Simultaneous equations: Grade 12 Mathematics

By the end of this lesson you should be able to

Solve pairs of linear simultaneous equations by elimination and by substitution
Solve a linear equation and a quadratic equation together
Form simultaneous equations from word problems

What are simultaneous equations?

Simultaneous equations are two or more equations that must be true at the same time. When there are two unknowns, such as $x$ and $y$ , you need two equations to find them both.

Graphically, each linear equation is a straight line. The solution is the point where the lines cross. That picture explains the three possible outcomes:

One solution: the lines cross at a single point.
No solution: the lines are parallel, so they never meet.
Infinitely many solutions: both equations describe the same line.

Method 1: elimination

Make the coefficients of one unknown equal (ignoring signs), then add or subtract the equations to remove it.

If the equal terms have different signs, add the equations.
If the equal terms have the same sign, subtract the equations.

Example 1. Solve $3x + 2y = 16$ and $5x - 2y = 16$ .

The $y$ terms are $+2y$ and $-2y$ , so add the equations:

8x = 32 \quad\Rightarrow\quad x = 4

Substitute into the first equation: $3(4) + 2y = 16$ , so $2y = 4$ and $y = 2$ .

Check in the second equation: $5(4) - 2(2) = 16$ . ✓

Example 2. Solve $2x + 3y = 7$ and $3x - 2y = 4$ .

No coefficients match, so make the $y$ coefficients equal. Multiply the first equation by $2$ and the second by $3$ :

4x + 6y = 14 \qquad 9x - 6y = 12

The $y$ terms have opposite signs, so add:

13x = 26 \quad\Rightarrow\quad x = 2

Substitute into $2x + 3y = 7$ : $4 + 3y = 7$ , so $y = 1$ .

Check in the second original equation: $3(2) - 2(1) = 4$ . ✓

Method 2: substitution

Use this method when one equation already gives one unknown in terms of the other.

Example 3. Solve $y = 2x - 1$ and $3x + y = 9$ .

Replace $y$ in the second equation with $2x - 1$ :

3x + (2x - 1) = 9 \quad\Rightarrow\quad 5x = 10 \quad\Rightarrow\quad x = 2

Then $y = 2(2) - 1 = 3$ .

A linear equation and a quadratic equation

When one equation is quadratic, such as a circle $x^2 + y^2 = 25$ , always substitute the linear equation into the quadratic one. This leaves a quadratic equation in one unknown, which gives two values. Then find the matching $y$ for each $x$ .

Example 4. Solve $y = x + 1$ and $x^2 + y^2 = 25$ .

Substitute $y = x + 1$ :

x^2 + (x + 1)^2 = 25 \quad\Rightarrow\quad 2x^2 + 2x + 1 = 25 \quad\Rightarrow\quad x^2 + x - 12 = 0

(x + 4)(x - 3) = 0 \quad\Rightarrow\quad x = -4 \text{ or } x = 3

Use $y = x + 1$ to find each $y$ : when $x = -4$ , $y = -3$ ; when $x = 3$ , $y = 4$ .

The solutions are $(-4, -3)$ and $(3, 4)$ . Check: $16 + 9 = 25$ and $9 + 16 = 25$ . ✓

Word problems

Choose letters for the two unknown quantities, write two equations from the information, then solve.

Example 5. Three pens and two books cost K34. Two pens and one book cost K19. Find the cost of one pen and one book.

Let a pen cost K $p$ and a book cost K $b$ .

3p + 2b = 34 \qquad 2p + b = 19

From the second equation, $b = 19 - 2p$ . Substitute into the first:

3p + 2(19 - 2p) = 34 \quad\Rightarrow\quad 3p + 38 - 4p = 34 \quad\Rightarrow\quad p = 4

Then $b = 19 - 8 = 11$ . A pen costs K4 and a book costs K11.

Check: $3(4) + 2(11) = 34$ . ✓

Common mistakes

Subtracting when you should add. When the terms to eliminate have opposite signs, add. Mixing this up is the most common error in elimination.

Multiplying only one term. When you multiply an equation, multiply every term, including the number on the right-hand side.

Stopping after finding one unknown. A solution is a pair of values. Always find both, then check them in the original equations.

Substituting into the wrong place in a linear-quadratic pair. Substitute the linear equation into the quadratic one, not the other way round. Remember that $(x + 1)^2 = x^2 + 2x + 1$ , not $x^2 + 1$ .

Practice questions

Solve $x + y = 10$ and $x - y = 4$ .
Solve $2x + y = 11$ and $x - y = 1$ .
Solve $3x - 2y = 1$ and $2x + 3y = 18$ .
Solve $y = 3x - 2$ and $2x + y = 13$ .
Solve $5x + 2y = 1$ and $3x + 4y = -5$ .
Solve $y = x - 2$ and $x^2 + y^2 = 20$ .
Two adults and three children pay K130 to enter a show. One adult and two children pay K75. Find the price of an adult ticket and of a child ticket.
How many solutions do $2x + y = 4$ and $4x + 2y = 9$ have? Give a reason.

Show answers

Adding gives $2x = 14$ , so $x = 7$ and $y = 3$ .
Adding gives $3x = 12$ , so $x = 4$ and $y = 3$ .
Multiply the first by $3$ and the second by $2$ : $9x - 6y = 3$ and $4x + 6y = 36$ . Adding gives $13x = 39$ , so $x = 3$ and $y = 4$ .
$2x + 3x - 2 = 13$ , so $5x = 15$ , giving $x = 3$ and $y = 7$ .
Multiply the first by $2$ : $10x + 4y = 2$ . Subtract the second equation: $7x = 7$ , so $x = 1$ and $y = -2$ .
$x^2 + (x - 2)^2 = 20$ gives $x^2 - 2x - 8 = 0$ , so $(x - 4)(x + 2) = 0$ . The solutions are $(4, 2)$ and $(-2, -4)$ .
Let an adult ticket cost K $a$ and a child ticket K $c$ . Then $2a + 3c = 130$ and $a + 2c = 75$ . From the second, $a = 75 - 2c$ , so $150 - 4c + 3c = 130$ , giving $c = 20$ and $a = 35$ . An adult ticket is K35 and a child ticket is K20.
None. Doubling the first equation gives $4x + 2y = 8$ , which contradicts $4x + 2y = 9$ . The lines are parallel, so they never meet.

Summary

Elimination works well when both equations are in the form $ax + by = c$ . Substitution works well when one unknown is already isolated.
Add the equations when the signs of the terms to eliminate differ, and subtract when they are the same.
For a linear and a quadratic equation, substitute the linear one into the quadratic, solve, then find both $y$ values.
Always check your pair of values in the original equations.

Now practise this topic

Real questions show you what still needs work. Try a past paper under timed conditions, or take a short quiz.

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