Question

How do you solve this equation for `c`: 2ab + 2ac + 2bc = A#?

Answer 1

See a solution process below:

Explanation:

First, subtract `color(red)(2ab)` from each side of the equation to isolate the `c` terms while keeping the equation balanced:

`A - color(red)(2ab) = 2ab - color(red)(2ab) + 2ac + 2bc`

`A - 2ab = 0 + 2ac + 2bc`

`A - 2ab = 2ac + 2bc`

Next, factor a `c` out of each term on the right side of the equation giving:

`A - 2ab = (2a * c) + (2b * c)`

`A - 2ab = (2a + 2b)c`

Now, divide each side of the equation by `color(red)(2a + 2b)` to solve for `c` while keeping the equation balanced:

`(A - 2ab)/color(red)(2a + 2b) = ((2a + 2b)c)/color(red)(2a + 2b)`

`(A - 2ab)/(2a + 2b) = (color(red)(cancel(color(black)((2a + 2b))))c)/cancel(color(red)(2a + 2b))`

`(A - 2ab)/(2a + 2b) = c`

`c = (A - 2ab)/(2a + 2b)`

Or

`c = (A - 2ab)/(2(a + b))`

Answer 2

`c=(A-2ab)/(2a+2b)`

Explanation:

`"we want to isolate the terms in c"`

`"subtract 2ab from both sides"`

`cancel(2ab)cancel(-2ab)+2ac+2bc=A-2ab`

`rArr2ac+2bc=A-2ab`

`"take out "color(blue)"common factor ""of c"`

`c(2a+2b)=A-2ab`

`"divide both sides by "(2a+2b)`

`(c cancel((2a+2b)))/cancel((2a+2b))=(A-2ab)/(2a+2b)`

`rArrc=(A-2ab)/(2a+2b)`