Mr Daniels Maths
Algebraic Fractions Simplification

Set 1

Set 2

Set 3

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

Q1) \({x+9}\over{x^2 -81}\) = [ \(1\over{x-9}\) ]

Q1) \({4x^2 +18x-36}\over{x+6}\) = [ \(4x-6\) ]

Q2) \({x-6\over{x^2 -11x+30}}\) = [ \(1\over{x-5}\) ]

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

Q2) \({2x^2 +2x-12}\over{x-2}\) = [ \(2x+6\) ]

Q3) \({x^2 +15x+56}\over{x+7}\) = [ \(x+8\) ]

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

Q3) \({3x^2 +5x-12}\over{x+3}\) = [ \(3x-4\) ]

Q4) \({x-5\over{x^2 -12x+35}}\) = [ \(1\over{x-7}\) ]

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

Q4) \({4x^2 -21x+20}\over{x-4}\) = [ \(4x-5\) ]

Q5) \({x-6\over{x^2 -2x-24}}\) = [ \(1\over{x+4}\) ]

Q5) \({x^2 -36}\over{x-6}\) = [ \(x+6\) ]

Q5) \({5x^2 -10x-15}\over{x-3}\) = [ \(5x+5\) ]

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

Q6) \({x^2 -100}\over{x-10}\) = [ \(x+10\) ]

Q6) \({3x^2 -3x-18}\over{x-3}\) = [ \(3x+6\) ]

Q7) \({x-5\over{x^2 +2x-35}}\) = [ \(1\over{x+7}\) ]

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

Q7) \({5x^2 -22x+8}\over{x-4}\) = [ \(5x-2\) ]

Q8) \({x-2\over{x^2 +x-6}}\) = [ \(1\over{x+3}\) ]

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

Q8) \({2x^2 -10x+12}\over{x-3}\) = [ \(2x-4\) ]

Q9) \({x+6\over{x^2 +11x+30}}\) = [ \(1\over{x+5}\) ]

Q9) \({x+6}\over{x^2 -36}\) = [ \(1\over{x-6}\) ]

Q9) \({4x^2 +11x+6}\over{x+2}\) = [ \(4x+3\) ]

Q10) \({x-3\over{x^2 -x-6}}\) = [ \(1\over{x+2}\) ]

Q10) \({x^2 -49}\over{x-7}\) = [ \(x+7\) ]

Q10) \({4x^2 +12x+8}\over{x+2}\) = [ \(4x+4\) ]