Linear Simultaneous Equations (Cambridge International IGCSE Mathematics)

Linear simultaneous equations

What are linear simultaneous equations?

• When there are two unknowns (x and y), we need two equations to find them both

  • For example, 3x + 2y = 11 and 2x - y = 5

    • The values that work are x = 3 and y = 1

• These are called linear simultaneous equations

  • Linear because there are no terms like x² or y²

How do I solve linear simultaneous equations by elimination?

• Elimination removes one of the variables, x or y

• To eliminate the x's from 3x + 2y = 11 and 2x - y = 5, make the number in front of the x (the coefficient) in both equations the same (the sign may be different)

  • Multiply every term in the first equation by 2

    • 6x + 4y = 22

  • Multiply every term in the second equation by 3

    • 6x - 3y = 15

  • Subtracting the second equation from the first eliminates x

    • When the sign in front of the term you want to eliminate is the same, subtract the equations

- bottom | 6 x + 4 y = 22 | | 6 x - 3 y = 15 | | 7 y = 7

• The y terms have become 4y - (-3y) = 7*y *(be careful with negatives)

  • Solve the resulting equation to find y
  • y = 1

• Then substitute y = 1 into one of the original equations to find x

  • 3x + 2 = 11, so 3*x *= 9, giving x = 3

• Write out both solutions together, x = 3 and y = 1

• Alternatively, you could have eliminated the y's from 3*x + 2y = 11 and 2x *- y = 5 by making the coefficient of y in both equations the same

  • Multiply every term in the second equation by 2

  • Adding this to the first equation eliminates y (and so on)

    • When the sign in front of the term you want to eliminate is different, add the equations

+ bottom | 3 x + 2 y = 11 | | 4 x - 2 y = 10 | | 7 x = 21

How do I solve linear simultaneous equations by substitution?

• Substitution means substituting one equation into the other

  • This is an alternative method to elimination

    • You can still use elimination if you prefer

• To solve 3x + 2y = 11 and 2x - y = 5 by substitution

  • Rearrange one of the equations into y = ... (or x = ...)

    • For example, the second equation becomes y = 2x - 5

  • Substitute this into the first equation

    • This means 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

How do I solve linear simultaneous equations graphically?

• 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)

  • The x and *y solutions to the simultaneous equations are the *x and y coordinates of the point of intersection

• For example, to solve 2x - y = 3 and 3x + y = 7 simultaneously

  • First plot them both on the same axes (see graph)
  • Find the point of intersection, (2, 1)
  • The solution is x = 2 and y = 1

Exam Hint Always check that your final solutions satisfy both original simultaneous equations!

Write out both solutions (x and y) together at the end to avoid examiners missing a solution in your working.

Worked Example Solve the simultaneous equations

$\begin{gathered} 5 x + 2 y & = 11 \\ 4 x - 3 y & = 18 \end{gathered}$

Answer:

It helps to number the equations

$\begin{gathered} 5 x + 2 y & = 11 \\ 4 x - 3 y & = 18 \end{gathered}\quad \begin{gathered} \text circled{1} \end{gathered} \begin{gathered} \text circled{2} \end{gathered}$

We will choose to eliminate the y terms Make the y terms equal by multiplying all parts of equation 1 by 3 and all parts of equation 2 by 2

$\begin{gathered} 15 x + 6 y & = 33 \\ 8 x - 6 y & = 36 \end{gathered}\quad \begin{gathered} \text circled{3} \end{gathered} \begin{gathered} \text circled{4} \end{gathered}$

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

15 x + 6 y = 33 > bottom + 8 x - 6 y = 36 > 23 x = 69

Solve the equation to find x(divide both sides by 23)

$\begin{gathered} x = \frac{69}{23} = 3 \end{gathered}$

Substitute x = 3 into either of the two original equations

$\text circled{1}$* *$5 ( 3 ) + 2 y = 11$

Solve this equation to find* y*

$\begin{gathered} 15 + 2 y & = 11 \\ 2 y & = 11 - 15 \\ 2 y & = - 4 \\ y = \frac{- 4}{2} = - 2 \end{gathered}$

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

$\begin{gathered} \text circled{2} \end{gathered}$* *$\begin{gathered} 4 x - 3 y & = 18 \end{gathered}$

• *$\begin{gathered} 4 ( 3 ) - 3 ( - 2 ) & = 18 \\ 12 - ( - 6 ) & = 18 \\ 18 = 18 \end{gathered}$

Write out both solutions together

• This question can also be done by eliminating x first (multiplying equation 1 by 4 and equation 2 by 5 then subtracting)

$x=3,y=-2$

How do I form simultaneous equations?

• Introduce two letters, x and y, to represent two unknowns

  • Make sure you know exactly what they stand for (and any units)

• Create two different equations from the words or contexts

  • 3 apples and 2 bananas cost £1.80, while 5 apples and 1 banana cost £2.30

    • 3x + 2y = 180 and 5x + y = 230 x is the price of an apple, in pence y is the price of a banana, in pence
    • This question could also be done in pounds, £

• Solve the equations simultaneously

• Give answers in context (relate them to the story, with units)

  • x = 40, y = 30
  • In context: an apple costs 40p and a banana costs 30p

• Some questions don't ask you to solve simultaneously, but you still need to

  • Two numbers have a sum of 19 and a difference of 5, what is their product?

    • x + y = 19 and x - y = 5
    • Solve simultaneously to get x = 12, y = 7
    • The product is xy = 12 × 7 = 84

Exam Hint Always check that you've answered the question! Sometimes finding x and y isn't the end.

E.g. you may have to state a conclusion related to the context of the problem.

Worked Example At a bakery a customer pays £9 in total for six bagels and twelve sausage rolls.

Another customer buys nine bagels and ten sausage rolls, which costs £12.30 in total.

Find the cost of 5 bagels and 15 sausage rolls.

Answer:

The two variables are the price of bagels, $b$, and the price of sausage rolls, $s$

Write an equation for the first customer's purchases, and label it equation 1

$\text circled{1} 6 b + 12 s = 9$

Write an equation for the second customer's purchases, and label it equation 2

$\text circled{2} 9 b + 10 s = 12.3$

We will choose to eliminate the $b$* *terms Make the $b$ terms equal by multiplying all parts of equation 1 by 3 and all parts of equation 2 by 2 Label these as equations 3 and 4

$\begin{gathered} \text circled{1} \times 3 18 b + 36 s & = 27 \text circled{3} \\ \text circled{2} \times 2 18 b + 20 s & = 24.6 \text circled{4} \end{gathered}$

To eliminate $b$, subtract equation 4 from equation 3

$\text circled{3} - \text circled{4} 16 s = 2.4$

Solve for $s$

$s=\frac{2.4}{16}=0.15$

Substitute this into either equation to find $b$, we will use equation 1

$\begin{gathered} \text circled{1} 6 b + 12 ( 0.15 ) & = 9 \\ 6 b + 1.8 & = 9 \\ 6 b & = 7.2 \\ b = 1.2 \end{gathered}$

So sausage rolls cost £0.15 each and bagels cost £1.20 each

Use these values to find the price of 5 bagels and 15 sausage rolls

$( 5 \times 1.2 ) + ( 15 \times 0.15 ) = 8.25$

Final answer: £8.25