How do you solve #5/8+7/16q=3/4+1/4q#?

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Answer 1 · 2015-08-07T05:33:10

#color(blue)(q=2/3#

#5/8 +color(blue)(7/16q) =3/4+color(blue)(1/4q#

#5/8 -3/4 =color(blue)(1/4q -color(blue)(7/16q)#

The L.C.M of the denominators, (#4,8,16#) is #16#

So,

#(5*2)/(8*2) -(3*4)/(4*4) =color(blue)((1*4)/(4*4)q -color(blue)(7/16q)#

#(10)/(16) -(12)/(16) =color(blue)((4)/(16)q -color(blue)(7/16q)#

#(-2)/(cancel16) =color(blue)(-3/cancel16q)#

#color(blue)(q=2/3#