Mr Daniels Maths
Algebraic Fractions Simplification

Set 1

Set 2

Set 3

Q1) \({x+2\over{x^2 -4}}\) = [ \(1\over{x-2}\) ]

Q1) \({x^2 -81}\over{x+9}\) = [ \(x-9\) ]

Q1) \({3x^2 +6x-24}\over{x+4}\) = [ \(3x-6\) ]

Q2) \({x-2\over{x^2 -4}}\) = [ \(1\over{x+2}\) ]

Q2) \({x^2 -4}\over{x-2}\) = [ \(x+2\) ]

Q2) \({5x^2 +14x+8}\over{x+2}\) = [ \(5x+4\) ]

Q3) \({x^2 +5x+6}\over{x+2}\) = [ \(x+3\) ]

Q3) \({x^2 -25}\over{x-5}\) = [ \(x+5\) ]

Q3) \({5x^2 +26x-24}\over{x+6}\) = [ \(5x-4\) ]

Q4) \({x^2 +8x+15}\over{x+5}\) = [ \(x+3\) ]

Q4) \({x+3}\over{x^2 -9}\) = [ \(1\over{x-3}\) ]

Q4) \({5x^2 +8x-4}\over{x+2}\) = [ \(5x-2\) ]

Q5) \({x^2 +3x-10}\over{x-2}\) = [ \(x+5\) ]

Q5) \({x^2 -9}\over{x-3}\) = [ \(x+3\) ]

Q5) \({2x^2 -10x+12}\over{x-3}\) = [ \(2x-4\) ]

Q6) \({x+2\over{x^2 +7x+10}}\) = [ \(1\over{x+5}\) ]

Q6) \({x-3}\over{x^2 -9}\) = [ \(1\over{x+3}\) ]

Q6) \({5x^2 +13x-6}\over{x+3}\) = [ \(5x-2\) ]

Q7) \({x^2 -3x-10}\over{x-5}\) = [ \(x+2\) ]

Q7) \({x+4}\over{x^2 -16}\) = [ \(1\over{x-4}\) ]

Q7) \({2x^2 -8x-24}\over{x-6}\) = [ \(2x+4\) ]

Q8) \({x^2 +4x+4}\over{x+2}\) = [ \(x+2\) ]

Q8) \({x^2 -49}\over{x-7}\) = [ \(x+7\) ]

Q8) \({2x^2 -8}\over{x+2}\) = [ \(2x-4\) ]

Q9) \({x+9\over{x^2 +4x-45}}\) = [ \(1\over{x-5}\) ]

Q9) \({x-2}\over{x^2 -4}\) = [ \(1\over{x+2}\) ]

Q9) \({3x^2 +12x-36}\over{x+6}\) = [ \(3x-6\) ]

Q10) \({x+9\over{x^2 +x-72}}\) = [ \(1\over{x-8}\) ]

Q10) \({x^2 -4}\over{x+2}\) = [ \(x-2\) ]

Q10) \({5x^2 +35x+30}\over{x+6}\) = [ \(5x+5\) ]