Simultaneous Equations — Substitution Method GCSE

When should you use substitution rather than elimination?

Both methods solve the same problems. Use substitution when:

• One equation already has a variable isolated (e.g. y = 3x − 1).
• One equation is easily rearranged to isolate a variable.
• You are solving a linear-quadratic pair (one linear, one quadratic) — elimination rarely simplifies these neatly, but substitution always works.

Use elimination when both equations are in the form ax + by = c and the coefficients match (or are easy to match) by multiplying.

How do you apply the substitution method — step by step?

The five-step process:

1. Choose the simpler equation and rearrange it to express one variable in terms of the other.
2. Substitute that expression into the other equation wherever that variable appears.
3. Solve the resulting single-variable equation.
4. Back-substitute the value found into the rearranged expression to find the second variable.
5. Check both values in the original equations.

Worked example 1 — basic substitution

Solve: y = 2x + 1 and 3x + y = 16.

Step 1: Equation 1 already gives y = 2x + 1. No rearrangement needed.

Step 2: Substitute into equation 2: 3x + (2x + 1) = 16

Step 3: Solve: 5x + 1 = 16 → 5x = 15 → x = 3

Step 4: Back-substitute into y = 2x + 1: y = 2(3) + 1 = y = 7

Step 5: Check in equation 2: 3(3) + 7 = 9 + 7 = 16 ✓

Solution: x = 3, y = 7

Worked example 2 — rearranging first

Solve: 2x + y = 11 and 5x − 3y = −1.

Step 1: Rearrange equation 1 for y (simpler — y has coefficient 1): y = 11 − 2x

Step 2: Substitute into equation 2: 5x − 3(11 − 2x) = −1

Step 3: Expand and solve: 5x − 33 + 6x = −1 11x − 33 = −1 11x = 32 → x = 32/11

Hmm — that is a non-integer. Let us instead try rearranging for x from equation 1:

Actually the first approach is correct. Let me redo cleanly:

2x + y = 11 ... (1) 5x − 3y = −1 ... (2)

From (1): y = 11 − 2x

Substitute into (2): 5x − 3(11 − 2x) = −1 5x − 33 + 6x = −1 11x = 32 → x = 32/11

This is an untidy answer suggesting we should use elimination for this pair. Let us swap to a cleaner worked example.

Better worked example 2: Solve 3x + 2y = 12 and y = x − 1.

Step 1: y = x − 1 (already rearranged).

Step 2: Substitute into equation 1: 3x + 2(x − 1) = 12

Step 3: Expand and solve: 3x + 2x − 2 = 12 5x = 14 → x = 2.8

Step 4: y = 2.8 − 1 = y = 1.8

Step 5: Check: 3(2.8) + 2(1.8) = 8.4 + 3.6 = 12 ✓

Worked example 3 — linear and quadratic pair (Higher)

This is where substitution is essential. Solve y = x + 3 and x² + y² = 29.

Step 1: y = x + 3 (linear equation, already rearranged).

Step 2: Substitute into the quadratic: x² + (x + 3)² = 29

Step 3: Expand and solve: x² + x² + 6x + 9 = 29 2x² + 6x − 20 = 0 x² + 3x − 10 = 0 (x + 5)(x − 2) = 0

So x = −5 or x = 2.

Step 4: Back-substitute:

• If x = −5: y = −5 + 3 = −2
• If x = 2: y = 2 + 3 = 5

Solution: (−5, −2) or (2, 5)

Step 5: Check (2, 5) in x² + y² = 29: 4 + 25 = 29 ✓

What is "back-substitution" and why must you do it?

After finding the first variable, you must use it to find the second. Do not substitute back into the equation you already rearranged to avoid an algebraic tautology (e.g. y = y). Instead, substitute into the other original equation — or into the rearranged expression, which is faster and equally valid.

Always write both values clearly: "x = 2, y = 5" (or as a coordinate pair "(2, 5)").

Frequently asked questions

Can I always use substitution rather than elimination?

Yes — substitution always works in principle. However, elimination is faster when coefficients are easy to match. For the pair 4x + 3y = 10 and 5x − 3y = 8, spotting that the y-coefficients are already equal and opposite makes elimination a two-line solution. Substitution would require rearranging, expanding and solving a longer expression. Recognise which method is quicker and use it.

What if neither equation has a variable with coefficient 1?

You can still substitute, but the rearranged expression will have fractions. For example, from 2x + 3y = 7: x = (7 − 3y)/2. Substituting this fraction is perfectly valid — multiply out carefully when you substitute. Alternatively, switch to elimination.

How do I know if a simultaneous pair has no solution or infinitely many solutions?

After substitution, if the variable cancels and you get a false statement (e.g. 0 = 5), the lines are parallel — no solution. If the variable cancels and you get a true statement (e.g. 0 = 0), the lines are identical — infinitely many solutions. A GCSE exam will not usually ask for these cases without signalling them, but it is worth knowing the sign.

In a linear-quadratic pair, why are there usually two solutions?

A straight line can intersect a circle (or parabola) in 0, 1 or 2 points. Two intersections give two solution pairs; one intersection (a tangent) gives one repeated solution; no intersection means no real solutions. You can predict the number of solutions using the discriminant of the resulting quadratic: if b² − 4ac > 0, there are two solutions; = 0, one; < 0, none.

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