Simultaneous Equations

Simultaneous equations are two equations that must both be satisfied by the same values of the unknowns at the same time. Finding these values is called solving simultaneously.

A single linear equation like $2x+y=10$ has infinitely many solutions — any point on that line. A second equation like $x-y=2$ narrows the answer to one point: the unique pair of values $(x,y)$ that satisfies both equations simultaneously.

The solution is where the two lines cross. For $2x+3y=12$ and $4x-3y=6$, that intersection is the point $(3,2)$ found algebraically below.

Number of equations	Unknowns	Solutions
1 linear equation	2 unknowns	Infinitely many
2 linear equations	2 unknowns	One pair (usually)

Two methods are required for GCSE: elimination and substitution. Both produce the same answer — the choice is down to which is faster for the structure of the equations given.

Linear simultaneous equations (two straight lines) are assessed at both Foundation and Higher. Linear/quadratic simultaneous equations are Higher tier only.

Elimination works by scaling one or both equations so that the coefficient of one variable becomes equal (or equal and opposite) in both equations. Adding or subtracting then removes that variable, leaving a single equation in one unknown.

Steps:

1. Multiply one (or both) equations by a constant so the coefficients of one variable match.
2. Add or subtract the equations to eliminate that variable.
3. Solve the resulting single-variable equation.
4. Substitute back to find the second variable.
5. Check in both original equations.

Worked example — Solve simultaneously: $2x+3y=12$ and $4x-3y=6$.

The coefficients of $y$ are already equal and opposite ($+3$ and $-3$). Add the equations:

$\begin{array}{rl}2x+3y& =12\\ 4x-3y& =6\\ 6x& =18\end{array}$

$x=3$

Substitute $x=3$ into the first equation: $6+3y=12\Rightarrow 3y=6\Rightarrow y=2$.

Check: $4(3)-3(2)=12-6=6$ ✓

Solution: $x=3$, $y=2$.

When coefficients do not match, multiply one or both equations before eliminating.

Worked example — Solve: $3x+2y=16$ and $5x-3y=7$.

To eliminate $y$, make the $y$-coefficients equal: multiply equation 1 by 3 and equation 2 by 2.

$3\times (3x+2y=16)\to 9x+6y=48$ $2\times (5x-3y=7)\to 10x-6y=14$

Add (coefficients are $+6$ and $-6$, so they cancel): $19x=62⟹x=\frac{62}{19}$

Substitute $x=\frac{62}{19}$ into equation 1:

$3\times \frac{62}{19}+2y=16⟹\frac{186}{19}+2y=\frac{304}{19}⟹2y=\frac{118}{19}⟹y=\frac{59}{19}$

Check in equation 2: $5\times \frac{62}{19}-3\times \frac{59}{19}=\frac{310-177}{19}=\frac{133}{19}=7$ ✓

Solution: $x=\frac{62}{19}\approx 3.26$, $y=\frac{59}{19}\approx 3.11$.

If the coefficients had the same sign (e.g., $+6$ and $+6$), subtract instead of add.

Eliminate the variable whose coefficients are easier to match. If one variable already appears with a simple coefficient (1, 2, 3), choose that one.

Substitution works by rearranging one equation to express one variable in terms of the other, then substituting that expression into the second equation.

Best used when one equation already has a variable with coefficient 1, making rearrangement simple.

Worked example — Solve: $y=2x-1$ and $3x+y=14$.

The first equation gives $y$ directly. Substitute $y=2x-1$ into the second:

$3x+(2x-1)=14⟹5x-1=14⟹5x=15⟹x=3$

Substitute $x=3$ back: $y=2(3)-1=5$.

Check: $3(3)+5=9+5=14$ ✓

Solution: $x=3$, $y=5$.

Worked example 2 — Solve: $2x+y=7$ and $x+3y=11$.

Rearrange the first equation: $y=7-2x$. Substitute:

$x+3(7-2x)=11⟹x+21-6x=11⟹-5x=-10⟹x=2$

Then $y=7-2(2)=3$. Check: $2+3(3)=2+9=11$ ✓

Many GCSE questions describe a real-world situation and ask you to form and solve simultaneous equations. The method is: define your variables clearly, write two equations, then solve.

Worked example — Two shops sell pens and rulers. Shop A charges £1.20 for 2 pens and 1 ruler. Shop B charges £1.70 for 3 pens and 1 ruler. Both shops charge the same price per pen and the same price per ruler. Find the price of one pen.

Let $p$ = price of a pen (£) and $r$ = price of a ruler (£).

$2p+r=1.20\text{(equation 1)}$ $3p+r=1.70\text{(equation 2)}$

Subtract equation 1 from equation 2: $p=0.50$.

Substitute: $2(0.50)+r=1.20\Rightarrow r=0.20$.

One pen costs 50p and one ruler costs 20p.

Define your variables at the start — write "let $p$ = ..." — and always state units. Examiners award marks for the definition, the equations, and the solution separately.

1. Subtracting equations incorrectly

When subtracting equation 2 from equation 1, subtract every term. A common error is subtracting only the $x$-terms and forgetting to subtract the constant. Write each equation on a separate line and work column by column.

2. Forgetting to find the second variable

Solving for $x$ completes only half the problem. You must substitute back to find $y$. Stopping after finding one variable loses half the solution marks.

3. Not checking the solution

Substituting your answer into the original equations takes 20 seconds and catches arithmetic slips before they cost marks. Always check in both equations — not just the one you used to find the second variable.

4. Sign errors when multiplying to match coefficients

Multiplying $3x-2y=5$ by $-2$ gives $-6x+4y=-10$. The negative must be distributed to every term, including the constant. Writing $-6x+4y=5$ is a common error.

5. Misidentifying which variable to eliminate

Both methods work on any pair of simultaneous equations. If elimination looks complicated, switch to substitution. Choose whichever method suits the structure of the equations, not habit.