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

2 Answers
Jan 18, 2017

Answer:

#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#

Jan 18, 2017

Answer:

#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"#