Simultaneous Equations

Mathematics (Core) revision notes · Cambridge (CIE) · 0580 · Core

1. What they are

A single equation with two unknowns has infinitely many solutions, so a second equation is required to pin them down
Simultaneous equations are solved together, and the answer is the pair of values satisfying both at once

They are linear when every term is a number or a number times one unknown, so nothing is squared and the unknowns are never multiplied together
Each solution should be checked in the equation that was not used to find it

Elimination removes one unknown by adding or subtracting the two equations
Multiply one or both equations so that one unknown has coefficients of equal size
Subtract when those coefficients have the same sign, and add when the signs are opposite
Solve the resulting single-unknown equation, then substitute back to find the other

Worked example

Elimination with both equations scaled

Solve 3x + 5y = 31 and 2x + 3y = 20.

Solution:

Elimination needs one letter to have the same coefficient in both equations, and here neither does
Match the y terms: 5 and 3 both divide into 15, so multiply the first equation by 3 and the second by 5
Multiply every term, the constant included, or the equation stops being true
9x + 15y = 93 and 10x + 15y = 100
The matched terms now have the same sign, so subtracting removes them
Subtracting the first from the second: 10x − 9x = 1x, 15y − 15y = 0, 100 − 93 = 7, so x = 7
Substitute back into either original equation: 3 × 7 + 5y = 31
21 + 5y = 31, so 5y = 10 and y = 2
The solution is x = 7, y = 2
Check in the equation not used for the substitution: 2 × 7 + 3 × 2 = 14 + 6 = 20
Had the matched terms carried opposite signs, they would be added instead

Worked example

Adding to eliminate, with a negative answer

Solve 2x + 3y = −1 and 5x − 6y = 38.

Solution:

Match the y terms: multiply every term of the first equation by 2, giving 4x + 6y = −2
Now +6y and −6y have opposite signs, so adding the two equations removes them
4x + 5x = 9x, 6y − 6y = 0 and −2 + 38 = 36, so 9x = 36 and x = 4
Substitute back into the first equation: 2 × 4 + 3y = −1
8 + 3y = −1, so 3y = −1 − 8 = −9 and y = −3
The solution is x = 4, y = −3
Check in the second equation: 5 × 4 − 6 × (−3) = 20 + 18 = 38

Rearrange one equation to make an unknown the subject, then substitute that expression into the other
This suits an equation such as 2x − y = 1, which rearranges easily to y = 2x − 1
Substituting gives one equation in one unknown
- With 4x + 3y = 27, substituting y = 2x − 1 gives 4x + 3(2x − 1) = 27, so 10x = 30 and x = 3, giving y = 5

Worked example

Solving by substitution

Solve the simultaneous equations y = 2x + 1 and 3x + y = 11.

Solution:

One equation already has y on its own, so substitute that expression into the other
3x + (2x + 1) = 11
Collect: 5x + 1 = 11
Subtract 1: 5x = 10, so x = 2
Substitute back into the isolated equation: y = 2 × 2 + 1 = 5
The solution is x = 2, y = 5
Check in the equation not used for the substitution: 3 × 2 + 5 = 11

Introduce a letter for each unknown and state precisely what each represents, including units
Each separate fact in the question becomes one equation
Solve, then answer in context rather than leaving the letters bare

Worked example

Forming and solving from a context

Two coffees and three teas cost £11.40. Four coffees and one tea cost £13.80. Find the cost of each drink.

Solution: