Simultaneous Equations Masterclass

1. What are Simultaneous Equations?

A normal equation like $2x = 10$ has one unknown ($x$) and one answer ($x = 5$).

Simultaneous equations have two unknowns (usually $x$ and $y$). Because there are two unknowns, we need two equations to solve them. "Simultaneous" means we are finding the specific value of $x$ and $y$ that makes both equations true at the exact same time.

The Visual Meaning:
If you draw a linear equation on a graph, it makes a straight line. If you draw two equations, they make two lines. The solution to the simultaneous equations is the exact coordinate $(x, y)$ where those two lines cross!

2. The Elimination Method

This is the most common method. The goal is to add or subtract the two equations together to make one of the letters "vanish" (eliminate).

The "SSS" Rule:
Look at the coefficients (the numbers in front) of the letter you want to eliminate.
If the signs are the Same, Subtract the equations.
If the signs are different, Add the equations.
Example 1: Simple Elimination (Adding)
Solve:
(1) $3x + y = 11$
(2) $2x - y = 4$

Step 1: Eliminate The $y$ in eq (1) is $+y$ and the $y$ in eq (2) is $-y$. The signs are different, so we Add the equations together.
$3x + 2x = 5x$
$+y + -y = 0$ (The $y$ vanishes!)
$11 + 4 = 15$
So, $5x = 15$. Divide by 5 to get $x = 3$. Step 2: Substitute Back Now we know $x = 3$, put it back into either original equation to find $y$. Let's use (1):
$3(3) + y = 11$
$9 + y = 11$
$y = 2$. Step 3: Check (Crucial!) Put $x=3$ and $y=2$ into equation (2): $2(3) - 2 = 6 - 2 = 4$. It works! Solution: $x = 3, y = 2$
Example 2: Scaling Equations First
Solve:
(1) $3x + 2y = 16$
(2) $x + 2y = 8$

The numbers in front of $y$ are both exactly $+2$. Since the signs are the Same, we Subtract eq (2) from eq (1).
$3x - x = 2x$
$2y - 2y = 0$
$16 - 8 = 8$
So, $2x = 8$, which means $x = 4$.

Substitute back into (2):
$4 + 2y = 8 \Rightarrow 2y = 4 \Rightarrow$ $y = 2$. Solution: $x = 4, y = 2$

3. The Substitution Method

This method is great when one of your equations already says $y = \dots$ or $x = \dots$

Example 3: Substitution
Solve:
(1) $y = 2x - 1$
(2) $3x + 2y = 12$

Step 1: Substitute Since eq (1) tells us exactly what $y$ is, we can replace the $y$ in eq (2) with $(2x - 1)$.
$3x + 2(2x - 1) = 12$
Step 2: Solve for x Expand the bracket: $3x + 4x - 2 = 12$
Collect terms: $7x - 2 = 12$
Add 2: $7x = 14$
Divide by 7: $x = 2$.
Step 3: Substitute back Put $x = 2$ into eq (1):
$y = 2(2) - 1 \Rightarrow y = 4 - 1 \Rightarrow$ $y = 3$. Solution: $x = 2, y = 3$

Practice Questions & Solutions

Try these on paper. Remember to write out your steps neatly and label your equations (1) and (2)! Click to reveal the solutions.

Q1: Solve $x + y = 10$ and $x - y = 2$
Add the equations to eliminate $y$ (since signs are different: $+y$ and $-y$).
$(x + x) = 2x$
$(10 + 2) = 12$
$2x = 12 \Rightarrow x = 6$

Sub back into eq 1: $6 + y = 10 \Rightarrow y = 4$
Answer: $x = 6, y = 4$
Q2: Solve $2x + y = 7$ and $x + y = 5$
Subtract eq 2 from eq 1 to eliminate $y$ (Same signs subtract).
$(2x - x) = x$
$(y - y) = 0$
$(7 - 5) = 2$
$x = 2$

Sub back into eq 2: $2 + y = 5 \Rightarrow y = 3$
Answer: $x = 2, y = 3$
Q3: Solve $3x + 2y = 12$ and $x - 2y = 4$
Add the equations to eliminate $y$ ($+2y$ and $-2y$).
$4x = 16 \Rightarrow x = 4$

Sub back into eq 1: $3(4) + 2y = 12$
$12 + 2y = 12 \Rightarrow 2y = 0 \Rightarrow y = 0$
Answer: $x = 4, y = 0$
Q4: Solve $4x + 3y = 17$ and $4x - y = 5$
The $x$'s have the same coefficient (+4). Same Signs Subtract.
$(4x - 4x) = 0$
Be careful here! $3y - (-y) = 3y + y = 4y$
$17 - 5 = 12$
$4y = 12 \Rightarrow y = 3$

Sub back into eq 2: $4x - 3 = 5 \Rightarrow 4x = 8 \Rightarrow x = 2$
Answer: $x = 2, y = 3$
Q5: Solve $2x + 3y = 11$ and $3x + y = 13$ (Scaling required)
We need to make the coefficients match. Let's multiply eq 2 by 3 to match the $y$'s.
Eq 2 becomes: $9x + 3y = 39$
Eq 1 is: $2x + 3y = 11$

Subtract Eq 1 from the new Eq 2:
$7x = 28 \Rightarrow x = 4$

Sub back into original eq 2: $3(4) + y = 13 \Rightarrow 12 + y = 13 \Rightarrow y = 1$
Answer: $x = 4, y = 1$
Q6: Solve $5x - 2y = 16$ and $3x + y = 14$
Multiply eq 2 by 2 to match the $y$'s.
Eq 2 becomes: $6x + 2y = 28$
Eq 1 is: $5x - 2y = 16$

Signs are different ($+2y$ and $-2y$), so ADD them:
$11x = 44 \Rightarrow x = 4$

Sub back into original eq 2: $3(4) + y = 14 \Rightarrow 12 + y = 14 \Rightarrow y = 2$
Answer: $x = 4, y = 2$
Q7: Use substitution: $y = x + 3$ and $2x + y = 15$
Sub the first equation into the $y$ of the second equation:
$2x + (x + 3) = 15$
$3x + 3 = 15$
$3x = 12 \Rightarrow x = 4$

Sub back into eq 1: $y = 4 + 3 \Rightarrow y = 7$
Answer: $x = 4, y = 7$
Q8: Word Problem - 2 apples and 3 bananas cost 80p. 2 apples and 1 banana cost 50p. Find the cost of each.
Let $a$ = apples and $b$ = bananas.
(1) $2a + 3b = 80$
(2) $2a + b = 50$

Subtract eq 2 from eq 1 to eliminate $a$:
$2b = 30 \Rightarrow b = 15$

Sub back into eq 2: $2a + 15 = 50 \Rightarrow 2a = 35 \Rightarrow a = 17.5$
Answer: Apples are 17.5p, Bananas are 15p.
Q9: Solve $y = 2x$ and $x + y = 18$
Use substitution. Replace $y$ with $2x$ in the second equation.
$x + (2x) = 18$
$3x = 18 \Rightarrow x = 6$

Sub back into eq 1: $y = 2(6) \Rightarrow y = 12$
Answer: $x = 6, y = 12$
Q10: Harder Scaling - Solve $3x + 4y = 24$ and $2x + 3y = 17$
We need to scale BOTH equations to make a coefficient match. Let's eliminate $x$ by making them both $6x$.
Multiply eq 1 by 2: $6x + 8y = 48$
Multiply eq 2 by 3: $6x + 9y = 51$

Subtract the top from the bottom:
$y = 3$

Sub back into original eq 1: $3x + 4(3) = 24$
$3x + 12 = 24 \Rightarrow 3x = 12 \Rightarrow x = 4$
Answer: $x = 4, y = 3$