Year 10 Maths - 10 Linear Relations

Simultaneous Equations Using Elimination

Q: What does it mean if both pronumerals disappear?

If you get a false statement such as 0 = 1, there is no solution and the lines are parallel. If you get 0 = 0, the two equations are the same line and there are infinitely many solutions.

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Theory

This Year 10 maths page shows how to solve simultaneous equations by elimination: line up the equations, add or subtract them to remove one pronumeral, multiply first when the numbers do not match, then substitute back and check. It covers decimals and money, answers that are fractions, pairs with no solution or infinitely many, and worded problems.

In the elimination method you add or subtract the two equations so that one pronumeral disappears (is eliminated). That leaves one equation with one pronumeral.

The number in front of a pronumeral is its coefficient. To eliminate $y$, its coefficients in the two equations must be the same number.

If the coefficients have opposite signs (such as $+4y$ and $-4y$), add the equations. If they have the same sign, subtract one equation from the other.

Multiplying every term of an equation by the same number gives an equivalent equation with the same solution. Use this to make the coefficients match.

Elimination works well when neither pronumeral is the subject, such as $3x + 4y = 5$ and $2x - 3y = 9$.

Line up the like terms. When the $y$ terms are opposites, adding the equations removes $y$.

Look at the numbers in front of one pronumeral to decide what to do. Multiplying first makes them match.

Both equations should be lined up in the same form before adding or subtracting:

\[\begin{array}{rcl} a_1x + b_1y &=& k_1 \\ a_2x + b_2y &=& k_2 \end{array}\]

a_{1} x + b_{1} y = k_{1}

Lining up. Move the terms so \(x\), \(y\) and the number are in the same places. \(y + 4 = 3x\) becomes \(3x - y = 4\), and \(x - 2y - 5 = 0\) becomes \(x - 2y = 5\).

Decimals and money. Multiply by \(10\) or \(100\) to work with whole numbers, or work in cents: $3.60 is \(360\) cents.

Choosing the multipliers. For \(2y\) and \(3y\), multiply to get \(6y\) (the lowest common multiple): the first equation by \(3\) and the second by \(2\).

No solution or infinitely many. If both pronumerals disappear, look at what is left. \(x + 2y = 4\) and \(2x + 4y = 9\) give \(0 = 1\): false, so no solution (the lines are parallel). \(x + 2y = 4\) and \(2x + 4y = 8\) give \(0 = 0\): always true, so infinitely many solutions (it is the same line).

Answers need not be whole numbers. Eliminating can leave an equation such as \(11x = 4\), so \(x = \dfrac{4}{11}\). Keep exact fractions while you work, substitute the fraction carefully, and give the answer as a fraction unless decimals are asked for.

How to solve by elimination

Line up both equations as $ax + by = k$ and number them. Clear fractions and decimals.
Choose a pronumeral to eliminate. If its coefficients differ, multiply one or both equations to make them the same number.
Add the equations if the signs are opposite; subtract if they are the same.
Solve for the pronumeral that is left.
Substitute back into one of the original equations to find the other.
Check both values in the other original equation.

Example 1 — Adding or subtracting

Solve simultaneously:
(i) $3x + 2y = 19$ and $x - 2y = 1$
(ii) $5x + 3y = 21$ and $2x + 3y = 12$

Solution

(i) The $y$ terms are $+2y$ and $-2y$: opposite signs, so add.

$(1) + (2):\quad 4x$	$=$	$20$
$x$	$=$	$5$
$5 - 2y$	$=$	$1 \quad \text{(in (2))}$
$y$	$=$	$2$

$\boldsymbol{x = 5,\ y = 2}$. Check in (1): $15 + 4 = 19$ ✓

(ii) The $y$ terms are both $+3y$: same sign, so subtract.

$(1) - (2):\quad 3x$	$=$	$9$
$x$	$=$	$3$
$2(3) + 3y$	$=$	$12 \quad \text{(in (2))}$
$y$	$=$	$2$

$\boldsymbol{x = 3,\ y = 2}$. Check in (1): $15 + 6 = 21$ ✓

x = 3, y = 2

Example 2 — Multiplying both equations

Solve simultaneously: $3x + 4y = 5$ and $2x - 3y = 9$.

Solution

Make the $y$ terms opposites: $12y$ and $-12y$.

$(1) \times 3:\quad 9x + 12y$	$=$	$15 \quad (3)$
$(2) \times 4:\quad 8x - 12y$	$=$	$36 \quad (4)$
$(3) + (4):\quad 17x$	$=$	$51$
$x$	$=$	$3$

Substitute $x = 3$ into (1):

$9 + 4y$	$=$	$5$
$4y$	$=$	$-4$
$y$	$=$	$-1$

$\boldsymbol{x = 3,\ y = -1}$. Check in (2): $6 - 3(-1) = 9$ ✓

x = 3, y = - 1

Example 3 — Line up and clear decimals first

Solve simultaneously: $2y = 3x - 1$ and $0.5x + 0.2y = 2.3$.

Solution

Write both equations as $ax + by = k$, with whole numbers.

$3x - 2y$	$=$	$1 \quad (1)$
$\times 10:\quad 5x + 2y$	$=$	$23 \quad (2)$

The $y$ terms are opposites, so add:

$8x$	$=$	$24$
$x$	$=$	$3$
$2y$	$=$	$3(3) - 1 = 8$
$y$	$=$	$4$

$\boldsymbol{x = 3,\ y = 4}$. Check: $0.5(3) + 0.2(4) = 1.5 + 0.8 = 2.3$ ✓

x = 3, y = 4

Example 4 — Snacks at the cinema

Two popcorns and three drinks cost $22.50. Four popcorns and one drink cost $25.00. Find the price of a popcorn and of a drink.

Solution

Let a popcorn cost $$p$ and a drink cost $$d$.

$2p + 3d$	$=$	$22.50 \quad (1)$
$4p + d$	$=$	$25.00 \quad (2)$

Make the $p$ terms match, then subtract:

$(1) \times 2:\quad 4p + 6d$	$=$	$45.00 \quad (3)$
$(3) - (2):\quad 5d$	$=$	$20.00$
$d$	$=$	$4.00$
$4p + 4$	$=$	$25$
$p$	$=$	$5.25$

A popcorn costs $5.25 and a drink costs $4.00. Check: $2(5.25) + 3(4) = 22.50$ ✓

p = 5.25, d = 4

Common pitfalls

Subtract every term. $(5x + 3y) - (2x - 3y) = 3x + 6y$, not $3x$. Subtracting $-3y$ gives $+3y$.

Multiply every term, including the number. $2x + y = 7$ times $3$ is $6x + 3y = 21$, not $6x + 3y = 7$.

Add for opposite signs, subtract for the same sign. Adding $+3y$ and $+3y$ gives $6y$; it does not eliminate $y$.

Line up before you start. $y + 5 = 2x$ has the $x$ term on the wrong side. Rewrite it as $2x - y = 5$ first.

Frequently asked questions

What is the elimination method?

You add or subtract the two equations so that one pronumeral cancels out. You then solve for the other pronumeral and substitute back.

When do you add and when do you subtract the equations?

Look at the coefficients of the pronumeral you want to remove. If they have opposite signs, add. If they have the same sign, subtract.

What if no coefficients match?

Multiply one or both equations by a number so that one pair of coefficients becomes the same, for example $2y$ and $3y$ both become $6y$.

Should I use substitution or elimination?

Use substitution when a pronumeral is already the subject. Use elimination when both equations are in the form $ax + by = k$. Both give the same answer.

What does it mean if both pronumerals disappear?

If you get a false statement such as $0 = 1$, there is no solution and the lines are parallel. If you get $0 = 0$, the two equations are the same line and there are infinitely many solutions.

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Simultaneous Equations Using Substitution

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Further Applications of Simultaneous Equations

\((1) + (2):\quad 4x\)	\(=\)	\(20\)
\(x\)	\(=\)	\(5\)
\(5 - 2y\)	\(=\)	\(1 \quad \text{(in (2))}\)
\(y\)	\(=\)	\(2\)

\((1) - (2):\quad 3x\)	\(=\)	\(9\)
\(x\)	\(=\)	\(3\)
\(2(3) + 3y\)	\(=\)	\(12 \quad \text{(in (2))}\)
\(y\)	\(=\)	\(2\)

\((1) \times 3:\quad 9x + 12y\)	\(=\)	\(15 \quad (3)\)
\((2) \times 4:\quad 8x - 12y\)	\(=\)	\(36 \quad (4)\)
\((3) + (4):\quad 17x\)	\(=\)	\(51\)
\(x\)	\(=\)	\(3\)

\(9 + 4y\)	\(=\)	\(5\)
\(4y\)	\(=\)	\(-4\)
\(y\)	\(=\)	\(-1\)

\(3x - 2y\)	\(=\)	\(1 \quad (1)\)
\(\times 10:\quad 5x + 2y\)	\(=\)	\(23 \quad (2)\)

\(8x\)	\(=\)	\(24\)
\(x\)	\(=\)	\(3\)
\(2y\)	\(=\)	\(3(3) - 1 = 8\)
\(y\)	\(=\)	\(4\)

\(2p + 3d\)	\(=\)	\(22.50 \quad (1)\)
\(4p + d\)	\(=\)	\(25.00 \quad (2)\)

\((1) \times 2:\quad 4p + 6d\)	\(=\)	\(45.00 \quad (3)\)
\((3) - (2):\quad 5d\)	\(=\)	\(20.00\)
\(d\)	\(=\)	\(4.00\)
\(4p + 4\)	\(=\)	\(25\)
\(p\)	\(=\)	\(5.25\)

Simultaneous Equations Using Elimination

📖 Theory

How to solve by elimination

Common pitfalls

Frequently asked questions

Theory