Question

How do you solve `3/5(15x+20)+1/6(18x-12)=38`?

Answer 1

See the full process below in the Explanation

Explanation:

Step 1) Expand the terms within the parenthesis:

`color(red)(3/5)(15x + 20) + color(blue)(1/6)(18x - 12) = 38`

`(color(red)(3/5) xx 15x) + (color(red)(3/5)xx 20) + (color(blue)(1/6) xx 18x) - (color(blue)(1/6) xx 12) = 38`

`(color(red)(3/cancel(5)) xx color(red)(cancel(color(black)(15)))3x) + (color(red)(3/cancel(5))xx color(black)(cancel(color(red)(20)))4) + (color(blue)(1/cancel(6)) xx color(blue)(cancel(color(black)(18)))3x) - (color(blue)(1/cancel(6)) xx color(blue)(cancel(color(black)(12)))2) = 38`

`(color(red)(3) xx 3x) + (color(red)(3)xx 4) + (color(blue)(1) xx 3x) - (color(blue)(1) xx 2) = 38`

`9x + 12 + 3x - 2 = 38`

We can now group and consolidate like terms:

`9x + 3x + 12 - 2 = 38`

`(9 + 3)x + 10 = 38`

`12x + 10 = 38`

We can next subtract `color(red)(10)` form each side of the equation to isolate the `x` term while keeping the equation balanced:

`12x + 10 - color(red)(10) = 38 - color(red)(10)`

`12x + 0 = 28`

`12x = 28`

Now we can divide each side of the equation by `color(red)(12)` to solve for `x` and keep the equation balanced:

`(12x)/color(red)(12) = 28/color(red)(12)`

`(color(black)(cancel(color(red)(12)))x)/cancel(color(red)(12)) = (4 xx 7)/(4 xx 3)`

`x = (cancel(4) xx 7)/(cancel(4) xx 3)`

`x = 7/3`