How to Solve Simultaneous Equations Using Algebra
Solving simultaneous equations is a core skill in Pearson International GCSE Maths and wider GCSE algebra. By using substitution or elimination, you can find values of variables that satisfy multiple equations simultaneously. This guide breaks down both methods clearly, with structured examples designed to support confident exam performance.
🔁 Substitution
- Rearrange one equation to make \(x\) or \(y\) the subject.
- Substitute that expression into the other equation.
- Solve the resulting single-variable equation.
- Substitute back to find the second variable.
➕ Elimination
- Multiply one or both equations so a variable has matching coefficients.
- Add or subtract the equations to eliminate that variable.
- Solve the resulting single-variable equation.
- Substitute back to find the second variable.
Substitution Method
💡 Example 1: Simple substitution
Step 1: Write the system:
\[ \begin{aligned} 3x + y &= 14 \quad (1)\\ y &= 2x - 1 \quad (2) \end{aligned} \]Step 2: Substitute (2) into (1):
\[3x + (2x - 1) = 14\]Step 3: Simplify and solve for \(x\):
\[ \begin{aligned} 5x - 1 &= 14\\ 5x &= 15\\ x &= 3 \end{aligned} \]Step 4: Substitute \(x = 3\) into (2):
\[ \begin{aligned} y &= 2(3) - 1\\ y &= 5 \end{aligned} \]Answer: \(x = 3,\quad y = 5\)
Elimination Method
💡 Example 2: Eliminating \(y\) by subtraction
Step 1: Write the equations:
\[ \begin{aligned} 4x + y &= 14 \quad (1)\\ 2x + y &= 8 \quad (2) \end{aligned} \]Step 2: Subtract (2) from (1) to eliminate \(y\):
\[ \begin{aligned} 4x + y &= 14\\ 2x + y &= 8\\ \hline 2x &= 6 \end{aligned} \] \[x = 3\]Step 3: Substitute \(x = 3\) into (2):
\[ \begin{aligned} 2(3) + y &= 8\\ 6 + y &= 8\\ y &= 2 \end{aligned} \]Answer: \(x = 3,\quad y = 2\)
💡 Example 3: Multiplying both equations to match coefficients
Step 1: Write the system:
\[ \begin{aligned} 2x + 3y &= 12 \quad (1)\\ 3x - 2y &= 5 \quad (2) \end{aligned} \]Step 2: Multiply (1) by 2 and (2) by 3 to match \(y\) coefficients:
\[ \begin{aligned} 4x + 6y &= 24 \quad (\times 2)\\ 9x - 6y &= 15 \quad (\times 3) \end{aligned} \]Step 3: Add to eliminate \(y\):
\[ \begin{aligned} 4x + 6y &= 24\\ 9x - 6y &= 15\\ \hline 13x &= 39 \end{aligned} \] \[x = 3\]Step 4: Substitute \(x = 3\) into (1):
\[ \begin{aligned} 2(3) + 3y &= 12\\ 6 + 3y &= 12\\ 3y &= 6\\ y &= 2 \end{aligned} \]Answer: \(x = 3,\quad y = 2\)
🔑 Key Points
- Line up \(x\), \(y\), and constant columns carefully before adding or subtracting.
- Write one operation per line with one equals sign per line throughout your working.
- Multiply entire equations (both sides) when matching coefficients for elimination.
- After finding one variable, substitute back into the simpler of the two original equations.
- Check your answer by substituting both values into the equation you did not use to find \(y\).
Ready to practise? Work through all four rooms below.
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