# Graphical solution

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## 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 difference 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:

$\begin{array}{c}\hfill y-2x=-4\\ \hfill {x}^{2}+y=4\end{array}$

## Graphical solution

The method of graphically finding the solution to one linear and one quadratic equation is identical to systems of linear simultaneous equations.

## 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 defined 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:

$\begin{array}{c}\hfill y=2x-4\\ \hfill y=4-{x}^{2}\end{array}$

Plotting the graph of each equation, gives a straight line for the first equation and a parabola for the second equation.

The parabola and the straight line intersect at two points: (2,0) and (-4,-12). Therefore, the solutions to the system of equations in [link] is $x=2,y=0$ and $x=-4,y=12$

Solve graphically:

$\begin{array}{ccc}\hfill y-{x}^{2}+9& =& 0\hfill \\ \hfill y+3x-9& =& 0\hfill \end{array}$
1. For the first equation:

$\begin{array}{ccc}\hfill y-{x}^{2}+9& =& 0\hfill \\ \hfill y& =& {x}^{2}-9\hfill \end{array}$

and for the second equation:

$\begin{array}{ccc}\hfill y+3x-9& =& 0\hfill \\ \hfill y& =& -3x+9\hfill \end{array}$
2. The graphs intersect at $\left(-6,27\right)$ and at $\left(3,0\right)$ .

3. The first solution is $x=-6$ and $y=27$ . The second solution is $x=3$ and $y=0$ .

## Graphical solution

Solve the following systems of equations graphically. Leave your answer in surd form, where appropriate.

1. ${b}^{2}-1-a=0,a+b-5=0$
2. $x+y-10=0,{x}^{2}-2-y=0$
3. $6-4x-y=0,12-2{x}^{2}-y=0$
4. $x+2y-14=0,{x}^{2}+2-y=0$
5. $2x+1-y=0,25-3x-{x}^{2}-y=0$

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