Simultaneous Equations | IEB Grade 12 Maths

Q: Which equation should I substitute back into?

The linear equation. It gives exactly one $y$ for each $x$, so you cannot pick up false pairs.

Simultaneous equations with one linear and one quadratic equation give up to two solution pairs. Each pair is a point where a line meets a curve, which links this algebra to graphs, circles and tangents later in the papers.

Solving one linear and one quadratic equation

Make $x$ or $y$ the subject of the linear equation.
Substitute into the quadratic equation.
Solve the quadratic.
Substitute each answer back into the linear equation to find the other variable.

Example. $y = x + 1$ and $x^{2} + y^{2} = 25$

$x^{2} + (x + 1)^{2} = 25 \Rightarrow 2 x^{2} + 2 x - 24 = 0 \Rightarrow (x + 4) (x - 3) = 0$

$x = 3$ , $y = 4$ or $x = - 4$ , $y = - 3$ .

Choose the variable that avoids fractions when you rearrange. Give answers as pairs.

Interpreting solutions as points of intersection

Each solution pair is a point where the two graphs meet.

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

The discriminant of the quadratic you get after substituting decides which case you have.

Example. $y = x^{2} - 4 x + 3$ and $y = 2 x - 6$

$x^{2} - 6 x + 9 = 0 \Rightarrow (x - 3)^{2} = 0$

One repeated solution, $x = 3$ , so the line touches the parabola at $(3; 0)$ .

For the line $y = 2 x + k$ to cut the parabola twice, $x^{2} - 6 x + 3 - k = 0$ needs $Δ > 0$ : $36 - 4 (3 - k) > 0$ , so $k > - 6$ .

Key terms

Simultaneous equations: Equations that must be true at the same time.
Solution pair: Values of $x$ and $y$ that satisfy both equations.
Tangent: A line that meets a curve at exactly one point without crossing there.

Common mistakes

Substituting back into the quadratic equation, which can give extra false $y$ -values.
Giving only $x$ -values without the matching $y$ -values.
Expanding $(x + 1)^{2}$ as $x^{2} + 1$ .
Mixing up the pairs, such as matching $x = 3$ with the $y$ -value for $x = - 4$ .

Practice questions

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

Question 1

5 marks total

routine

Solve simultaneously: $y = x + 1$ and $x^{2} + y^{2} = 25$

Worked solution

Hidden - try the question first

Solution

$x^{2} + (x + 1)^{2} = 25$
$2 x^{2} + 2 x - 24 = 0 \Rightarrow x^{2} + x - 12 = 0$
$(x + 4) (x - 3) = 0$
$x = 3$ , $y = 4$
$x = - 4$ , $y = - 3$

Question 2

6 marks total

routine

Solve simultaneously: $2 x + y = 3$ and $x^{2} - x y + y^{2} = 7$

Worked solution

Hidden - try the question first

Solution

$x^{2} + (x + 1)^{2} = 25$
$x^{2} - x (3 - 2 x) + (3 - 2 x)^{2} = 7$
$7 x^{2} - 15 x + 2 = 0$
$(7 x - 1) (x - 2) = 0$
$x = 2$ , $y = - 1$
$x = \frac{1}{7}$ , $y = \frac{19}{7}$

Question 3

6 marks total

complex

Given $f (x) = x^{2} - 4 x + 3$ and $g (x) = 2 x - 6$ .

(a) Solve $f (x) = g (x)$ . (3)

(b) What does your answer tell you about the graphs of $f$ and $g$ ? (1)

Worked solution

Hidden - try the question first

Solution

(a) $x^{2} - 4 x + 3 = 2 x - 6$
$x^{2} - 6 x + 9 = 0 \Rightarrow (x - 3)^{2} = 0$
$x = 3$
(b) The line is a tangent to the parabola at $(7 x - 1) (x - 2) = 0$
(c) $x^{2} - 6 x + 3 - k = 0$ needs $Δ > 0$ : $36 - 4 (3 - k) > 0$
$k > - 6$

Question 4

9 marks total

problem-solving

A rectangular garden has a perimeter of 34 m and a diagonal of 13 m.

(a) Determine the length and breadth of the garden. (5)

(b) Calculate its area. (1)

Worked solution

Hidden - try the question first

Solution

(a) $x^{2} - 4 x + 3 = 2 x - 6$ and $x^{2} + y^{2} = 169$
$x^{2} + (17 - x)^{2} = 169$
$2 x^{2} - 34 x + 120 = 0 \Rightarrow x^{2} - 17 x + 60 = 0$
$y = - 1$
The garden is 12 m by 5 m
(b) $12 \times 5 = 60$ m²
(c) $x^{2} + (17 - x)^{2} = 144 \Rightarrow 2 x^{2} - 34 x + 145 = 0$
$Δ = (- 34)^{2} - 4 (2) (145) = 1156 - 1160 = - 4$
$x^{2} + (x + 1)^{2} = 25 \Rightarrow 2 x^{2} + 2 x - 24 = 0 \Rightarrow (x + 4) (x - 3) = 0$ , so there are no real dimensions

FAQs

Which equation should I substitute back into?

The linear equation. It gives exactly one $y$ for each $x$ , so you cannot pick up false pairs.

What does it mean if the quadratic has one repeated root?

The line touches the curve at one point, so it is a tangent.

Will simultaneous equations come up in analytical geometry too?

Yes. Finding where a line meets a circle uses exactly the same method.

Simultaneous equations (linear and quadratic)

Solving one linear and one quadratic equation

Interpreting solutions as points of intersection

Key terms

Common mistakes

Practice questions

FAQs

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