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

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#

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"#