OpenStax-CNX module: m38854
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Simultaneous equations: graphical
solution
∗
Free High School Science Texts Project
This work is produced by OpenStax-CNX and licensed under the
†
Creative Commons Attribution License 3.0
1 Introduction
In grade 10, you learnt how to solve sets of simultaneous equations where both equations were linear (i.e.
had the highest power equal to 1). In this chapter, you will learn how to solve sets of simultaneous equations
where one is linear and one is quadratic. As in Grade 10, the solution will be found both algebraically and
graphically.
The only dierence between a system of linear simultaneous equations and a system of simultaneous
equations with one linear and one quadratic equation, is that the second system will have at most two
solutions.
An example of a system of simultaneous equations with one linear equation and one quadratic equation
is:
y − 2x = −4
x2 + y = 4
(1)
2 Graphical Solution
The method of graphically nding the solution to one linear and one quadratic equation is identical to
systems of linear simultaneous equations.
2.1 Method: Graphical solution to a system of simultaneous equations with one linear and one
quadratic equation
1. Make y the subject of each equation.
2. Draw the graphs of each equation as dened above.
3. The solution of the set of simultaneous equations is given by the intersection points of the two graphs.
For this example, making y the subject of each equation, gives:
y = 2x − 4
y = 4 − x2
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OpenStax-CNX module: m38854
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Plotting the graph of each equation, gives a straight line for the rst equation and a parabola for the second
equation.
Figure 1
The parabola and the straight line intersect at two points: (2,0) and (-4,-12). Therefore, the solutions to
the system of equations in (1) is x = 2, y = 0 and x = −4, y = 12
Exercise 1: Simultaneous Equations
(Solution on p. 3.)
Solve graphically:
y − x2 + 9
=
0
y + 3x − 9
=
0
(3)
2.1.1 Graphical Solution
Solve the following systems of equations graphically. Leave your answer in surd form, where appropriate.
1.
2.
3.
4.
5.
b2 − 1 − a = 0, a + b − 5 = 0
x + y − 10 = 0, x2 − 2 − y = 0
6 − 4x − y = 0, 12 − 2x2 − y = 0
x + 2y − 14 = 0, x2 + 2 − y = 0
2x + 1 − y = 0, 25 − 3x − x2 − y = 0
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OpenStax-CNX module: m38854
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Solutions to Exercises in this Module
Solution to Exercise (p. 2)
Step 1. For the rst equation:
y − x2 + 9
y
=
=
0
2
x −9
(4)
and for the second equation:
y + 3x − 9
y
=
0
= −3x + 9
Step 2.
Figure 2
Step 3. The graphs intersect at (−6, 27) and at (3, 0).
Step 4. The rst solution is x = −6 and y = 27. The second solution is x = 3 and y = 0.
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