How do you divide and simplify $\frac { x + 1} { 4x + y } \div \frac { 3x ^ { 2} - x - 4} { 16x ^ { 2} - y ^ { 2} }$ ?

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Answer 1 · 2017-10-07T07:18:07

(4x+y)/(3x-4)

When dividing by a factor $a/b$ , it is the same as multiplying by $b/a$ . That in mind...

$(x+1)/(4x+y)div(3x^2-x-4)/(16x^2-y^2)= (x+1)/(4x+y)*(16x^2-y^2)/(3x^2-x-4)$ .

Factoring the right hand term gives us:

$(x+1)/(4x+y)times(16x^2-y^2)/(3x^2-x-4)$ $= (x+1)/(4x+y)times((4x+y)(4x-y))/((3x-4)(x+1)$ .

The $x+1$ and $4x+y$ terms cancel out, leaving...

$(x+1)/(4x+y)times((4x+y)(4x-y))/((3x-4)(x+1)) = (4x-y)/(3x-4)$

How do you divide and simplify \frac { x + 1} { 4x + y } \div \frac { 3x ^ { 2} - x - 4} { 16x ^ { 2} - y ^ { 2} }\frac { x + 1} { 4x + y } \div \frac { 3x ^ { 2} - x - 4} { 16x ^ { 2} - y ^ { 2} }?