Simultaneous Equations: Substitution vs Elimination

Simultaneous equations ask you to find values that satisfy two (or more) equations at the same time. They appear in Matric, O Level, FSc and A Level papers, and in physics and chemistry problems whenever two unknowns are linked. There are two main algebraic methods, substitution and elimination, and knowing when to use each will save you time and marks.

What Are Simultaneous Equations?

A pair of linear simultaneous equations looks like this:

$$ \begin{aligned} a_1x + b_1y &= c_1 \\ a_2x + b_2y &= c_2 \end{aligned} $$

Each equation represents a straight line. The solution is the point \((x, y)\) where the two lines intersect.

Method 1: Substitution

Steps

1. Make one variable the subject of one equation.
2. Substitute that expression into the other equation.
3. Solve the resulting equation in one variable.
4. Substitute back to find the second variable.
5. Check in both original equations.

Worked Example 1

Solve:

$$ \begin{aligned} y &= 2x - 3 \quad (1) \\ 3x + 2y &= 15 \quad (2) \end{aligned} $$

Step 1: Equation (1) already gives \(y\) in terms of \(x\).

Step 2: Substitute into (2):

$$ 3x + 2(2x - 3) = 15 $$

Step 3: Expand and solve:

$$ 3x + 4x - 6 = 15 \quad \Rightarrow \quad 7x = 21 \quad \Rightarrow \quad x = 3 $$

Step 4: Substitute into (1):

$$ y = 2(3) - 3 = 3 $$

Check in (2): \(3(3) + 2(3) = 9 + 6 = 15\). Correct.

Answer: \(x = 3\), \(y = 3\).

Method 2: Elimination

Steps

1. Multiply one or both equations so that one variable has equal (or opposite) coefficients.
2. Add or subtract the equations to eliminate that variable.
3. Solve for the remaining variable.
4. Substitute back to find the other.
5. Check in both original equations.

A handy memory aid: same signs subtract, different signs add.

Worked Example 2

Solve:

$$ \begin{aligned} 3x + 4y &= 10 \quad (1) \\ 5x - 2y &= 21 \quad (2) \end{aligned} $$

Step 1: Make the \(y\) coefficients opposites. Multiply (2) by 2:

$$ 10x - 4y = 42 \quad (3) $$

Step 2: The \(y\) terms have different signs, so add (1) and (3):

$$ 13x = 52 $$

Step 3: Solve:

$$ x = 4 $$

Step 4: Substitute into (1):

$$ 3(4) + 4y = 10 \quad \Rightarrow \quad 4y = -2 \quad \Rightarrow \quad y = -\frac{1}{2} $$

Check in (2): \(5(4) - 2\left(-\frac{1}{2}\right) = 20 + 1 = 21\). Correct.

Answer: \(x = 4\), \(y = -\frac{1}{2}\).

Worked Example 3: Multiplying Both Equations

Solve:

$$ \begin{aligned} 4x + 3y &= 1 \quad (1) \\ 3x + 5y &= -9 \quad (2) \end{aligned} $$

Step 1: Eliminate \(x\) by making both coefficients 12. Multiply (1) by 3 and (2) by 4:

$$ \begin{aligned} 12x + 9y &= 3 \quad (3) \\ 12x + 20y &= -36 \quad (4) \end{aligned} $$

Step 2: Same signs, so subtract (3) from (4):

$$ 11y = -39 \quad \Rightarrow \quad y = -\frac{39}{11} $$

Step 3: Rather than substituting a fraction, find \(x\) by eliminating \(y\) instead. Multiply (1) by 5 and (2) by 3:

$$ \begin{aligned} 20x + 15y &= 5 \\ 9x + 15y &= -27 \end{aligned} $$

Subtract: \(11x = 32\), so \(x = \frac{32}{11}\).

Check in (1): \(4\left(\frac{32}{11}\right) + 3\left(-\frac{39}{11}\right) = \frac{128 - 117}{11} = \frac{11}{11} = 1\). Correct.

Answer: \(x = \frac{32}{11}\), \(y = -\frac{39}{11}\). Not every answer is a neat integer, and that is fine as long as it checks out.

Substitution vs Elimination: Which Is Better?

Elimination is often faster for two linear equations with "messy" coefficients, because it avoids fractions until the end. Substitution is essential when one equation is quadratic.

Worked Example 4: One Linear and One Quadratic

Solve \(y = x + 1\) and \(x^2 + y^2 = 25\).

Step 1: Substitute \(y = x + 1\):

$$ x^2 + (x + 1)^2 = 25 $$

Step 2: Expand and simplify:

$$ 2x^2 + 2x + 1 = 25 \quad \Rightarrow \quad 2x^2 + 2x - 24 = 0 \quad \Rightarrow \quad x^2 + x - 12 = 0 $$

Step 3: Factorize:

$$ (x + 4)(x - 3) = 0 \quad \Rightarrow \quad x = -4 \text{ or } x = 3 $$

Step 4: Find \(y\): when \(x = 3\), \(y = 4\); when \(x = -4\), \(y = -3\).

Answer: \((3, 4)\) and \((-4, -3)\). Both satisfy \(x^2 + y^2 = 25\).

Common Mistakes

• Subtracting incorrectly. When subtracting equations, subtract every term, including the constants on the right.
• Multiplying only one side. Scale the entire equation, both left and right.
• Stopping after one variable. You must find both values.
• Pairing solutions wrongly in non-linear systems. Always substitute each \(x\) into the linear equation to get its matching \(y\).

Key Takeaways

• Simultaneous equations find the point(s) where graphs intersect.
• Substitution works best when a variable is already isolated or one equation is non-linear.
• Elimination is usually faster for two linear equations with larger coefficients.
• Always check your solution in both original equations.

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