Solving complex linear equations | Maths | Quizalize & Oak National Academy

Starter quiz

$1\over7$ is the ______ of $7$
- fraction
- division
- reciprocal ✓
- divisor
- multiplier
Solve $6x+9=27$
- $x=3$ ✓
- $x=6$
- $x=9$
- $x=18$
Solve ${x\over3}=15$
- $x=45$ ✓
- $x=18$
- $x=5$
- $x={1\over5}$
Calculate ${2\over3}\times{4\over5}$
- $6\over15$
- $6\over20$
- $8\over15$ ✓
- $10\over12$
- $12\over10$
Which of these expressions are equivalent to ${{x+1}\over3}\times{2\over3}$ ?
- ${2x+1}\over9$
- ${2{(x+1)}\over9}$ ✓
- ${2{(x+1)}\over3}$
- ${2x+2}\over3$
- ${2x+2}\over9$ ✓
Expand and simplify $8({{x}\over8})$
- $8$
- $x$ ✓
- $8x$
- ${{8x}\over8}$
- $x=8$

${{56}\over{y}}=7$ is an example of a ______ equation.
- variable
- rational ✓
- irrational
- special
The solution to ${{6x−2}\over{4}}=7$ is $x=$ ______
- '5' ✓
Solve ${{2x}\over{3}}+1=11$
- $x=15$ ✓
- $x=16$
- $x=17$
- $x=18$
Which is the most efficient first step when solving the equation ${{x+7}\over{9}}={{x-1}\over{3}}$ ?
- Subtract $7$ from both sides.
- Add $1$ to both sides.
- Multiply both sides by $9$ ✓
- Multiply both sides by $3$
- Multiply both sides by $27$
Solve ${{x+7}\over{9}}={{x−1}\over{3}}$ . The solution is $x=$ ______
- '5' ✓
Solve ${{32}\over{3x}}=8$
- $x={3\over4}$
- $x=3$
- $x=4$
- $x={4\over3}$ ✓

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Any linear equation written in the form Ax + B = C can be solved.
Equations involving division by a constant can be manipulated to reach the form Ax + B = C.
Equations involving division by a variable can be manipulated to reach the form Ax + B = C.

Pupils often \"multiply through by $5$ \" and turn ${{2x+1}\\over5}={10}$ into ${{10x+5}}={50}$

Use equivalent fractions as opposed to multiplying through. In one simple step ${{2x+1}\\over5}={10}$ becomes ${{2x+1}\\over5}={50\\over5}$

Equation - An equation is used to show two expressions that are equal to each other.