Ever wondered how to find the exact point where two roads cross on a map? That’s the same idea behind solving simultaneous linear equations – finding the one spot that satisfies both equations at once.
In simple words, a pair of equations that share the same x and y are like two riddles that point to the same secret number. By using a clever trick, we uncover that number without guessing.
What are Simultaneous Linear Equations?
A simultaneous linear equation (or simply a system of equations) is a set of two or more equations that involve the same variables. "Linear" means each equation draws a straight line when you plot it on a graph. The word "simultaneous" tells you we need to solve them together, because the answer must work for every equation at the same time.
For example,
2x + 3y = 12
4x - y = 5
Both equations talk about x and y. The pair of numbers (x, y) that makes the left‑hand side equal the right‑hand side in each line is the solution.
Why learn these methods?
Beyond the exam, simultaneous equations pop up in everyday puzzles – budgeting where two expenses add up to a total, mixing ingredients in a recipe, or even figuring out the speed of two cars moving towards each other.
Method 1: Substitution Method
The word substitution means "put one thing inside another". First, we solve one of the equations for a single variable, then we plug (substitute) that expression into the other equation.
Steps
- Pick the easier equation to isolate a variable.
- Rewrite that equation so the chosen variable stands alone (e.g., y = ...).
- Replace the isolated variable in the second equation with the expression you just found.
- Solve the resulting single‑variable equation.
- Back‑substitute the found value into the expression from step 2 to get the other variable.
Worked Example
Solve:
3x + 2y = 16
x - y = 1
1. From the second equation, isolate x: x = y + 1.
2. Substitute x in the first equation: 3(y + 1) + 2y = 16.
3. Expand: 3y + 3 + 2y = 16 → 5y + 3 = 16.
4. Subtract 3: 5y = 13 → y = 13/5 = 2.6.
5. Back‑substitute: x = 2.6 + 1 = 3.6.
Solution: (x, y) = (3.6, 2.6).
Method 2: Elimination (Addition) Method
The elimination method means we add or subtract the equations so that one variable disappears, leaving a single‑variable equation.
Steps
- Look at the coefficients (the numbers in front of the variables) of x or y in both equations.
- Multiply one or both equations by suitable numbers so that the coefficients of the chosen variable become opposites (e.g., +4 and -4).
- Add or subtract the equations; the chosen variable cancels out.
- Solve the remaining single‑variable equation.
- Plug the found value back into one of the original equations to get the second variable.
Worked Example
Solve:
5x - 2y = 9
3x + 2y = 7
1. Notice the y‑coefficients are -2 and +2 – they’re already opposites!
2. Add the two equations: (5x - 2y) + (3x + 2y) = 9 + 7 → 8x = 16.
3. Divide by 8: x = 2.
4. Substitute x into the second original equation: 3(2) + 2y = 7 → 6 + 2y = 7 → 2y = 1 → y = 0.5.
Solution: (x, y) = (2, 0.5).
Method 3: Cross‑Multiplication (for two‑variable equations)
This shortcut works when both equations are written in the form ax + by = c. We cross‑multiply the coefficients and constants to directly find one variable.
Quick Steps
- Write the equations as a1x + b1y = c1 and a2x + b2y = c2.
- Compute x = (c1b2 - c2b1) / (a1b2 - a2b1).
- Compute y = (a1c2 - a2c1) / (a1b2 - a2b1).
While fast, it’s best to double‑check the denominator isn’t zero (that would mean the lines are parallel or the same).
When to Use Which Method?
| Method | Best For | Pros | Cons |
|---|---|---|---|
| Substitution | One equation already solved for a variable or easy to isolate. | Clear steps, works well with fractions. | Can create messy fractions if coefficients are large. |
| Elimination | Coefficients can be made equal or opposite easily. | Usually quicker, avoids fractions until the end. | May need extra multiplication, watch sign errors. |
| Cross‑Multiplication | Both equations in standard form and you need a fast answer. | Very fast for 2‑variable systems. | Easy to mis‑apply; denominator zero means no unique solution. |
Step‑by‑Step Flowchart (Elimination Method)
📝 Likely Exam Questions
- Q1. Solve by substitution: 2x + y = 7, 3x - 2y = 4.
Answer: x = 2, y = 3. - Q2. Using elimination, find x and y: 4x + 5y = 20, 2x - 5y = -2.
Answer: x = 2, y = 2. - Q3. A word problem: A shop sells pens at $2 each and notebooks at $5 each. If a student buys 3 pens and 2 notebooks spending $19, how many pens and notebooks did they actually buy? Set up simultaneous equations and solve.
Answer: 4 pens and 3 notebooks. - Q4. State one advantage of the elimination method over substitution for the system: 6x - y = 11, 2x + y = 5.
Answer: Eliminates the variable without creating fractions, leading to a quicker solution. - Q5. Verify whether the pair (x, y) = (1, 2) satisfies the system: 3x + 4y = 11, 5x - y = 3.
Answer: Substituting gives 3(1)+4(2)=11 ✔ and 5(1)-2=3 ✔, so it is a correct solution.