Why learn to solve linear equations?
Imagine you’re trying to find the exact point where two roads intersect – that’s exactly what a pair of linear equations does on a graph. Knowing how to locate that point is a super‑useful skill for maths exams and real life.
In simple words, solving a pair of linear equations means finding the values of two unknowns that make both equations true at the same time. It’s like figuring out the exact spot where two straight lines cross.
What is a pair of linear equations?
A linear equation is an algebraic statement where each term is either a constant or the product of a constant and a single variable (like 2x or ‑3y). When you have two such equations with the same two variables, you call it a pair of linear equations. The goal is to discover the single pair of numbers (x, y) that satisfy both equations simultaneously.
Substitution method – step by step
The substitution method works best when one of the equations can be easily rearranged to isolate a variable.
Steps
- Pick an equation and solve it for one variable.
- Replace that variable in the other equation with the expression you just found.
- Solve the resulting single‑variable equation.
- Plug the found value back into the expression from step 1 to get the second variable.
- Check both original equations to be safe.
Worked example
Solve:
1) 3x + 2y = 16
2) x ‑ y = 1
Step 1: From equation 2, isolate x:
x = y + 1
Step 2: Substitute x in equation 1:
3(y + 1) + 2y = 16
Expand: 3y + 3 + 2y = 16 → 5y + 3 = 16
Step 3: Solve for y:
5y = 13 → y = 13/5 = 2.6
Step 4: Find x using x = y + 1:
x = 2.6 + 1 = 3.6
Step 5: Quick check – plug (3.6, 2.6) into both original equations; they both hold true.
Elimination method – step by step
The elimination method (also called addition method) removes one variable by adding or subtracting the equations after adjusting their coefficients.
Steps
- Arrange both equations in the form ax + by = c.
- Make the coefficients of one variable equal (or opposite) by multiplying one or both equations.
- Add or subtract the equations so that the chosen variable cancels out.
- Solve the resulting single‑variable equation.
- Substitute this value back into one of the original equations to find the second variable.
- Verify the pair in both equations.
Worked example
Solve the same pair as before:
1) 3x + 2y = 16
2) x ‑ y = 1
Step 1: Write both equations clearly.
Step 2: Make the coefficients of y opposite. Multiply equation 2 by 2:
2x ‑ 2y = 2
Now add this to equation 1:
(3x + 2y) + (2x ‑ 2y) = 16 + 2 → 5x = 18
Step 3: Solve for x:
x = 18/5 = 3.6
Step 4: Substitute x into equation 2:
3.6 ‑ y = 1 → y = 2.6
Step 5: Check – both equations are satisfied.
When to use which method?
| Substitution | Elimination |
|---|---|
| One variable already isolated or easy to isolate. | Coefficients can be quickly made equal or opposite. |
| Works well with fractions or decimals in one equation. | Better for whole‑number coefficients and larger systems. |
Quick reference flowchart
📝 Likely Exam Questions
- Solve by substitution: 2x + 3y = 12, x ‑ y = 2.
Answer: x = 4, y = 2. - Solve by elimination: 5x ‑ 2y = 9, 3x + 4y = 7.
Answer: x = 1, y = 1. - Explain why the elimination method is preferred when both equations have large coefficients.
Answer: It avoids fractions early, keeping calculations tidy. - Given the system 4x + y = 20 and 2x ‑ 3y = ‑4, solve and verify.
Answer: x = 4, y = 4; both equations hold. - State two situations where substitution might be faster than elimination.
Answer: When one equation already has a variable isolated, or when one variable’s coefficient is 1.