Mr Daniels Maths
Algebraic Fractions Simplification

Set 1

Set 2

Set 3

Q1) \({x-5\over{x^2 -9x+20}}\) = [ \(1\over{x-4}\) ]

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

Q1) \({4x^2 -21x+20}\over{x-4}\) = [ \(4x-5\) ]

Q2) \({x^2 -5x+6}\over{x-2}\) = [ \(x-3\) ]

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

Q2) \({2x^2 +9x+10}\over{x+2}\) = [ \(2x+5\) ]

Q3) \({x-6\over{x^2 -11x+30}}\) = [ \(1\over{x-5}\) ]

Q3) \({x+2}\over{x^2 -4}\) = [ \(1\over{x-2}\) ]

Q3) \({5x^2 -7x-6}\over{x-2}\) = [ \(5x+3\) ]

Q4) \({x^2 -13x+42}\over{x-7}\) = [ \(x-6\) ]

Q4) \({x+4}\over{x^2 -16}\) = [ \(1\over{x-4}\) ]

Q4) \({2x^2 -9x-18}\over{x-6}\) = [ \(2x+3\) ]

Q5) \({x^2 -4x-12}\over{x+2}\) = [ \(x-6\) ]

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

Q5) \({5x^2 -35x+30}\over{x-6}\) = [ \(5x-5\) ]

Q6) \({x-4\over{x^2 -6x+8}}\) = [ \(1\over{x-2}\) ]

Q6) \({x+5}\over{x^2 -25}\) = [ \(1\over{x-5}\) ]

Q6) \({4x^2 +29x+30}\over{x+6}\) = [ \(4x+5\) ]

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

Q7) \({x^2 -9}\over{x+3}\) = [ \(x-3\) ]

Q7) \({2x^2 -13x+20}\over{x-4}\) = [ \(2x-5\) ]

Q8) \({x-7\over{x^2 -13x+42}}\) = [ \(1\over{x-6}\) ]

Q8) \({x+4}\over{x^2 -16}\) = [ \(1\over{x-4}\) ]

Q8) \({3x^2 -17x+20}\over{x-4}\) = [ \(3x-5\) ]

Q9) \({x-5\over{x^2 -10x+25}}\) = [ \(1\over{x-5}\) ]

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

Q9) \({5x^2 -32x+12}\over{x-6}\) = [ \(5x-2\) ]

Q10) \({x^2 -10x+24}\over{x-6}\) = [ \(x-4\) ]

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

Q10) \({5x^2 -27x+10}\over{x-5}\) = [ \(5x-2\) ]