Mr Daniels Maths
Algebraic Fractions Multiplication and Division

Set 1

Set 2

Set 3

Q1) \(x + 7\over 10\) x \(x + 3\over 4\) = [ \(x^2 + 10 x + 21\over 40\) ]

Q1) \(x + 9\over 2\) x \(3 \over{ x + 1}\) = [ \(3( x + 9) \over 2 ( x + 1)\) ]

Q1) \(x + 3\over 6\) x \(x + 2\over x + 3\) = [ \(x + 2\over 6\) ]

Q2) \(x + 8\over 10\) x \(x + 2\over 4\) = [ \(x^2 + 10 x + 16\over 40\) ]

Q2) \(x + 6\over 2\) x \(9 \over{ x + 9}\) = [ \(9( x + 6) \over 2 ( x + 9)\) ]

Q2) \(x + 10\over 8\) ÷ \( x + 10\over x + 10\) = [ \(x + 10\over 8\) ]

Q3) \(x + 9\over 4\) x \(x + 10\over 3\) = [ \(x^2 + 19 x + 90\over 12\) ]

Q3) \(x + 8\over 7\) x \(2 \over{ x + 1}\) = [ \(2( x + 8) \over 7 ( x + 1)\) ]

Q3) \(x + 9\over 3\) x \(x + 2\over x + 9\) = [ \(x + 2\over 3\) ]

Q4) \(x + 6\over 10\) x \(x + 2\over 10\) = [ \(x^2 + 8 x + 12\over 100\) ]

Q4) \(x + 9\over 5\) x \(2 \over{ x + 8}\) = [ \(2( x + 9) \over 5 ( x + 8)\) ]

Q4) \(x + 1\over 4\) x \(x + 7\over x + 1\) = [ \(x + 7\over 4\) ]

Q5) \(x + 10\over 6\) ÷ \(10 \over {x + 3}\) = [ \(x^2 + 13 x + 30\over 60\) ]

Q5) \(x + 5\over 5\) x \(3 \over{ x + 3}\) = [ \(3( x + 5) \over 5 ( x + 3)\) ]

Q5) \(x + 3\over 5\) x \(x + 3\over x + 3\) = [ \(x + 3\over 5\) ]

Q6) \(x + 2\over 4\) ÷ \(7 \over {x + 1}\) = [ \(x^2 + 3 x + 2\over 28\) ]

Q6) \(x + 8\over 10\) ÷ \({ x + 1} \over 9 \) = [ \(9( x + 8) \over 10 ( x + 1)\) ]

Q6) \(x + 2\over 5\) ÷ \( x + 2\over x + 10\) = [ \(x + 10\over 5\) ]

Q7) \(x + 4\over 3\) x \(x + 5\over 6\) = [ \(x^2 + 9 x + 20\over 18\) ]

Q7) \(x + 8\over 5\) x \(8 \over{ x + 7}\) = [ \(8( x + 8) \over 5 ( x + 7)\) ]

Q7) \(x + 6\over 5\) ÷ \( x + 6\over x + 3\) = [ \(x + 3\over 5\) ]

Q8) \(x + 1\over 9\) x \(x + 9\over 7\) = [ \(x^2 + 10 x + 9\over 63\) ]

Q8) \(x + 6\over 4\) ÷ \({ x + 2} \over 5 \) = [ \(5( x + 6) \over 4 ( x + 2)\) ]

Q8) \(x + 2\over 7\) x \(x + 7\over x + 2\) = [ \(x + 7\over 7\) ]

Q9) \(x + 5\over 8\) ÷ \(2 \over {x + 4}\) = [ \(x^2 + 9 x + 20\over 16\) ]

Q9) \(x + 5\over 4\) x \(3 \over{ x + 8}\) = [ \(3( x + 5) \over 4 ( x + 8)\) ]

Q9) \(x + 5\over 10\) x \(x + 8\over x + 5\) = [ \(x + 8\over 10\) ]

Q10) \(x + 6\over 4\) ÷ \(4 \over {x + 1}\) = [ \(x^2 + 7 x + 6\over 16\) ]

Q10) \(x + 2\over 1\) x \(1 \over{ x + 5}\) = [ \(1( x + 2) \over( x + 5)\) ]

Q10) \(x + 9\over 6\) x \(x + 10\over x + 9\) = [ \(x + 10\over 6\) ]