Checking and securing understanding of solving with simple algebraic fractions
I can solve equations involving simple algebraic fractions.
- Year 11
Checking and securing understanding of solving with simple algebraic fractions
I can solve equations involving simple algebraic fractions.
Lesson details
Key learning points
- The product of a number and its reciprocal is 1
- The product of a term and its reciprocal is 1
- Equivalence can be maintained by multiplying the connected statements by the same reciprocal.
- This step can help you simplify an equation so you can solve it.
Keywords
Solution - A solution to an equality with one variable is a value which, when substituted, maintains the equality between the expressions.
Reciprocal - The reciprocal is the multiplicative inverse of any non-zero number. Any non-zero number multiplied by its reciprocal is equal to 1
Common misconception
When solving equations with fractional coefficients, pupils may divide by the reciprocal rather than multiply. E.g. $$\\frac{1}{2}x = 8$$ pupils may just halve $$8$$
Start with integer coefficients, and get pupils used to the idea of multiplying by the reciprocal (not just dividing by the coefficient). This step should be written as part of the working. Answers can be checked by substitution.
Licence
Prior knowledge starter quiz
6 Questions
Q1.Any non-zero number multiplied by its is equal to 1
Q2.What is the correct calculation to solve $$7x=8$$ in one step?
Q3.Match each number to its reciprocal.
$$2$$ -Â $$1\over 2$$
$$1\over 2$$ -Â $$2$$
$$3\over 2$$ -Â $$2\over 3$$
$$-{1\over 2}$$ -Â $$-2$$
$$-{2\over 3}$$ -Â $$-{3\over 2}$$
$$1\over 6$$ -Â $$6$$
Q4.The solution to the equation $$4x + 8 = 12$$ is when $$x = $$ .
Q5.What is the value of $$\frac{x-10}{5}$$ when $$x = 20$$?
Q6.Match each expression containing brackets to its expanded form.
$$2(3x-6)$$ -Â $$6x-12$$
$$6(x-1)$$ -Â $$6x-6$$
$$2(x-6)$$ -Â $$2x-12$$
$$-6(-2-x)$$ -Â $$6x+12$$
$$3(2x-6)$$ -Â $$6x-18$$
$$6(3-x)$$ -Â $$-6x+18$$
Assessment exit quiz
6 Questions
Q1.The solution to the equation $${1\over 3}a = 6$$ is when $$a =$$ .
Q2.Match each equation to its solution.
$${1\over 4}a + 3 = 10$$ -Â $$a = 28$$
$${1\over 4}(a + 3)=10$$ -Â $$a = 37$$
$${4\over 7} a = 1$$ -Â $$a = {7\over 4}$$
$${7\over 4}a=14$$ -Â $$a = 8$$
$$\frac{a+3}{7} = 2$$ -Â $$a = 11$$
Q3.Match each expression to an equivalent expression.
$${2\over 3}x$$ -Â $$2x\over 3$$
$$2\over 3x$$ -Â $$2({1\over 3x})$$
$${2\over 3}\times {1\over 3x}$$ -Â $$2\over 9x$$
$$2x \times {1\over 9}$$ -Â $$2x\over 9$$
$${3\over 2}\times {1\over x}$$ -Â $$3\over 2x$$