Why learn to solve linear equations?
Imagine you’re trying to find the exact point where two roads cross. Those roads are like two equations – the crossing point is the solution. Mastering the tricks to locate that point makes the whole math journey smoother and scores you extra marks in the exam.
💡 In Simple Words: A pair of linear equations is just two straight‑line formulas that share the same x‑ and y‑values. Solving them means finding the one (or none) set of numbers that makes both formulas true at the same time. You can do it by swapping one equation into the other (substitution) or by adding/subtracting them to cancel a variable (elimination).
Substitution method – step‑by‑step
Step 1: Choose the easier equation. Look for the one where a variable already has a coefficient of 1 (or -1). That makes the next step painless.
Step 2: Isolate the variable. Rearrange the chosen equation so that one variable stands alone, like x = 3y + 2.
Step 3: Plug it into the other equation. Replace the isolated variable in the second equation with the expression you just found. It’s like swapping a puzzle piece for a new shape that fits.
Step 4: Solve for the remaining variable. You now have a single‑variable equation – solve it just as you would a normal algebra problem.
Step 5: Back‑substitute. Put the value you found back into the expression from Step 2 to get the other variable.
Step 6: Check. Plug both numbers into the original pair to confirm they satisfy both equations.
Worked example (substitution)
Solve: \(2x + 3y = 12\) and \(x - y = 1\).
- Step 1 & 2: The second equation already has a 1 in front of x, so isolate x: \(x = y + 1\).
- Step 3: Substitute \(x\) in the first equation: \(2(y+1) + 3y = 12\).
- Step 4: Expand and combine: \(2y + 2 + 3y = 12 \Rightarrow 5y = 10 \Rightarrow y = 2\).
- Step 5: Back‑substitute into \(x = y + 1\): \(x = 2 + 1 = 3\).
- Step 6: Check: \(2(3)+3(2)=6+6=12\) ✔️ and \(3-2=1\) ✔️.
Elimination method – step‑by‑step
Step 1: Write both equations in standard form. That means all variable terms on the left, constants on the right, like \(ax + by = c\).
Step 2: Make the coefficients of one variable opposites. Multiply one or both equations by suitable numbers so that adding them cancels that variable. Think of it as turning two opposite‑facing gears to lock together.
Step 3: Add or subtract the equations. The chosen variable disappears, leaving a single‑variable equation.
Step 4: Solve for the remaining variable.
Step 5: Substitute back. Put the value you just found into either original equation to get the other variable.
Step 6: Verify. As always, plug the pair back into both original equations.
Worked example (elimination)
Solve: \(4x - 2y = 6\) and \(3x + y = 7\).
- Step 1: Equations are already in standard form.
- Step 2: Make the y‑coefficients opposites. Multiply the second equation by 2: \(6x + 2y = 14\).
- Step 3: Add the modified second equation to the first: \((4x-2y) + (6x+2y) = 6 + 14\) → \(10x = 20\).
- Step 4: \(x = 2\).
- Step 5: Substitute into \(3x + y = 7\): \(3(2) + y = 7 \Rightarrow y = 1\).
- Step 6: Check: \(4(2)-2(1)=8-2=6\) ✔️ and \(3(2)+1=7\) ✔️.
When to use which method?
| Situation | Best method | Why? |
|---|---|---|
| A variable already has coefficient 1 (or -1) | Substitution | Isolating the variable is trivial, saving time. |
| Coefficients are easy to make opposites | Elimination | Multiplying small numbers avoids fractions. |
| Both equations contain fractions | Elimination | Clear fractions first, then cancel a variable. |
| One equation is already solved for a variable | Substitution | Direct plug‑in works fastest. |
Quick visual guide
📝 Likely Exam Questions
- Solve by substitution: \(x + 2y = 9\), \(3x - y = 4\).
Answer: \(x = 3, y = 3\). - Solve by elimination: \(5x + 3y = 27\), \(2x - 3y = 4\).
Answer: \(x = 5, y = 2\). - Check the solution: Verify that \((x, y) = (4, -1)\) satisfies \(2x - y = 9\) and \(x + 3y = 1\).
Answer: Both equations hold true. - Word problem: A theater sells 120 tickets. Adult tickets cost \₹150 and child tickets \₹100. If the total collection is \₹15,500, how many adult tickets were sold?
Answer: 70 adult tickets and 50 child tickets (use elimination). - Conceptual: Explain why elimination works even if you have to multiply both equations by fractions.
Answer: Multiplying keeps the equality true; opposite coefficients guarantee the variable cancels when added.