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Year 11
Higher

Solving equations with algebraic fractions

I can manipulate and solve equations involving algebraic fractions.

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New
New
Year 11
Higher

Solving equations with algebraic fractions

I can manipulate and solve equations involving algebraic fractions.

Download all resources
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Share activities with pupils
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Lesson details

Key learning points

  1. Before performing any operation, equivalent fractions should be considered.
  2. Algebraic fractions follow the same rules as fractions.
  3. It is important to maintain equivalence between two algebraic statements.
  4. Multiplying both connected statements by the reciprocal can simplify.

Keywords

  • Factorise - To factorise is to express a term as the product of its factors.

  • Quadratic formula - The quadratic formula is a formula for finding the solutions to any quadratic equation of the form $$ax^2 + bx + c = 0$$

Common misconception

Assuming that 'cross multiplying' is always the best way to solve equations with fractions on both sides.

Examples are included in the lesson where this results in unnecessary quadratic equations. Factorising and looking for common factors allows pupils to spot efficient methods. This also links with fraction skills of finding a common denominator.

Solving quadratics is an essential skill for this lesson. This is a good opportunity to review choices of methods to solve quadratics as well as the use of technology to check answers. Some pupils may wish to use the equation tool on their calculators to solve quadratic equations.
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Licence

This content is © Oak National Academy Limited (2024), licensed on Open Government Licence version 3.0 except where otherwise stated. See Oak's terms & conditions (Collection 2).

Lesson video

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Starter quiz

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6 Questions

Q1.
The solution to the equation $${2x\over 5}=4$$ is when $$x=$$ .
Correct Answer: 10
Q2.
Solve $${12\over 5x} = 8 $$.
$$2\over 15$$
Correct answer: $$3\over 10$$
$$4\over 5$$
$$10\over 3$$
$$15\over 2$$
Q3.
Factorise $$x^2 + 3x -4$$.
$$(x-2)(x-2)$$
$$(x-1)(x+3)$$
Correct answer: $$(x-1)(x+4)$$
$$(x-4)(x+3)$$
Q4.
Find all the solutions to the equation $$x^2 + 5x +6 = 2$$.
$$x=-2$$
$$x=-2 $$ and $$x=-3$$
Correct answer: $$x=-1 $$ and $$x=-4$$
$$x=1 $$ and $$x=4$$
$$x=2 $$ and $$x=3$$
Q5.
Which of these is equivalent to $${x+3\over 2}\times {4\over x+4}$$?
$$2x+12\over 8$$
Correct answer: $$2x+6\over x+4$$
$$x+3\over 2x+2$$
$$4x+3\over 2x+4$$
$$x^2 + 7x +12\over 8$$
Q6.
Simplify $${x+1\over 3x}+{x+2\over 6x}$$.
$$2x+3\over 6x$$
$$2x+3\over 9x$$
$$3x+3\over 18x$$
Correct answer: $$3x+4\over 6x$$
$$3x^2+1\over 6x^2$$

Exit quiz

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6 Questions

Q1.
What would be the most efficient first step to solve $${x+2\over 4x}+{3\over 6x}=4$$?
Correct answer: Convert to fractions over a common denominator.
Factorise all expressions.
Multiply all terms by $$6x$$.
Subtract 2 from both sides.
Subtract 4 from both sides.
Q2.
Which of these show all the solutions to the equation $${x+2\over 4x}+{3\over 6x}=4$$?
$$x=-{8\over 3}$$
$$x=0$$ or $$x={4\over 7}$$
$$x={5\over 39}$$
Correct answer: $$x={4\over 15}$$
$$x={1\over 3}$$
Q3.
Which of these is a correct step when solving $${4\over x-2}-{6\over x-4}=-2$$?
$${4\over x-4}=-2$$
$${4\over x-2}-{4\over x-4} =0$$
$${-2x-28\over (x-4)(x-2)} =-2$$
$${-2x-6\over (x-4)(x-2)} =-2$$
Correct answer: $${-2x-4\over (x-4)(x-2)} =-2$$
Q4.
Find all the solutions to $${4\over x-2}-{6\over x-4}=-2$$.
$$x=-6$$ or $$x=1$$
$$x=-2$$ or $$x=5$$
$$x=2$$ or $$x=3$$
$$x=2$$ or $$x=5$$
Correct answer: $$x=6$$ or $$x=1$$
Q5.
Which of these is a correct step when solving $${x+3\over 2x-3}={5\over x-1}$$?
$$x^2-3=10x-15$$
$${-1(x-3)\over 2(x-3)}={5\over x-1}$$
$${x+3\over 2x-3}={5+x+2\over 2x-3}$$
Correct answer: $${x^2 +2x-3\over (2x-3)(x-1)}={10x-15\over (2x-3)(x-1)}$$
$${(x+3)(x-1)\over 2(x-3)(x-1)}={10(x-3)\over 2(x-3)(x-1)}$$
Q6.
Find all the solutions to $${x+3\over 2x-3}={5\over x-1}$$.
Correct answer: $$x=2$$ or $$x=6$$
$$x=2$$ or $$x=-6$$
$$x=-2$$ or $$x=6$$
$$x=-2$$ or $$x=-6$$