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13 Nonlinear relationships
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Stage 5.1-3

13.10 Nonlinear simultaneous equations

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Nonlinear simultaneous equations

We solve non-linear simultaneous equations the same way as we solve linear simultaneous equations. The most significant difference is that non-linear simultaneous equations can have more than one solution.

Examples

Example 1

Consider the system of equations: \begin{aligned} x^2 +y^2&=10\\ x-y&=2 \end{aligned}

a

(x,y) is a solution to the system of equations. First solve for x.

Worked Solution
Create a strategy

Use the substitution method.

Apply the idea

First lets label the equations to make them easier to refer to:

\displaystyle x^2 +y^2\displaystyle =\displaystyle 10Equation 1
\displaystyle x-y\displaystyle =\displaystyle 2Equation 2

We can make y the subject of Equation 2 and then substitute this into Equation 1 to solve for x.

\displaystyle x\displaystyle =\displaystyle 2+yAdd y to both sides
\displaystyle y\displaystyle =\displaystyle x-2Subtract 2 from both sides
\displaystyle x^2+(x-2)^2\displaystyle =\displaystyle 10Substitute y=x-2 into Equation 1
\displaystyle x^2+x^2-4x+4\displaystyle =\displaystyle 10Expand the brackets
\displaystyle 2x^2-4x+4\displaystyle =\displaystyle 10Add like terms
\displaystyle 2x^2-4x-6\displaystyle =\displaystyle 0Subtract 10 from both sides
\displaystyle x^2-2x-3\displaystyle =\displaystyle 0Divide both sides by 2
\displaystyle (x-3)(x+1)\displaystyle =\displaystyle 0Factorise the quadratic
\displaystyle x\displaystyle =\displaystyle -1,\,3Solve for x
Reflect and check

We see that we get 2 values of x since one of our equations is non-linear.

b

Complete the statement:

Therefore, the solutions are (3,⬚) and (-1,⬚).

Worked Solution
Create a strategy

Substitute each x-value from part (a) into one of the equations to find the corresponsing y-values.

Apply the idea

The simplest equation is Equation 2, x-y=2, so we will substitute x=-1 and x=3 into that equation.

\displaystyle -1-y\displaystyle =\displaystyle 2Substitute x=-1 into Equation 2
\displaystyle -y\displaystyle =\displaystyle 3Add 1 to both sides
\displaystyle y\displaystyle =\displaystyle -3Multiply both sides by -1
\displaystyle 3-y\displaystyle =\displaystyle 2Substitute x=3 into Equation 2
\displaystyle -y\displaystyle =\displaystyle -1Subtract 3 from both sides
\displaystyle y\displaystyle =\displaystyle 1Multiply both sides by -1

Therefore, the solutions are (3,1) and (-1,-3).

Example 2

Consider the hyperbola xy=2 and the straight line y=x+6

a

The graph of the hyperbola xy=2 is given. On the same set of axes, graph y=x+6.

-8
-6
-4
-2
2
4
6
8
x
-8
-6
-4
-2
2
4
6
8
y
Worked Solution
Create a strategy

Draw the line by finding the intercepts.

Apply the idea
-8
-6
-4
-2
2
4
6
8
x
-8
-6
-4
-2
2
4
6
8
y

For the line y=x+6, when x=0: y=0+6=6. So the y-intercept is (0,6).

When y=0: 0=x+6, so x=-6. So the x-intercept is (-6,0)).

By plotting these points we can draw the graph of the line as shown.

b

How many points of intersection are there between the hyperbola xy=2 and the line y=x+6?

Worked Solution
Apply the idea

By considering the graph from part (a), we can see that the line and the hyperbola instersect twice. So there are 2 points of intersection.

Idea summary

We solve non-linear simultaneous equations the same way as we solve linear simultaneous equations. Substitution and graphical methods are the most appropriate methods.

The most significant difference is that non-linear simultaneous equations can have more than one solution.

Outcomes

MA5.3-7NA

solves complex linear, quadratic, simple cubic and simultaneous equations, and rearranges literal equations

MA5.3-9NA

sketches and interprets a variety of non-linear relationships

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