Non-linear simultaneous equations

In a nutshell

Simultaneous equations can be solved where one of the equations is non-linear. This means that the one or both variables have powers. Questions will involve solving a pair of simultaneous equations where one equation is quadratic and the other is linear, or where one is the equation of a circle and the other is linear. Non-linear simultaneous equations can be solved using the substitution method.

Solve simultaneous equations with quadratic and linear equations

A quadratic equation of the form $y=ax^2+bx+c$ can be solved simultaneously with a linear equation $ax+by=c$ by substituting the linear equation into the quadratic equation.

PROCEDURE

$1.$	Rearrange the linear equation for either $x$ or $y$ . Label this as equation $1$ , label the quadratic equation $2$ .
$2.$	Substitute equation $1$ into equation $2$ to obtain a quadratic.
$3.$	Solve the quadratic, there should usually be two answers.
$4.$	Substitute the answers from step $3$ into equation $1$ to find two answers for the other variable.
$5.$	Check both pairs of answers in equation $2$ .

Example 1

Solve simultaneously

$y-7x=10 \\y= x^2+4x+6$

Rearrange the linear equation and label both equations.

$\begin {aligned}y&=7x+10 \quad &\textcircled {1} \\y&=x^2+4x+6 \quad &\textcircled{2}\\\end {aligned}$

Substitute equation $1$ into equation $2$ and solve.

$\begin {aligned}7x+10 &= x^2+4x+6 \\0 &= x^2-3x-4 \\0 &= (x-4)(x+1) \\\end {aligned}\\x= 4, x=-1$

Substitute these answers for $x$ into equation $1$ to find the answers for $y$ .

$\begin{aligned}x=-1 \\y = 7x+10 \\y = 7 \times -1 +10 \\y = 3 \\\\x=4y = 7x+10 \\y = 7 \times 4 +10 \\y = 38 \\\end{aligned}$

Check both pairs of answers in equation $2$ .

$\begin {aligned}y&=x^2+4x+6 \\\\x=-1, y=3 \\y &= (-1)^2+4(-1)+6 = 3 \\\\x=4, y=38 \\y &= 4^2+4(4)+6 = 38 \\\end {aligned}$

The answers are

$\underline{x=-1,y=3 \space or \space x=4,y=38}$

Solve simultaneous equations with circle and linear equations

A circle equation of the form $x^2+y^2=r^2$ can be solved simultaneously with a linear equation $ax+by=c$ by substituting the linear equation into the circle equation.

PROCEDURE

$1.$	Rearrange the linear equation for either $x$ or $y$ . Label this as equation $1$ , label the circle equation $2$ .
$2.$	Substitute equation $1$ into equation $2$ to obtain a quadratic.
$3.$	Solve the quadratic, there should usually be two answers.
$4.$	Substitute the answers from step $3$ into equation $1$ to find two answers for the other variable.
$5.$	Check both pairs of answers in equation $2$ .

Example 2

Solve simultaneously

$y-x=-2 \\x^2+y^2=20$

Rearrange the linear equation and label both equations.

$\begin {aligned}y&=x-2 \quad &\textcircled {1} \\x^2+y^2&=20 \quad &\textcircled{2}\\\end {aligned}$

Substitute equation $1$ into equation $2$ and solve.

$\begin {aligned}x^2 + (x-2)^2 &= 20\\x^2 + (x-2)(x-2) &= 20 \\x^2+x^2-2x-2x+4 &=20 \\2x^2-4x+4&=20 \\2x^2 - 4x -16 &=0 \\x^2-2x-8 &=0 \\(x+2)(x-4) &=0 \\\end {aligned}\\x= -2, x=4$

Substitute these answers for $x$ into equation $1$ to find the answers for $y$ .

$\begin{aligned}x=-2 \\y &= x-2 \\y &= -2-2 \\y &= -4 \\\\x=4\\y &= x-2 \\y &= 4-2 \\y &= 2 \\\end{aligned}$

Check both pairs of answers in equation $2$ .

$\begin {aligned}x^2+y^2&=20 \\\\x=-2, y=-4 \\(-2)^2+(-4)^2&=20 \\\\x=4, y=2 \\4^2+2^2&=20 \\\end {aligned}$

The answers are

$\underline{x=-2,y=-4 \space or \space x=4,y=2}$