How do you solve #4(y-2/4)=9(y+1/3) #?

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Answer 1 · 2015-12-20T08:33:15

#y=-1#

#4(y-2/4)=9(y+1/3)#

According to BEDMAS, work on the brackets first. Reduce #2/4# to #1/2#.

#4(y-1/2)=9(y+1/3)#

Make the bracketed terms have the same denominator.

#4((2y)/2-1/2)=9((3y)/3+1/3)#

Subtract the bracketed terms.

#4((2y-1)/2)=9((3y+1)/3)#

Reduce the fractions by cancelling.

#color(red)cancelcolor(black)4^2((2y-1)/color(red)cancelcolor(black)2)=color(blue)cancelcolor(black)9^3((3y+1)/color(blue)cancelcolor(black)3)#

Rewrite the equation.

#2(2y-1)=3(3y+1)#

Multiply.

#4y-2=9y+3#

Isolate for #y# by bringing all terms with #y# to the left and all without to the right.

#4y-9y=3+2#

Solve.

#-5y=5#

#y=-1#