How do you solve #(5/(y+5)) = (4/(y-3)) #?

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Answer 1 · 2016-02-13T05:35:35

#color(green)(y=35#

#(5/(y+5))=(4/(y-3))#

Remove the brackets:

#rarr5/(y+5)=4/(y-3)#

Cross multiply #color(blue)(a/b=c/d,ad=bc#

#rarr5(y-3)=4(y+5)#

Use distributive property #color(orange)(a(b+c)=ab+ac#

#rarr5y-15=4y+20#

Subtract #4y# both sides:

#rarry-15=20#

Add #15# both sides:

#rArrcolor(green)(y=35#