How do you solve #5/4s-1/2=4/3#?

1 Answer
Jim G.
Oct 11, 2017

#s=22/15#

Explanation:

#"we can eliminate the fractions by multiplying each term "#
#"by the "color(blue)"lowest common multiple"" of 4,2 and 3"#

#"the lowest common multiple of 4, 2 and 3 is "12#

#"note "5/4s-=(5s)/4#

#(cancel(12)^3xx(5s)/cancel(4)^1)-(cancel(12)^6xx1/cancel(2)^1)=(cancel(12)^4xx4/cancel(3)^1)#

#rArr15s-6=16larrcolor(blue)" no fractions"#

#"add 6 to both sides"#

#15scancel(-6)cancel(+6)=16+6#

#rArr15s=22#

#"divide both sides by 15"#

#(cancel(15) s)/cancel(15)=22/15#

#rArrs=22/15#

#color(blue)"As a check"#

Substitute this value into the left side of the equation and if equal to the right side then it is the solution.

#"left "=(5/4xx22/15)-1/2=11/6-1/2=4/3=" right side"#

#rArrs=22/15" is the solution"#

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