Elimination method

The second algebraic method, elimination, is often faster than substitution when no coefficient is $\pm 1$. The idea is to align the coefficients of one variable in the two equations so that adding or subtracting kills that variable.

Algorithm

1. Pick the variable you want to eliminate. Multiply each equation by a suitable non-zero number so that the chosen variable has equal (or opposite) coefficients in both equations.
2. Add the equations (if the coefficients are opposite) or subtract them (if equal). The chosen variable disappears, leaving a single linear equation in the other variable.
3. Solve that equation.
4. Substitute back into one of the original equations to find the eliminated variable.
5. Verify with both original equations.

A common shortcut: to eliminate $x$, multiply the first equation by $a_2$ and the second by $a_1$, then subtract. To eliminate $y$, multiply by $b_2$ and $b_1$ and subtract.

If after elimination you obtain the true statement $0 = 0$, the system has infinitely many solutions. If you obtain a false statement $0 = k$ (with $k \ne 0$), the system has no solution.

Why this is often the cleanest method

For typical exam systems with integer or simple fractional coefficients, the LCM of the coefficients of $x$ (or of $y$) is small, so multiplying by them keeps the arithmetic manageable. The advantage of elimination over substitution is that you avoid fractions until the last step.

It is also adaptable to simultaneous equations in disguise: equations involving $\dfrac{1}{x}$ and $\dfrac{1}{y}$, age problems, fractions, ratios , all become linear after a substitution like $u = 1/x, v = 1/y$, and then elimination is the natural tool.

Worked examples

Example 1. Solve $2 x + 3 y = 11$ and $5 x - 4 y = -3$.

Eliminate $x$. Multiply first by $5$ and second by $2$: $10 x + 15 y = 55$ and $10 x - 8 y = -6$.

Subtract: $23 y = 61 \Rightarrow y = 61/23$.

Hmm, that doesn't look clean. Let me re-check the system. The system should have a clean integer solution. Let me try eliminating $y$ instead.

Multiply first by $4$ and second by $3$: $8 x + 12 y = 44$ and $15 x - 12 y = -9$.

Add: $23 x = 35 \Rightarrow x = 35/23$. Same non-integer result. So the answer is genuinely $(35/23, 61/23)$.

(Some board-style problems do have rational, not integer, solutions , the method still works fine.)

Example 2. Solve $3 x + 4 y = 10$ and $2 x - 2 y = 2$.

Eliminate $y$. Multiply the second by $2$: $4 x - 4 y = 4$. Add to the first: $(3 x + 4 y) + (4 x - 4 y) = 10 + 4 \Rightarrow 7 x = 14 \Rightarrow x = 2.$ Substitute into the first: $6 + 4 y = 10 \Rightarrow y = 1$.

Solution $(2, 1)$.

Example 3. Solve $9 x - 4 y = 2 0$ and $7 x - 3 y = 1 5$.

Eliminate $y$. Multiply first by $3$ and second by $4$: $27 x - 12 y = 60$ and $28 x - 12 y = 60$.

Subtract: $-x = 0 \Rightarrow x = 0$. Then from the first: $-4 y = 20 \Rightarrow y = -5$.

Solution $(0, -5)$. Verify: $7(0) - 3(-5) = 15$ ✓.

Example 4. Solve $\dfrac{5}{x - 1} + \dfrac{1}{y - 2} = 2$ and $\dfrac{6}{x - 1} - \dfrac{3}{y - 2} = 1$.

Let $u = 1/(x - 1), v = 1/(y - 2)$. System: $5 u + v = 2$, $6 u - 3 v = 1$.

Eliminate $v$. Multiply the first by $3$: $15 u + 3 v = 6$. Add to the second: $21 u = 7 \Rightarrow u = 1/3$.

Then $v = 2 - 5/3 = 1/3$. So $1/(x - 1) = 1/3 \Rightarrow x = 4$, and $1/(y - 2) = 1/3 \Rightarrow y = 5$.

Solution $(4, 5)$.

Example 5. Solve $4 x + 7 y = 16$ and $4 x + 7 y = 18$.

Same coefficients of $x$ and $y$ but different constants $\Rightarrow$ no solution.

Subtracting: $0 = -2$. False $\Rightarrow$ inconsistent.

Try it yourself

1. Solve by elimination: $x + y = 5, 2 x - 3 y = 4$.
2. Solve: $2 x + 3 y = 8, 4 x + 6 y = 7$.
3. Solve: $3 x - 5 y = 4, 9 x - 2 y = 7$.
4. Solve: $\dfrac{x}{3} + \dfrac{y}{2} = 8, \dfrac{5 x}{2} - \dfrac{y}{3} = 7$.
5. Solve: $2(x + y) = 5 (x - y), 8 x - 3 y = 19$.
6. Solve: $11 x + 15 y = -23, 7 x - 2 y = 20$.
7. Solve: $\dfrac{2}{x + y} + \dfrac{3}{x - y} = 17/5, \dfrac{4}{x + y} - \dfrac{5}{x - y} = -23/15$.
8. Solve: $a x + b y = a + b, b x + a y = a + b$ ($a + b \ne 0$).
9. Solve: $5(x + 1) - 2(y + 2) = 3, 3 x - 2 y = 4$.
10. Solve and discuss: $3 x + 4 y = 12, 6 x + 8 y = 24$.

Pitfalls / Insight

• Multiply each equation in full , students often forget to multiply the constant term.
• Sign of the result of subtraction matters: $(8 x) - (15 x) = -7 x$, not $7 x$.
• Always verify in both original equations; elimination is more error-prone than substitution due to multiplied coefficients.

Insight. Elimination is matrix algebra in disguise: making one column equal so you can "row reduce" the system. Class XI and XII will pick this up formally as Gaussian elimination.