Linear Simultaneous Equations (Cambridge International IGCSE Further Maths)
Elimination Method
What are simultaneous linear equations?
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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
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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
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If they just have x and y in them (no x2 or y2 or xy etc) then they are linear simultaneous equations
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Solving a pair of simultaneous equations means finding pairs of values that make both equations true at the same time
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A linear equation in two unknowns will produce a straight line if you graph it... linear = line
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A pair of simultaneous equations will produce lines that will cross each other (if there is a solution!)
How do I use elimination to solve simultaneous linear equations?
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"Elimination" completely removes one of the variables, x or y
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Begin by multiplying one (or both) of the equations by a constant (or constants)to get the numbers in front of one of the unknowns to match
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For example 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
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If the matching numbers have the same sign, then subtract one equation from the other
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If the matching numbers have different signs then add the equations together
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Subtract the second result from the first to eliminate the 6x's
- 4y - (-3y) = 22 - 15
- 7y = 7
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Solve the new equation to find the value of one of the unknowns
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Solve to find y
- y = 1
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Substitute the value into one of the original equations and solve to find the value of the other unknown
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Substitute y = 1 back into either original equation
- 3*x *+ 2(1) = 11
- x = 3
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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
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Check your final solutions satisfy both equations
- 3(3) + 2(1) = 11 and 2(3) - (1) = 5
How do I solve linear simultaneous equations from worded contexts?


Exam Hint
- Don't skip the checking step (it only takes a few seconds) – there are many places to go wrong when solving simultaneous equations!
- Mishandling minus signs is probably the single biggest cause of student error in simultaneous equations questions
Worked Example
Substitution Method
How do I use substitution to solve simultaneous linear equations?
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"Substitution" means substituting one equation into the other
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Rearrange one of the equations to make one of the unknowns the subject
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To solve 3x + 2y = 11 and 2x - y = 5 by substitution
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Rearrange one of the equation into y = ... (or x = ...)
- For example, the second equation becomes y = 2x - 5
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Substitute the expression found for x or y into the equation not used to rearrange
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Replace all y's with 2x - 5 in brackets
- 3x + 2(2x - 5) = 11
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Solve the new equation to find the value of one of the unknowns
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Solve this equation to find x
- x = 3
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Substitute the value found for x or y into the rearranged equation from to find the value of the other unknown
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Substitute x = 3 into y = 2x - 5 to find y
- y = 2(3) - 5
- y = 1
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Check your final solutions satisfy both equations
- 3(3) + 2(1) = 11 and 2(3) - (1) = 5
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
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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
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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

Exam Hint
- Although elimination will always work to solve simultaneous linear equations, sometimes substitution can be easier and quicker
- Knowing both methods can help you a lot in the exam (plus you will need substitution to solve quadratic simultaneous equations)
Worked Example
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