 + r)/7`

`14 = cancel(color(red)(7)) xx (10 + r)/color(red)(cancel(color(black)(7)))`

`14 = 10 + r`

Now, subtract `color(red)(10)` from each side of the equation to solve for `r` while keeping the equation balanced:

`14 - color(red)(10) = 10 - color(red)(10) + r`

`4 = 0 + r`

`4 = r`

`r = 4`

Answer 2

`r = 4`

Explanation:

The aim is to manipulate the equation so that the only term on the left hand side is `r`, so that this is equal to whatever is on the right hand side.

The right hand side consists of a fraction. The top part of the fraction (the numerator) is an expression and the bottom part (the denominator) is `7`. This means that everything on the top of the fraction has been divided by `7`. This can be "undone" by multiplying the right hand side by `7`. To keep the equation true, the left hand side must also be divided by `7`.

So

`2 = (10 + r)/7`

implies

`7 xx 2 = 7 (10 + r)/7`

that is,

`14 = 10 + r`

At this point, it becomes easier to see what is happening by switching the sides, that is

`14 = 10 + r`

implies

`10 + r = 14`

By inspection, `r` can now be isolated and left on its own if `10` is subtracted from the (new) left hand side. To keep the equality of both sides of the equation, the same must be done to the (new) right hand side.

That is,

`10 + r = 14`

implies

`10 + r -10 = 14 - 10`

that is

`r = 4`

That is the solution to the equation.

It is always a good idea to check the result in the original equation.

You might note that the (original) right hand side is

`(10 + r)/7`

Plugging in the candidate solution `r = 4`, the right hand side becomes

`(10 + 4)/7 = 14/7 = 2`

This equals the (original) left hand side of the equation, as expected, corroborating the solution.