What is ((2x^2+5x-7))/((10x+35)) ?

The solution to ((2x^2+5x-7))/((10x+35)) is (x-1)/5

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simplify ((2x^2+5x-7))/((10x+35))

simplify

\frac{( 2 x ^{2} + 5 x - 7 )}{( 10 x + 35 )}

\frac{x - 1}{5}

Solution steps

\frac{2 x ^{2} + 5 x - 7}{10 x + 35}

Factor

2 x^{2} + 5 x - 7 : (x - 1) (2 x + 7)

= \frac{( x - 1 ) ( 2 x + 7 )}{10 x + 35}

Factor

10 x + 35 : 5 (2 x + 7)

= \frac{( x - 1 ) ( 2 x + 7 )}{5 ( 2 x + 7 )}

Cancel the common factor:

2 x + 7

= \frac{x - 1}{5}

\frac{( y ^{3} - y ^{2} + y - 1 )}{( y ^{3} + y ^{2} + y + 1 )}

(\frac{5}{x}) + 2 (y - 1)

x y + x y^{2}

25 x^{2} - 5 x - 49 y^{2} - 7 y

4 x^{2} + 6 x y

What is ((2x^2+5x-7))/((10x+35)) ?
The solution to ((2x^2+5x-7))/((10x+35)) is (x-1)/5