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CIE IGCSE Maths: Extended Your notes

Simultaneous Equations
Contents
Simultaneous Equations

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Simultaneous Equations
Your notes
Linear Simultaneous Equations
What are linear simultaneous equations?
When there are two unknowns (say x and y) in a problem, we need two equations to be able to find
them both: these are called simultaneous equations
you solve two equations to find two unknowns, x and y
for example, 3x + 2y = 11 and 2x - y = 5
the solutions are x = 3 and y = 1
If they just have x and y in them (no x2 or y2 or xy etc) then they are linear simultaneous equations
How do I solve linear simultaneous equations by elimination?
"Elimination" completely removes one of the variables, x or y
To eliminate the x's from 3x + 2y = 11 and 2x - y = 5
Multiply every term in the first equation by 2
6x + 4y = 22
Multiply every term in the second equation by 3
6x - 3y = 15
Subtract the second result from the first to eliminate the 6x's, leaving 4y - (-3y) = 22 - 15, i.e. 7y = 7
Solve to find y (y = 1) then substitute y = 1 back into either original equation to find x (x = 3)
Alternatively, to eliminate the y's from 3x + 2y = 11 and 2x - y = 5
Multiply every term in the second equation by 2
4x - 2y = 10
Add this result to the first equation to eliminate the 2y's (as 2y + (-2y) = 0)
The process then continues as above
Check your final solutions satisfy both equations
How do I solve linear simultaneous equations by substitution?
"Substitution" means substituting one equation into the other
Solve 3x + 2y = 11 and 2x - y = 5 by substitution
Rearrange one of the equation into y = ... (or x = ...)
For example, the second equation becomes y = 2x - 5
Substitute this into the first equation (replace all y's with 2x - 5 in brackets)
3x + 2(2x - 5) = 11
Solve this equation to find x (x = 3), then substitute x = 3 into y = 2x - 5 to find y (y = 1)
Check your final solutions satisfy both equations
How do you use graphs to solve linear simultaneous equations?

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Plot both equations on the same set of axes

to do this, you can use a table of values or rearrange into y = mx + c if that helps
Find where the lines intersect (cross over) Your notes
The x and y solutions to the simultaneous equations are the x and y coordinates of the point of
intersection
e.g. to solve 2x - y = 3 and 3x + y = 4 simultaneously, first plot them both (see graph)
find the point of intersection, (2, 1)
the solution is x = 2 and y = 1

How do I solve linear simultaneous equations from worded contexts?

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Your notes

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Your notes

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Exam Tip
Your notes
Always check that your final solutions satisfy the original simultaneous equations - you will know
immediately if you've got the right solutions or not.

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Worked example
Your notes
Solve the simultaneous equations
5x + 2y = 11
4x - 3y = 18
Number the equations.

Make the y terms equal by multiplying all parts of equation (1) by 3 and all parts of equation (2) by 2.
This will give two 6y terms with different signs. The question could also be done by making the x terms
equal by multiplying all parts of equation (1) by 4 and all parts of equation (2) by 5, and subtracting the
equations.

The 6y terms have different signs, so they can be eliminated by adding equation (4) to equation (3).

Solve the equation to find x by dividing both sides by 23.

Substitute into either of the two original equations.

Solve this equation to find y.

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Your notes

Substitute x = 3 and y = - 2 into the other equation to check that they are correct

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Quadratic Simultaneous Equations

What are quadratic simultaneous equations? Your notes
When there are two unknowns (say x and y) in a problem, we need two equations to be able to find them
both: these are called simultaneous equations
If there is an x2 or y2 or xy in one of the equations then they are quadratic (or non-linear) simultaneous
equations
How do I solve quadratic simultaneous equations?
Use the method of substitution
Substitute the linear equation, y = ... (or x = ...), into the quadratic equation
Do not try to substitute the quadratic equation into the linear equation
Solve x + y2 = 25 and y - 2x = 5
2
Rearrange the linear equation into y = 2x + 5
Substitute this into the quadratic equation, replacing all y's with (2x + 5) in brackets
x2 + (2x + 5)2 = 25
Expand and solve this quadratic equation (x = 0 and x = -4)
Substitute each value of x into the linear equation, y = 2x + 5, to get their value of y
Present your solutions in a way that makes it obvious which x belongs to which y
x = 0, y = 5 or x = -4, y = -3
Check your final solutions satisfy both equations
How do you use graphs to solve quadratic simultaneous equations?
Plot both equations on the same set of axes
to do this, you can use a table of values (or, for straight lines, rearrange into y = mx + c if it helps)
Find where the lines intersect (cross over)
The x and y solutions to the simultaneous equations are the x and y coordinates of the point of
intersection
e.g. to solve y = x2 + 3x + 1 and y = 2x + 1 simultaneously, first plot them both (see graph)
find the points of intersection, (-1, -1) and (0, 1)
the solutions are x = -1 and y = -1 or x = 0 and y = 1

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Your notes

Exam Tip
If the resulting quadratic has a repeated root then the line is a tangent to the curve. If the resulting
quadratic has no roots then the line does not intersect with the curve – or you have made a
mistake!
When giving your final answer, make sure you indicate which x and y values go together. If you
don’t make this clear you can lose marks for an otherwise correct answer.
Don't make the common mistake of thinking each squared term in x2 + y2 = 25 can be square-
rooted to give x + y = 5 (they can't, the most you can do is , but you shouldn't
be making x or y the subject of this anyway!)

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Worked example
Your notes
Solve the equations
x2 + y2 = 36
x = 2y + 6
Number the equations.

There is one quadratic equation and one linear equation so this must be done by substitution.

Equation (2) is equal to so this can be eliminated by substituting it into the part for equation (1).
Substitute into equation (1).

Expand the brackets, remember that a bracket squared should be treated the same as double
brackets.

Simplify.

Rearrange to form a quadratic equation that is equal to zero.

The question does not give a specified degree of accuracy, so this can be factorised.
Take out the common factor of .

Solve to find the values of .

Let each factor be equal to 0 and solve.

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Your notes

Substitute the values of into one of the equations (the linear equation is easier) to find the values of
.

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