Question

How do you solve `2/5x+1/4=-7/10` by clearing the fractions?

Answer 1

`x = -19/8`

Explanation:

If you have an equation with fractions, you can get rid of the denominators by multiplying each term by the LCM of the denominators. IN this case it is `color(blue)(20)`

`color(blue)(20xx)2/5x +color(blue)(20xx)1/4 =(-7color(blue)(xx20))/10`

`8x+5 =-14" "larr` no fractions !

`8x = -14-5`

`8x = -19`

`x = -19/8`

Answer 2

`x=-19/8`

Explanation:

First note that `2/5x=(2x)/5`

To 'clear' the fractions we require to multiply ALL terms on both sides of the equation by the `color(blue)"lowest common multiple"` (LCM ) of the denominators 5 , 4 and 10.

#the LCM of 5 , 4 and 10 is 20

so multiply all terms by 20

`(cancel(20)^4xx(2x)/cancel(5)^1)+(cancel(20)^5xx1/cancel(4)^1)=cancel(20)^2xx(-7)/cancel(10)^1`

`rArr(4xx2x)+(5xx1)=(2xx-7)larr" no fractions"`

`rArr8x+5=-14`

subtract 5 from both sides.

`8xcancel(+5)cancel(-5)=-14-5`

`rArr8x=-19`

To solve for x, divide both sides by 8

`(cancel(8) x)/cancel(8)=(-19)/8`

`rArrx=-19/8" is the solution"`

`color(blue)"As a check"`

Substitute `x=-19/8` into the left side and if it is a solution then it should equal the right side.

`"left side " =(2/5xx-19/8)+1/4=-38/40+1/4`

`=-38/40+10/40=-28/40=-7/10color(white)(xx)✔︎`

`rArrx=-19/8" is the solution"`