What is (6c^2+13c-63)/(6c^2-17c+7)*(2c^2-9c+4)/(12c+54) ?

The solution to (6c^2+13c-63)/(6c^2-17c+7)*(2c^2-9c+4)/(12c+54) is (c-4)/6

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simplify (6c^2+13c-63)/(6c^2-17c+7)*(2c^2-9c+4)/(12c+54)

simplify

\frac{6 c ^{2} + 13 c - 63}{6 c ^{2} - 17 c + 7} \cdot \frac{2 c ^{2} - 9 c + 4}{12 c + 54}

\frac{c - 4}{6}

Solution steps

\frac{6 c ^{2} + 13 c - 63}{6 c ^{2} - 17 c + 7} \cdot \frac{2 c ^{2} - 9 c + 4}{12 c + 54}

Cancel

\frac{6 c ^{2} + 13 c - 63}{6 c ^{2} - 17 c + 7} : \frac{2 c + 9}{2 c - 1}

= \frac{2 c + 9}{2 c - 1} \cdot \frac{2 c ^{2} - 9 c + 4}{12 c + 54}

Apply the fraction rule:

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

= \frac{( 2 c + 9 ) ( 2 c ^{2} - 9 c + 4 )}{( 2 c - 1 ) ( 12 c + 54 )}

Factor

2 c^{2} - 9 c + 4 : (2 c - 1) (c - 4)

= \frac{( 2 c + 9 ) ( 2 c - 1 ) ( c - 4 )}{( 2 c - 1 ) ( 12 c + 54 )}

Cancel the common factor:

2 c - 1

= \frac{( 2 c + 9 ) ( c - 4 )}{12 c + 54}

Factor

12 c + 54 : 6 (2 c + 9)

= \frac{( 2 c + 9 ) ( c - 4 )}{6 ( 2 c + 9 )}

Cancel the common factor:

2 c + 9

= \frac{c - 4}{6}

- 19 \leq 3 x - 5 \leq 1

5 < ∣ 4 - x ∣ < 7

x^{2} + 8 x - 7 < 0

lo g_{\frac{2}{3}} (\frac{9}{4}) = x

(4 x - 1)^{2} = 11

What is (6c^2+13c-63)/(6c^2-17c+7)*(2c^2-9c+4)/(12c+54) ?
The solution to (6c^2+13c-63)/(6c^2-17c+7)*(2c^2-9c+4)/(12c+54) is (c-4)/6