In short
You just sweated through elimination and got x = 2. Stop. Don't run elimination AGAIN to find y — just SUBSTITUTE x = 2 into any one of the original equations and solve. Back-substitution takes 5 seconds. Re-running elimination takes another 30 seconds and gives you a fresh chance to mess up arithmetic. Same answer, double the risk.
You're sitting in a CBSE Class 10 board exam. You've got a system of two equations, you eliminated y, you found x = 2. Now you need y. You have two options:
- Back-substitute: Plug x = 2 into one of the original equations. Solve a one-variable equation. Done.
- Re-eliminate: Set up the multipliers AGAIN, this time to kill x instead of y. Multiply, subtract, simplify, divide.
Option 2 looks "fair" — it treats x and y symmetrically. But it's a trap. You're doing the same problem twice. Every multiplication is another chance to drop a sign. Every subtraction is another chance to mess up 13 - 8. Why: each arithmetic operation has a small probability of error; the more operations you do, the higher the chance at least one slips.
A worked system, the right way
Eliminate y, then back-substitute
Solve:
Step 1 — Eliminate y. Multiply equation (2) by 3 to match the y-coefficient:
Subtract (1) from (2'):
Step 2 — Back-substitute into the simpler equation. Equation (2) has the smaller coefficients, so use it:
Step 3 — Check in the OTHER equation. Plug both into (1): 2(2) + 3(3) = 4 + 9 = 13. ✓
Solution: (x, y) = (2, 3). Total operations after finding x: one multiplication, one subtraction. About 5 seconds.
The same system, the wasteful way
Re-eliminate to find y (don't do this)
Same system:
You already have x = 2. But suppose you ignore that and set up elimination AGAIN, this time to kill x.
Multiply (1) by 2:
Subtract (2) from (1'):
Same answer, y = 3. But count the work: one multiplication of a whole equation, one subtraction of a whole equation, one division. Why redundant: you already KNOW x = 2. The new elimination derives y from scratch, ignoring information you've already paid for.
Worse: if you slip and get y = 4 here, now you have a contradiction with the back-substituted answer — and no idea which one is right. Two answers, no tiebreaker.
Why back-substituting is genuinely safer
Counting the chances of error
Back-substitution after x = 2 into 4x + y = 11:
- Multiply: 4 \times 2 = 8. (1 operation)
- Subtract: 11 - 8 = 3. (1 operation)
Total: 2 arithmetic operations.
Re-elimination to find y:
- Multiply equation (1) by 2: three terms updated → 4x, 6y, 26. (3 operations)
- Subtract equation (2) term by term: 4x - 4x, 6y - y, 26 - 11. (3 operations)
- Divide: 15 / 5 = 3. (1 operation)
Total: 7 arithmetic operations.
If your per-operation error rate is even 2%, the chance of a clean back-substitution is 0.98^2 \approx 96\%. The chance of a clean re-elimination is 0.98^7 \approx 87\%. Why this matters: in a 3-hour board exam with 30+ calculations, those extra 5 operations per problem add up to several wrong answers across the paper.
The post-x workflow
Pick the easier equation
After you have x, you can substitute into either original equation. They both work. But one is usually easier than the other — and "easier" always wins. Look for:
- Smaller coefficients. 4x + y = 11 beats 7x - 5y = -1. Multiplying by 4 is faster than multiplying by 7.
- A coefficient of 1 on y. If one equation has y standing alone (like 4x + y = 11), that's free — no division needed.
- Smaller constants on the right. 4x + y = 11 beats 4x + y = 137 because 11 - 8 is cleaner than 137 - 8.
For the system in our examples, equation (2) is 4x + y = 11 — coefficient 1 on y, small numbers everywhere. Pick it. Equation (1) is 2x + 3y = 13, which would force you to divide by 3 at the end. Why this matters: dividing 13 - 4 = 9 by 3 gives a clean 3 here, but in many problems the division produces fractions you have to simplify. Avoid it when you can.
The exam-time payoff
Every CBSE Class 10 paper has at least one pair-of-equations problem, often two. If you back-substitute correctly, each saves you about 30 seconds and one possible arithmetic slip. Across a paper, that's a free 1–2 minutes for re-checking other questions, plus one fewer silly mistake. Why this compounds: in a board exam, every saved minute is a minute you can spend on the geometry proof you weren't sure about.
The same rule scales up. In three-equation systems (CBSE Class 11, JEE), once you've found z, back-substitute into a two-variable equation to get y, then back-substitute both into a one-variable equation for x. Each step is a one-variable solve. Never re-do the whole elimination.
The one-line takeaway
Elimination is expensive. Substitution is cheap. The moment you have one variable, switch from expensive to cheap.
References
- NCERT, Mathematics Class 10, Chapter 3: Pair of Linear Equations in Two Variables — the official syllabus source.
- Khan Academy, Solving systems by elimination — practice problems with worked back-substitution.
- Paul's Online Math Notes, Linear Systems with Two Variables — clear treatment of substitution after elimination.
- Art of Problem Solving, Systems of equations — competition-level technique notes.