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Year 9 Maths Linear and Simultaneous Equations

Solving Linear Equations

20 practice questions 1 video lesson Theory + worked examples
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Theory

To solve an equation, find the value of the pronumeral that makes both sides equal. This Year 9 maths page shows how to undo the operations in reverse order, doing the same to both sides each time. It covers decimals, a negative coefficient, equations with a fraction, and checking the answer by substitution.

An equation says that two expressions are equal, such as \(3x + 5 = 20\). To solve it, find the value of the pronumeral that makes it true.

Think of an equation as a balance. Whatever you do to one side, do to the other. Then it stays balanced.

Each operation has an inverse that undoes it. Adding is undone by subtracting. Multiplying is undone by dividing.

Undo the operations in reverse order. In \(3x + 5\), the \(x\) was multiplied by \(3\) and then \(5\) was added. So subtract \(5\) first, then divide by \(3\).

The number in front of the pronumeral is its coefficient. In \(-2x\), the coefficient is \(-2\).

An equation as a balanceLeft: a balance with 3x plus 5 on one side and 20 on the other. Right: after taking 5 from both sides, the balance holds 3x and 15 and is still level. 3x + 520 3x + 5 = 20 βˆ’5 3x15 3x = 15
An equation is a balance. Take the same amount from both sides and it stays level.
Undoing the operations in reverse orderTop row: x is multiplied by 3, then 5 is added, giving 3x plus 5. Bottom row: starting from 20, subtract 5 to get 15, then divide by 3 to get 5. x Γ— 3 3x + 5 3x + 5 5 Γ· 3 15 βˆ’ 5 20
To solve \(3x + 5 = 20\), undo the steps in reverse order: subtract \(5\) first, then divide by \(3\). So \(x = 5\).

The inverse operations:

OperationUndo it by
\(+\) a number\(-\) the same number
\(-\) a number\(+\) the same number
\(\times\) a number\(\div\) the same number
\(\div\) a number\(\times\) the same number

Clearing a fraction: multiply both sides by the denominator.

\[\dfrac{x}{4} = 3 \quad\Rightarrow\quad x = 3 \times 4 = 12\]
x4=3⇒x=12

A whole expression over a denominator is cleared the same way:

\[\dfrac{2x + 1}{3} = 5 \quad\Rightarrow\quad 2x + 1 = 15\]
2x+13=5
Decimals work the same way. For \(n + 2.6 = 1.4\), subtract \(2.6\) from both sides: \(n = 1.4 - 2.6 = -1.2\).
Checking. Substitute your answer into the original equation. If both sides come out the same, the answer is right.

How to solve an equation with the pronumeral on one side

  1. Undo any adding or subtracting outside the fraction or term, on both sides.
  2. Clear a fraction by multiplying both sides by its denominator.
  3. Divide both sides by the coefficient, keeping its sign.
  4. Check by substituting the answer into the original equation.
Example 1 β€” Two steps
Solve
(i) \(4x + 7 = 31\)
(ii) \(6y - 5 = -2\).
Solution

Undo the adding or subtracting first. Then undo the multiplying.

(i)

\(4x + 7 - 7\)\(=\)\(31 - 7\)
\(4x\)\(=\)\(24\)
\(x\)\(=\)\(\dfrac{24}{4}\)
\(x\)\(=\)\(\boldsymbol{6}\)
x=6

Check: \(4 \times 6 + 7 = 31\) \(\checkmark\)

(ii)

\(6y - 5 + 5\)\(=\)\(-2 + 5\)
\(6y\)\(=\)\(3\)
\(y\)\(=\)\(\dfrac{3}{6}\)
\(y\)\(=\)\(\boldsymbol{\dfrac{1}{2}}\)
y=12

The answer does not have to be a whole number. \(y = \dfrac{1}{2}\) is the same as \(y = 0.5\).

Example 2 β€” A negative coefficient
Solve \(5 - 2m = 17\).
Solution

The \(5\) is added on, so subtract \(5\) from both sides. Then divide by \(-2\).

\(5 - 2m - 5\)\(=\)\(17 - 5\)
\(-2m\)\(=\)\(12\)
\(m\)\(=\)\(\dfrac{12}{-2}\)
\(m\)\(=\)\(\boldsymbol{-6}\)
m=−6

Check: \(5 - 2 \times (-6) = 5 + 12 = 17\) \(\checkmark\)

A positive divided by a negative is negative.

Example 3 β€” A fraction
Solve \(\dfrac{3y}{5} = 12\).
Solution

Multiply both sides by \(5\) to clear the fraction. Then divide by \(3\).

\(\dfrac{3y}{5} \times 5\)\(=\)\(12 \times 5\)
\(3y\)\(=\)\(60\)
\(y\)\(=\)\(\dfrac{60}{3}\)
\(y\)\(=\)\(\boldsymbol{20}\)
y=20

Check: \(\dfrac{3 \times 20}{5} = \dfrac{60}{5} = 12\) \(\checkmark\)

Example 4 β€” A fraction with a number outside
Solve \(\dfrac{3a - 1}{4} + 2 = 7\).
Solution

The \(+\,2\) is outside the fraction, so deal with it first. Then clear the fraction.

\(\dfrac{3a - 1}{4} + 2 - 2\)\(=\)\(7 - 2\)
\(\dfrac{3a - 1}{4}\)\(=\)\(5\)
\(\dfrac{3a - 1}{4} \times 4\)\(=\)\(5 \times 4\)
\(3a - 1\)\(=\)\(20\)
\(3a\)\(=\)\(21\)
\(a\)\(=\)\(\boldsymbol{7}\)
a=7

Check: \(\dfrac{3 \times 7 - 1}{4} + 2 = \dfrac{20}{4} + 2 = 7\) \(\checkmark\)

Common pitfalls

Watch the order. Dividing \(3x + 5 = 20\) by \(3\) first means dividing the \(5\) as well. Undo the \(+\,5\) first.
Both sides, every time. Changing only one side breaks the balance and gives a wrong answer.
Watch the sign when dividing by a negative. \(-2x = 12\) gives \(x = -6\), and \(-2x = -12\) gives \(x = 6\). The same goes for a negative denominator: in \(\dfrac{x + 1}{-2} = 3\), multiply both sides by \(-2\).
Multiply the whole side. To clear \(\dfrac{3a - 1}{4} + 2\), take away the \(2\) first. Multiplying only part of a side by \(4\) breaks the equation.

Frequently asked questions

How do you solve a two-step equation?

Undo the adding or subtracting first, then undo the multiplying or dividing. Do the same to both sides each time. For \(4x + 7 = 31\), subtract \(7\) to get \(4x = 24\), then divide by \(4\) to get \(x = 6\).

What is an inverse operation?

It is the operation that undoes another one. Subtracting undoes adding, and dividing undoes multiplying.

How do you solve an equation with a fraction?

Multiply both sides by the denominator to clear the fraction. Then solve the equation that is left. If a number is added outside the fraction, take it away first.

What do you do when the pronumeral has a negative in front?

Divide both sides by the negative number. For \(-2m = 12\), divide by \(-2\) to get \(m = -6\).

How do you check the answer to an equation?

Substitute the answer into the original equation. Work out each side. If they are equal, the answer is correct.