Simultaneous Equations | IEB Grade 11 Maths

In Grade 11 you solve a linear and a quadratic equation together. Each solution pair is a point where a straight line meets a curve, so the algebra and the graphs tell the same story.

The substitution method

Make one variable the subject of the linear equation.
Substitute into the quadratic equation.
Solve the quadratic.
Substitute each answer into the linear equation for the other variable.

Example. $y = 2 x - 3$ and $y = x^{2} - 4 x + 5$

$x^{2} - 4 x + 5 = 2 x - 3 \Rightarrow x^{2} - 6 x + 8 = 0 \Rightarrow x = 2 or x = 4$

The solutions are $(2; 1)$ and $(4; 5)$ .

Choose the variable that avoids fractions when you make it the subject.

Solutions as points of intersection

Each solution pair $(x; y)$ is a point on both graphs.

Two solutions: the line cuts the curve twice.
One repeated solution: the line touches the curve (a tangent).
No real solutions: the line misses the curve.

Example. $x + y = 5$ and $x y = 6$ : substituting gives $x^{2} - 5 x + 6 = 0$ , so $(2; 3)$ and $(3; 2)$ . The line $y = 5 - x$ cuts the hyperbola $y = \frac{6}{x}$ at these two points.

Graphical solutions

Sketch both graphs on the same axes. The solutions are the coordinates of the intersection points.

Graphical solutions are useful to check how many solutions to expect, but they are only as accurate as the drawing. Use algebra for exact values.

Example. Two numbers differ by 3, and the sum of their squares is 89. Then $x - y = 3$ and $x^{2} + y^{2} = 89$ . Graphically, the line $y = x - 3$ cuts the circle $x^{2} + y^{2} = 89$ twice, so there are two pairs: $(8; 5)$ and $(- 5; - 8)$ .

Key terms

Simultaneous equations: Two or more equations that must be true at the same time.
Substitution: Replacing a variable with an equal expression.
Point of intersection: A point that lies on both graphs.

Common mistakes

Substituting back into the quadratic, which can give extra wrong pairs.
Giving $x$ -values without matching $y$ -values.
Mixing up which $y$ goes with which $x$ .
Expanding brackets like $(2 y - 1)^{2}$ incorrectly.

Practice questions

Reveal steps one at a time, then tick the marks you earned.

Question 1

5 marks total

routine

Solve simultaneously: $y = 2 x - 3$ and $y = x^{2} - 4 x + 5$

Worked solution

Hidden - try the question first

Solution

$x^{2} - 4 x + 5 = 2 x - 3$
$x^{2} - 6 x + 8 = 0 \Rightarrow (x - 2) (x - 4) = 0$
$x = 2$ or $x = 4$
$x = 2$ : $y = 1$
$x = 4$ : $y = 5$

Question 2

6 marks total

routine

Solve simultaneously: $x + y = 5$ and $x y = 6$

Worked solution

Hidden - try the question first

Solution

$x^{2} - 4 x + 5 = 2 x - 3$
$x (5 - x) = 6$
$x^{2} - 5 x + 6 = 0$
$(x - 2) (x - 3) = 0$
$x = 2$ , $y = 3$
$x = 3$ , $y = 2$

Question 3

6 marks total

complex

Solve simultaneously: $2 y - x = 1$ and $x^{2} + y^{2} - x y = 7$

Worked solution

Hidden - try the question first

Solution

$x^{2} - 4 x + 5 = 2 x - 3$
$(2 y - 1)^{2} + y^{2} - (2 y - 1) y = 7$
$3 y^{2} - 3 y - 6 = 0 \Rightarrow y^{2} - y - 2 = 0$
$(y - 2) (y + 1) = 0$
$y = 2$ , $x = 3$
$y = - 1$ , $x = - 3$

Question 4

8 marks total

problem-solving

Two numbers differ by 3. The sum of their squares is 89.

(a) Write down two equations and solve them to find the numbers. (5)

(b) Interpret your answer graphically. (3)

Worked solution

Hidden - try the question first

Solution

(a) $x - y = 3$ and $x^{2} + y^{2} = 89$
$(y + 3)^{2} + y^{2} = 89$
$2 y^{2} + 6 y - 80 = 0 \Rightarrow y^{2} + 3 y - 40 = 0$
$(y + 8) (y - 5) = 0$
The numbers are 8 and 5, or $- 5$ and $- 8$
(b) $x - y = 3$ is a straight line
$x^{2} + y^{2} = 89$ is a circle centred at the origin
The line cuts the circle at $(8; 5)$ and $(- 5; - 8)$

FAQs

Which variable should I make the subject?

Choose one from the linear equation that has a coefficient of 1, to avoid fractions.

How many solutions can a line and a parabola have?

Two, one (if the line is a tangent) or none.

Can I solve simultaneous equations by graphing?

Yes, to see how many solutions there are, but exact values need algebra.

Simultaneous equations (one linear, one quadratic)

The substitution method

Solutions as points of intersection

Graphical solutions

Key terms

Common mistakes

Practice questions

FAQs

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