Mr Daniels Maths
Algebraic Fractions Simplification

Set 1

Set 2

Set 3

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

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

Q1) \({3x^2 -17x+20}\over{x-4}\) = [ \(3x-5\) ]

Q2) \({x^2 -12x+35}\over{x-5}\) = [ \(x-7\) ]

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

Q2) \({2x^2 -10x+12}\over{x-3}\) = [ \(2x-4\) ]

Q3) \({x-2\over{x^2 +7x-18}}\) = [ \(1\over{x+9}\) ]

Q3) \({x^2 -9}\over{x-3}\) = [ \(x+3\) ]

Q3) \({2x^2 -17x+30}\over{x-6}\) = [ \(2x-5\) ]

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

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

Q4) \({5x^2 -22x-15}\over{x-5}\) = [ \(5x+3\) ]

Q5) \({x+2\over{x^2 -6x-16}}\) = [ \(1\over{x-8}\) ]

Q5) \({x+2}\over{x^2 -4}\) = [ \(1\over{x-2}\) ]

Q5) \({5x^2 +15x+10}\over{x+2}\) = [ \(5x+5\) ]

Q6) \({x+3\over{x^2 +5x+6}}\) = [ \(1\over{x+2}\) ]

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

Q6) \({2x^2 -16x+24}\over{x-6}\) = [ \(2x-4\) ]

Q7) \({x+2\over{x^2 -4x-12}}\) = [ \(1\over{x-6}\) ]

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

Q7) \({4x^2 -22x+10}\over{x-5}\) = [ \(4x-2\) ]

Q8) \({x+7\over{x^2 +3x-28}}\) = [ \(1\over{x-4}\) ]

Q8) \({x^2 -9}\over{x+3}\) = [ \(x-3\) ]

Q8) \({4x^2 -14x-8}\over{x-4}\) = [ \(4x+2\) ]

Q9) \({x+2\over{x^2 +5x+6}}\) = [ \(1\over{x+3}\) ]

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

Q9) \({5x^2 -27x-18}\over{x-6}\) = [ \(5x+3\) ]

Q10) \({x^2 -6x+8}\over{x-4}\) = [ \(x-2\) ]

Q10) \({x^2 -36}\over{x-6}\) = [ \(x+6\) ]

Q10) \({3x^2 -9x-30}\over{x-5}\) = [ \(3x+6\) ]