How do you subtract $\frac{6 a^{2} - 7 a + 8}{2 a - 1} - \frac{a^{2} - 4 a + 6}{2 a - 1}$ $\frac { 6a ^ { 2} - 7a + 8} { 2a - 1} - \frac { a ^ { 2} - 4a + 6} { 2a - 1}$ ?

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Answer 1 · 2017-05-15T06:41:14

Since the denominators are the same you can get right to subtracting.

There is a minus sign splitting the two parts of the equation

Multiply the second half of the equation by -1 because that negative applies to ALL terms

$\frac{- 1 \cdot (a^{2} - 4 a + 6)}{2 a - 1}$ $(-1*(a^2 - 4a +6))/(2a-1)$

That will then result in

$(- a^{2} + 4 a - 6)$ $(-a^2 + 4a -6)$

Then combine the equation under one denominator like so

$\frac{6 a^{2} - 7 a + 8 - a^{2} + 4 a - 6}{2 a - 1}$ $(6a^2-7a+8-a^2+4a-6)/(2a -1)$

Simplify

$\frac{5 a^{2} - 3 a + 2}{2 a - 1}$ $(5a^2-3a+2)/(2a-1)$

Since the numerator cannot be simplified further the final answer is

$\frac{5 a^{2} - 3 a + 2}{2 a - 1}$ $(5a^2-3a+2)/(2a-1)$

How do you subtract 6a2−7a+82a−1−a2−4a+62a−1\frac { 6a ^ { 2} - 7a + 8} { 2a - 1} - \frac { a ^ { 2} - 4a + 6} { 2a - 1}?