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

Question

1518 views around the world

You can reuse this answer
Creative Commons License

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

$(-1*(a^2 - 4a +6))/(2a-1)$

That will then result in

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

Then combine the equation under one denominator like so

$(6a^2-7a+8-a^2+4a-6)/(2a -1)$

Simplify

$(5a^2-3a+2)/(2a-1)$

Since the numerator cannot be simplified further the final answer is

$(5a^2-3a+2)/(2a-1)$

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