Why learn substitution and elimination?
Imagine you have two secret codes that share a hidden number. Cracking both codes at once feels like a cool puzzle – that’s exactly what solving a pair of linear equations does.
In simple words, you have two straight‑line equations with two unknowns. By using substitution or elimination, you turn the puzzle into a single‑number answer that works for both equations.
What is a pair of linear equations?
A linear equation is an expression where the variable(s) are only to the first power – no squares or cubes. When you have two such equations together, we call them a pair of linear equations (also known as simultaneous equations). Think of them like two road maps that intersect at a single point; the intersection is the solution you’re after.
Substitution method – step by step
The substitution method is like solving a mystery by first finding a clue in one equation and then plugging that clue into the other.
- Step 1: Choose the equation that looks easiest to isolate a variable (make that variable alone on one side).
- Step 2: Express that variable in terms of the other variable.
- Step 3: Substitute this expression into the second equation. Now you have only one variable.
- Step 4: Solve for the remaining variable.
- Step 5: Plug the found value back into the expression from Step 2 to get the other variable.
- Step 6: Check both original equations to be sure.
Example: Solve 2x + y = 7 and x – 3y = -4.
Step 1: The first equation looks easy for y. Rearrange: y = 7 – 2x.
Step 2: Substitute y in the second equation: x – 3(7 – 2x) = -4.
Step 3: Expand: x – 21 + 6x = -4 → 7x – 21 = -4.
Step 4: 7x = 17 → x = 17/7 ≈ 2.43.
Step 5: Put x back into y = 7 – 2x → y = 7 – 2(17/7) = 7 – 34/7 = (49 – 34)/7 = 15/7 ≈ 2.14.
Step 6: Verify: 2(17/7) + 15/7 = 34/7 + 15/7 = 49/7 = 7 ✔️ and 17/7 – 3(15/7) = 17/7 – 45/7 = -28/7 = -4 ✔️.
Elimination method – step by step
The elimination method works like a tug‑of‑war where you add or subtract the equations to cancel one variable.
- Step 1: Write both equations in standard form (ax + by = c).
- Step 2: Make the coefficients (numbers in front) of one variable opposites. You may need to multiply one or both equations.
- Step 3: Add or subtract the equations so that the chosen variable disappears.
- Step 4: Solve the resulting single‑variable equation.
- Step 5: Substitute this value back into one of the original equations to find the other variable.
- Step 6: Verify both original equations.
Example: Solve 3x + 2y = 16 and 5x – 2y = 4.
Step 1: Both are already in standard form.
Step 2: Notice the coefficients of y are +2 and -2 – they are already opposites, so we can add the equations directly.
Step 3: Add: (3x + 2y) + (5x – 2y) = 16 + 4 → 8x = 20.
Step 4: x = 20/8 = 5/2 = 2.5.
Step 5: Put x into the first equation: 3(2.5) + 2y = 16 → 7.5 + 2y = 16 → 2y = 8.5 → y = 4.25.
Step 6: Check with the second equation: 5(2.5) – 2(4.25) = 12.5 – 8.5 = 4 ✔️.
Comparison of Substitution vs Elimination
| Aspect | Substitution | Elimination |
|---|---|---|
| When to use | One equation easily isolates a variable. | Coefficients can be made opposite with simple multiplication. |
| Number of steps | Usually 5–6 algebraic manipulations. | Often fewer steps if coefficients already match. |
| Risk of error | Higher if fractions appear early. | Multiplying equations can introduce large numbers. |
| Best for | Equations with small coefficients or clear isolation. | Equations where variables have similar coefficients. |
Common mistakes to avoid
- Forgetting to flip the sign when moving a term to the other side.
- Multiplying only one side of an equation instead of the whole equation.
- Skipping the verification step – a tiny arithmetic slip can give a wrong pair.
- Mixing up which variable you are eliminating; double‑check coefficients.
📝 Likely Exam Questions
- Question: Solve 4x – y = 5 and 2x + 3y = 12 using elimination.
Answer: Multiply first equation by 3 → 12x – 3y = 15. Add to second equation → 14x = 27 → x = 27/14. Substitute back → y = 4x – 5 = 4(27/14) – 5 = 108/14 – 70/14 = 38/14 = 19/7. - Question: Using substitution, solve 5a + b = 9 and a – 2b = -3.
Answer: From first, b = 9 – 5a. Substitute into second: a – 2(9 – 5a) = -3 → a – 18 + 10a = -3 → 11a = 15 → a = 15/11. Then b = 9 – 5(15/11) = 9 – 75/11 = (99 – 75)/11 = 24/11. - Question: Explain when you would prefer elimination over substitution.
- Answer: Choose elimination when the coefficients of one variable are already opposites or can be made opposites with a small multiplier, because it usually requires fewer steps and avoids early fractions.
- Question: Verify that x = 2, y = 3 satisfies the system 2x + y = 7 and x – y = -1.
- Answer: Plug in: 2(2)+3 = 4+3 = 7 ✔️; 2 – 3 = -1 ✔️.