What is (5y-20)/(3y+15)*(7y+35)/(10y+40) ?

The solution to (5y-20)/(3y+15)*(7y+35)/(10y+40) is (7(y-4))/(6(y+4))

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simplify (5y-20)/(3y+15)*(7y+35)/(10y+40)

simplify

\frac{5 y - 20}{3 y + 15} \cdot \frac{7 y + 35}{10 y + 40}

\frac{7 ( y - 4 )}{6 ( y + 4 )}

Solution steps

\frac{5 y - 20}{3 y + 15} \cdot \frac{7 y + 35}{10 y + 40}

Apply the fraction rule:

\frac{a}{b} \cdot \frac{c}{d} = \frac{a \cdot c}{b \cdot d}

= \frac{( 5 y - 20 ) ( 7 y + 35 )}{( 3 y + 15 ) ( 10 y + 40 )}

Factor

5 y - 20 : 5 (y - 4)

= \frac{5 ( y - 4 ) ( 7 y + 35 )}{( 3 y + 15 ) ( 10 y + 40 )}

Factor

10 y + 40 : 10 (y + 4)

= \frac{5 ( y - 4 ) ( 7 y + 35 )}{( 3 y + 15 ) \cdot 10 ( y + 4 )}

Factor the number:

10 = 5 \cdot 2

= \frac{5 ( y - 4 ) ( 7 y + 35 )}{( 3 y + 15 ) \cdot 5 \cdot 2 ( y + 4 )}

Cancel the common factor:

5

= \frac{( y - 4 ) ( 7 y + 35 )}{( 3 y + 15 ) \cdot 2 ( y + 4 )}

Factor

7 y + 35 : 7 (y + 5)

= \frac{( y - 4 ) \cdot 7 ( y + 5 )}{( 3 y + 15 ) \cdot 2 ( y + 4 )}

Factor

3 y + 15 : 3 (y + 5)

= \frac{( y - 4 ) \cdot 7 ( y + 5 )}{3 ( y + 5 ) \cdot 2 ( y + 4 )}

Cancel the common factor:

y + 5

= \frac{( y - 4 ) \cdot 7}{3 \cdot 2 ( y + 4 )}

Multiply the numbers:

3 \cdot 2 = 6

= \frac{( y - 4 ) \cdot 7}{6 ( y + 4 )}

= \frac{7 ( y - 4 )}{6 ( y + 4 )}

\frac{x + 2}{x - 2} - \frac{x - 2}{x + 2}

54 \cdot 6

(4 a - 2 b)^{2}

\frac{x ^{9} y ^{8} z ^{5}}{x ^{3} y ^{4} z ^{5}}

What is (5y-20)/(3y+15)*(7y+35)/(10y+40) ?
The solution to (5y-20)/(3y+15)*(7y+35)/(10y+40) is (7(y-4))/(6(y+4))