Question 178805
{{{((5y-20)/(3y+15))((7y+35)/(10y+40))}}} Start with the given expression. 



{{{((5(y-4))/(3y+15))((7y+35)/(10y+40))}}} Factor {{{5y-20}}} to get {{{5(y-4)}}}.



{{{((5(y-4))/(3(y+5)))((7y+35)/(10y+40))}}} Factor {{{3y+15}}} to get {{{3(y+5)}}}.



{{{((5(y-4))/(3(y+5)))((7(y+5))/(10y+40))}}} Factor {{{7y+35}}} to get {{{7(y+5)}}}.



{{{((5(y-4))/(3(y+5)))((7(y+5))/(10(y+4)))}}} Factor {{{10y+40}}} to get {{{10(y+4)}}}.



{{{(5*7(y-4)(y+5))/(3*10(y+5)(y+4))}}} Combine the fractions. 



{{{(35(y-4)(y+5))/(30(y+5)(y+4))}}} Multiply



{{{(35(y-4)highlight((y+5)))/(30*highlight((y+5))(y+4))}}} Highlight the common terms. 



{{{(35(y-4)cross((y+5)))/(30*cross((y+5))(y+4))}}} Cancel out the common terms. 



{{{(35(y-4))/(30(y+4))}}} Simplify. 



{{{(35y-140)/(30y+120)}}} Distribute



So {{{((5y-20)/(3y+15))((7y+35)/(10y+40))}}} simplifies to {{{(35y-140)/(30y+120)}}}.



In other words, {{{((5y-20)/(3y+15))((7y+35)/(10y+40))=(35y-140)/(30y+120)}}} where {{{y<>-5}}} or {{{y<>-4}}} (these are the restrictions since these values will cause a division by zero)