Systems of Equations • Topic 1 of 4

Substitution

Substitution shines when one equation already has a variable alone, or is one easy step from it. Solve that equation for a single variable, then replace that variable everywhere in the other equation — now you have one equation in one unknown, which you already know how to solve. Once you have the first value, plug it back into either original equation to get the second. The method is most efficient when a coefficient is 1, because isolating is clean and no fractions appear. Always substitute into the OTHER equation, never the one you just rearranged (that only gives an identity). Finish by checking the pair in both originals; a quick check catches an arithmetic slip before it costs the question.

✅ Solved examples

1. Solve y = 2x + 1 and 3x + y = 11.
Substitute y: 3x + (2x + 1) = 11 ⇒ 5x + 1 = 11 ⇒ x = 2. Then y = 2(2) + 1 = 5. Solution (2, 5).
2. Solve x = y − 3 and 2x + 3y = 19.
Substitute x: 2(y − 3) + 3y = 19 ⇒ 2y − 6 + 3y = 19 ⇒ 5y = 25 ⇒ y = 5. Then x = 5 − 3 = 2. Solution (2, 5).
3. Solve y = 4 − x and x − y = 2.
Substitute y: x − (4 − x) = 2 ⇒ 2x − 4 = 2 ⇒ x = 3. Then y = 4 − 3 = 1. Solution (3, 1).
4. Solve 2x + y = 7 and y = x − 2. Then find xy.
Substitute: 2x + (x − 2) = 7 ⇒ 3x = 9 ⇒ x = 3, y = 1. So xy = 3.

✏️ Practice — try these, take hints as needed

1. Solve y = 3x and x + y = 8.
Substitute 3x for y.
x + 3x = 8.
4x = 8.
x = 2, y = 6
2. Solve x = 2y and 3x − y = 10.
Replace x with 2y.
6y − y = 10.
5y = 10.
x = 4, y = 2
3. Solve y = x + 4 and 2x + y = 13.
Substitute for y.
2x + x + 4 = 13.
3x = 9.
x = 3, y = 7
4. Solve y = 5 − 2x and 4x + y = 9.
Substitute for y.
4x + 5 − 2x = 9.
2x = 4.
x = 2, y = 1
5. Solve x = 3y − 1 and x + 2y = 9. Then find x + y.
Substitute for x.
3y − 1 + 2y = 9.
5y = 10.
x = 5, y = 2, x + y = 7

📝 Topic test — 8 questions

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