2.5 Simultaneous equations by elimination and substitution — Practise with help

Work each one first, closed book. Before you start, look at the two equations and decide: are they both linear, or is one of them a curve? That choice decides your method.

Warm-up: remember

Solve $2x + 1 = 9$ .
Expand $(x + 3)^2$ .
Factorise $x^2 - 4x - 12$ .

Answers

$2x = 8$ , so $x = 4$ .
$(x + 3)^2 = x^2 + 6x + 9$ .
$x^2 - 4x - 12 = (x - 6)(x + 2)$ .

If question 3 was hard, revisit factor pairs: you need two numbers that multiply to $-12$ and add to $-4$ .

Practise with help

1. Solve $x + y = 7$ and $x - y = 3$ by elimination.

Hint

The coefficients of $y$ are $+1$ and $-1$ , so adding the equations removes $y$ .

Next step

Add the two equations to get $2x = 10$ .

Full working

Add the equations:

(x + y) + (x - y) = 7 + 3, \qquad 2x = 10, \qquad x = 5.

Substitute into $x + y = 7$ :

5 + y = 7, \qquad y = 2.

The solution is $(5, 2)$ .

2. Solve $4x + 3y = 18$ and $2x - y = 4$ by elimination.

Hint

Scale the second equation so that the $y$ terms match in size. Multiplying by $3$ gives $6x - 3y = 12$ .

Next step

Add $4x + 3y = 18$ and $6x - 3y = 12$ to remove $y$ .

Full working

Multiply the second equation by $3$ :

6x - 3y = 12.

Add the two equations:

(4x + 3y) + (6x - 3y) = 18 + 12, \qquad 10x = 30, \qquad x = 3.

Substitute into $2x - y = 4$ :

6 - y = 4, \qquad y = 2.

The solution is $(3, 2)$ .

3. Solve $y = x - 1$ and $x^2 + y^2 = 25$ by substitution.

Hint

The first equation already gives $y$ in terms of $x$ . Replace $y$ in the second equation with $x - 1$ .

Next step

Substitute to get $x^2 + (x - 1)^2 = 25$ , expand, and collect everything on the left.

Full working

Substitute:

x^2 + (x - 1)^2 = 25.

Expand and collect:

2x^2 - 2x + 1 = 25, \qquad 2x^2 - 2x - 24 = 0, \qquad x^2 - x - 12 = 0.

Factorise:

(x - 4)(x + 3) = 0,

so $x = 4$ or $x = -3$ . Then $y = x - 1$ gives $y = 3$ or $y = -4$ . The solutions are $(4, 3)$ and $(-3, -4)$ .

4. Solve $y = x^2 + 1$ and $y = 3x - 1$ .

Hint

Both expressions equal $y$ , so set them equal to each other and rearrange.

Next step

Write $x^2 + 1 = 3x - 1$ , then move every term to the left to get $x^2 - 3x + 2 = 0$ .

Full working

Set the right sides equal:

x^2 + 1 = 3x - 1, \qquad x^2 - 3x + 2 = 0, \qquad (x - 1)(x - 2) = 0.

So $x = 1$ or $x = 2$ . Use $y = 3x - 1$ to find each $y$ : when $x = 1$ , $y = 2$ ; when $x = 2$ , $y = 5$ . The solutions are $(1, 2)$ and $(2, 5)$ .

5. Solve $x + y = 6$ and $xy = 8$ .

Hint

Make $y$ the subject of the linear equation, then substitute into $xy = 8$ .

Next step

With $y = 6 - x$ , substitute to get $x(6 - x) = 8$ and rearrange to a quadratic in $x$ .

Full working

From $x + y = 6$ , $y = 6 - x$ . Substitute:

x(6 - x) = 8, \qquad 6x - x^2 = 8, \qquad x^2 - 6x + 8 = 0.

Factorise:

(x - 2)(x - 4) = 0,

so $x = 2$ or $x = 4$ . Then $y = 6 - x$ gives $(2, 4)$ and $(4, 2)$ .

6. The line $y = 3x - 2$ and the curve $y = x^2$ may meet twice, once, or not at all. Solve the equations and state what happens.

Hint

Set $x^2 = 3x - 2$ and look at the discriminant of the resulting quadratic.

Next step

Rearrange to $x^2 - 3x + 2 = 0$ and work out $b^2 - 4ac$ .

Full working

Set the expressions equal:

x^2 = 3x - 2, \qquad x^2 - 3x + 2 = 0.

The discriminant is $(-3)^2 - 4 \times 1 \times 2 = 1 > 0$ , so there are two distinct solutions. Factorise as $(x - 1)(x - 2) = 0$ , so $x = 1$ or $x = 2$ , giving the points $(1, 1)$ and $(2, 4)$ . The line and the curve meet twice.