Year 10 Maths - 10 Linear Relations

Simultaneous Equations Using Substitution

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Theory

Simultaneous equations are two equations that must be true at the same time. This Year 10 maths page shows how to solve a pair by substitution: replace one pronumeral with an expression from the other equation, solve, then substitute back and check. It also shows why the answer is the point where two lines cross, how fractions are handled, and how to set up and solve worded problems such as comparing two plans.

Simultaneous equations are two (or more) equations that must be true at the same time. With two pronumerals, such as $x$ and $y$, you need two equations.

The solution is the pair of values that makes both equations true, written as $x = 2,\ y = 4$ or as the point $(2, 4)$.

On a graph, each linear equation is a straight line. The solution is the point of intersection, where the lines cross.

In the substitution method, one equation gives an expression for one pronumeral, such as $y = 2x + 1$. That expression is put in place of the pronumeral in the other equation, leaving one equation with one pronumeral.

Substitution is quickest when a pronumeral is already the subject, or has no number in front of it.

Substitution: one equation tells you what $y$ equals, so put that expression, in brackets, in place of $y$ in the other equation.

Each equation is a straight line. The solution is the point where the lines cross: here $x = 2$ and $y = 4$.

The idea in symbols: if $y = mx + c$, replace every $y$ in the other equation with $(mx + c)$.

\[ax + by = k \quad \text{and} \quad y = mx + c\]\[\Rightarrow \quad ax + b(mx + c) = k\]

a x + b (m x + c) = k

Both equations start with y =. Make the right-hand sides equal. With a fraction, multiply every term to clear it: \(4 - x = \dfrac{1}{2}x + 1\) becomes \(8 - 2x = x + 2\), so \(x = 2\).

Fractions in an equation. Clear them first by multiplying by the lowest common denominator. For \(\dfrac{x}{3} + \dfrac{y}{2} = 4\) and \(y = x - 2\): multiply by \(6\) to get \(2x + 3y = 24\), then \(2x + 3(x - 2) = 24\), so \(x = 6\) and \(y = 4\).

Reading from a graph. Read the crossing point, then check it in both equations. A graph answer can be off if the point is not on the grid lines.

Parallel lines. \(y = 2x + 1\) and \(y = 2x - 3\) never meet. Substituting gives \(1 = -3\), which is false, so there is no solution.

A fraction as the subject. If \(y = \dfrac{x + 2}{3}\) and \(2x - 3y = 5\), substitute the whole fraction in brackets: \(2x - 3\left(\dfrac{x + 2}{3}\right) = 5\). The \(3\)s cancel, leaving \(2x - (x + 2) = 5\), so \(x = 7\) and \(y = \dfrac{7 + 2}{3} = 3\).

How to solve by substitution

Number the equations (1) and (2).
Make a pronumeral the subject of one equation, if it is not already. Choose one with no number in front.
Substitute that expression, in brackets, into the other equation.
Solve the new equation, which has only one pronumeral.
Substitute back into the simplest equation to find the other pronumeral.
Check both values in the equation you did not use last.

How to solve a worded problem

Define the pronumerals, with units.
Write two equations, one for each fact (for example a count and a cost).
Solve and answer the question in words.

Example 1 — One pronumeral is already the subject

Solve simultaneously: $y = 3x - 1$ and $2x + y = 14$.

Solution

Number the equations. Equation (1) says what $y$ equals.

$y$	$=$	$3x - 1 \quad (1)$
$2x + y$	$=$	$14 \quad (2)$

Substitute (1) into (2), using brackets:

$2x + (3x - 1)$	$=$	$14$
$5x - 1$	$=$	$14$
$5x$	$=$	$15$
$x$	$=$	$3$

Substitute $x = 3$ into (1):

$y$

$=$

$3(3) - 1 = 8$

$\boldsymbol{x = 3,\ y = 8}$. Check in (2): $2(3) + 8 = 14$ ✓

x = 3, y = 8

Example 2 — Rearrange first

Solve simultaneously using substitution: $3x + y = 11$ and $5x - 2y = 11$.

Solution

Make $y$ the subject of the equation where it has no number in front.

$3x + y$	$=$	$11 \quad (1)$
$y$	$=$	$11 - 3x \quad (3)$

Substitute (3) into $5x - 2y = 11$. The bracket matters, because $-2$ multiplies both terms.

$5x - 2(11 - 3x)$	$=$	$11$
$5x - 22 + 6x$	$=$	$11$
$11x$	$=$	$33$
$x$	$=$	$3$

$y$

$=$

$11 - 3(3) = 2$

$\boldsymbol{x = 3,\ y = 2}$. Check: $5(3) - 2(2) = 11$ ✓

x = 3, y = 2

Example 3 — Two expressions for y

Find the point where the lines $y = 2x - 3$ and $y = -x + 6$ meet.

Solution

At the meeting point both lines have the same $y$, so make the right-hand sides equal.

$2x - 3$	$=$	$-x + 6$
$3x - 3$	$=$	$6$
$3x$	$=$	$9$
$x$	$=$	$3$

$y$

$=$

$2(3) - 3 = 3$

Check in the other line: $-3 + 6 = 3$ ✓. The lines meet at $\boldsymbol{(3, 3)}$.

(3, 3)

Example 4 — Comparing two gyms

Gym A charges a $40 joining fee plus $15 a week. Gym B charges a $10 joining fee plus $20 a week. After how many weeks do the two gyms cost the same, and what is that cost?

Solution

Let $C$ be the total cost in dollars after $w$ weeks.

$C$	$=$	$40 + 15w \quad (\text{A})$
$C$	$=$	$10 + 20w \quad (\text{B})$

The costs are equal, so make the right-hand sides equal:

$40 + 15w$	$=$	$10 + 20w$
$30$	$=$	$5w$
$w$	$=$	$6$

$C$

$=$

$40 + 15(6) = 130$

The gyms cost the same after $6$ weeks, when each costs $130. After that, Gym A is cheaper, because it adds less each week.

w = 6, C = 130

Common pitfalls

Use brackets when you substitute. $5x - 2y$ with $y = 11 - 3x$ is $5x - 2(11 - 3x)$, not $5x - 22 - 3x$.

Find both values. Finding $x$ is only half the answer. Substitute back to find $y$.

Substitute into the other equation. Putting $y = 3x - 1$ back into itself gives $3x - 1 = 3x - 1$, which tells you nothing.

Check in both equations. A pair that works in only one equation is not the solution. Checking also catches sign slips.

Frequently asked questions

What does it mean to solve simultaneous equations?

It means finding the values of the pronumerals that make both equations true at the same time. On a graph, this is the point where the two lines cross.

When should you use the substitution method?

Use it when one equation already has a pronumeral as the subject, such as $y = 2x + 1$, or when a pronumeral has no number in front of it.

Why do you need brackets when substituting?

The whole expression replaces the pronumeral. Brackets make sure any number or minus sign in front multiplies every term of the expression.

How do you check the answer to simultaneous equations?

Substitute both values into each original equation. Both equations must come out true.

Can simultaneous equations have no solution?

Yes. If the two lines are parallel they never meet, so there is no solution. Solving gives a false statement such as $1 = -3$.

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Simultaneous Equations Using Elimination

\(y\)	\(=\)	\(3x - 1 \quad (1)\)
\(2x + y\)	\(=\)	\(14 \quad (2)\)

\(2x + (3x - 1)\)	\(=\)	\(14\)
\(5x - 1\)	\(=\)	\(14\)
\(5x\)	\(=\)	\(15\)
\(x\)	\(=\)	\(3\)

\(3x + y\)	\(=\)	\(11 \quad (1)\)
\(y\)	\(=\)	\(11 - 3x \quad (3)\)

\(5x - 2(11 - 3x)\)	\(=\)	\(11\)
\(5x - 22 + 6x\)	\(=\)	\(11\)
\(11x\)	\(=\)	\(33\)
\(x\)	\(=\)	\(3\)

\(2x - 3\)	\(=\)	\(-x + 6\)
\(3x - 3\)	\(=\)	\(6\)
\(3x\)	\(=\)	\(9\)
\(x\)	\(=\)	\(3\)

\(C\)	\(=\)	\(40 + 15w \quad (\text{A})\)
\(C\)	\(=\)	\(10 + 20w \quad (\text{B})\)

\(40 + 15w\)	\(=\)	\(10 + 20w\)
\(30\)	\(=\)	\(5w\)
\(w\)	\(=\)	\(6\)

Simultaneous Equations Using Substitution

📖 Theory

How to solve by substitution

How to solve a worded problem

Common pitfalls

Frequently asked questions

Theory