Mr Daniels Maths
Algebraic Fractions Multiplication and Division

Set 1

Set 2

Set 3

Q1) \(x + 5\over 8\) x \(x + 4\over 6\) = [ \(x^2 + 9 x + 20\over 48\) ]

Q1) \(x + 6\over 2\) ÷ \({ x + 7} \over 3 \) = [ \(3( x + 6) \over 2 ( x + 7)\) ]

Q1) \(x + 1\over 4\) x \(x + 10\over x + 1\) = [ \(x + 10\over 4\) ]

Q2) \(x + 9\over 6\) ÷ \(2 \over {x + 4}\) = [ \(x^2 + 13 x + 36\over 12\) ]

Q2) \(x + 10\over 2\) x \(1 \over{ x + 9}\) = [ \(1( x + 10) \over 2 ( x + 9)\) ]

Q2) \(x + 9\over 2\) x \(x + 10\over x + 9\) = [ \(x + 10\over 2\) ]

Q3) \(x + 6\over 8\) x \(x + 1\over 6\) = [ \(x^2 + 7 x + 6\over 48\) ]

Q3) \(x + 8\over 3\) ÷ \({ x + 3} \over 1 \) = [ \(1( x + 8) \over 3 ( x + 3)\) ]

Q3) \(x + 2\over 4\) x \(x + 5\over x + 2\) = [ \(x + 5\over 4\) ]

Q4) \(x + 4\over 4\) x \(x + 2\over 8\) = [ \(x^2 + 6 x + 8\over 32\) ]

Q4) \(x + 9\over 1\) ÷ \({ x + 6} \over 2 \) = [ \(2( x + 9) \over( x + 6)\) ]

Q4) \(x + 10\over 6\) x \(x + 6\over x + 10\) = [ \(x + 6\over 6\) ]

Q5) \(x + 6\over 2\) ÷ \(9 \over {x + 3}\) = [ \(x^2 + 9 x + 18\over 18\) ]

Q5) \(x + 2\over 1\) ÷ \({ x + 5} \over 2 \) = [ \(2( x + 2) \over( x + 5)\) ]

Q5) \(x + 9\over 4\) x \(x + 8\over x + 9\) = [ \(x + 8\over 4\) ]

Q6) \(x + 7\over 4\) ÷ \(4 \over {x + 6}\) = [ \(x^2 + 13 x + 42\over 16\) ]

Q6) \(x + 6\over 7\) x \(3 \over{ x + 5}\) = [ \(3( x + 6) \over 7 ( x + 5)\) ]

Q6) \(x + 1\over 2\) ÷ \( x + 1\over x + 2\) = [ \(x + 2\over 2\) ]

Q7) \(x + 8\over 6\) ÷ \(4 \over {x + 6}\) = [ \(x^2 + 14 x + 48\over 24\) ]

Q7) \(x + 8\over 2\) ÷ \({ x + 9} \over 1 \) = [ \(1( x + 8) \over 2 ( x + 9)\) ]

Q7) \(x + 6\over 5\) x \(x + 5\over x + 6\) = [ \(x + 5\over 5\) ]

Q8) \(x + 7\over 10\) ÷ \(5 \over {x + 4}\) = [ \(x^2 + 11x + 28\over 50\) ]

Q8) \(x + 6\over 4\) ÷ \({ x + 2} \over 9 \) = [ \(9( x + 6) \over 4 ( x + 2)\) ]

Q8) \(x + 6\over 5\) x \(x + 1\over x + 6\) = [ \(x + 1\over 5\) ]

Q9) \(x + 1\over 6\) ÷ \(7 \over {x + 9}\) = [ \(x^2 + 10 x + 9\over 42\) ]

Q9) \(x + 8\over 2\) x \(1 \over{ x + 7}\) = [ \(1( x + 8) \over 2 ( x + 7)\) ]

Q9) \(x + 1\over 8\) x \(x + 3\over x + 1\) = [ \(x + 3\over 8\) ]

Q10) \(x + 1\over 8\) x \(x + 5\over 2\) = [ \(x^2 + 6 x + 5\over 16\) ]

Q10) \(x + 7\over 6\) x \(7 \over{ x + 2}\) = [ \(7( x + 7) \over 6 ( x + 2)\) ]

Q10) \(x + 3\over 2\) ÷ \( x + 3\over x + 4\) = [ \(x + 4\over 2\) ]