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Non-linear simultaneous equations - Higher

Non-linear simultaneous equations - Higher

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Tutor: Bilal

Summary

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=ax2+bx+cy=ax^2+bx+c can be solved simultaneously with a linear equation ax+by=cax+by=c by substituting the linear equation into the quadratic equation.


PROCEDURE

1.1.​​

Rearrange the linear equation for either xx or yy. Label this as equation 11, label the quadratic equation 22.​

2.2.​​

Substitute equation 11 into equation 22 to obtain a quadratic.​

3.3.​​

Solve the quadratic, there should usually be two answers.

4.4.​​

Substitute the answers from step 33 into equation 11 to find two answers for the other variable.​

5.5.​​

Check both pairs of answers in equation 22.​


Example 1

Solve simultaneously

y7x=10y=x2+4x+6y-7x=10 \\y= x^2+4x+6​​


Rearrange the linear equation and label both equations.

y=7x+101y=x2+4x+62\begin {aligned}y&=7x+10 \quad &\textcircled {1} \\y&=x^2+4x+6 \quad &\textcircled{2}\\\end {aligned}​​


Substitute equation 11 into equation 22 and solve.

7x+10=x2+4x+60=x23x40=(x4)(x+1)x=4,x=1\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 xx into equation 11 to find the answers for yy.

x=1y=7x+10y=7×1+10y=3x=4y=7x+10y=7×4+10y=38\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 22.

y=x2+4x+6x=1,y=3y=(1)2+4(1)+6=3x=4,y=38y=42+4(4)+6=38\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

x=1,y=3 or x=4,y=38\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 x2+y2=r2x^2+y^2=r^2 can be solved simultaneously with a linear equation ax+by=cax+by=c by substituting the linear equation into the circle equation.


PROCEDURE

1.1.​​

Rearrange the linear equation for either xx or yy. Label this as equation 11, label the circle equation 22.​

2.2.​​

Substitute equation 11 into equation 22 to obtain a quadratic.​

3.3.​​

Solve the quadratic, there should usually be two answers.

4.4.​​

Substitute the answers from step 33 into equation 11 to find two answers for the other variable.​

5.5.​​

Check both pairs of answers in equation 22.​


Example 2

Solve simultaneously


yx=2x2+y2=20y-x=-2 \\x^2+y^2=20​​


Rearrange the linear equation and label both equations.

y=x21x2+y2=202\begin {aligned}y&=x-2 \quad &\textcircled {1} \\x^2+y^2&=20 \quad &\textcircled{2}\\\end {aligned}​​


Substitute equation 11 into equation 22 and solve.

x2+(x2)2=20x2+(x2)(x2)=20x2+x22x2x+4=202x24x+4=202x24x16=0x22x8=0(x+2)(x4)=0x=2,x=4\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 xx into equation 11 to find the answers for yy.

x=2y=x2y=22y=4x=4y=x2y=42y=2\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 22.

x2+y2=20x=2,y=4(2)2+(4)2=20x=4,y=242+22=20\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

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


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FAQs - Frequently Asked Questions

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