Algebraic Method — Elimination

The Elimination Idea

The elimination method is often the fastest way to solve a pair, especially when no variable has coefficient 1. The idea is to make the coefficients of one variable equal (in size), then add or subtract the equations so that variable cancels out.

The idea: Multiply the equations by suitable numbers so one variable has matching coefficients. Add (if opposite signs) or subtract (if same signs) to eliminate it, leaving one equation in one variable.

Think of it this way: we 'line up' one variable and knock it out, leaving a simple equation to solve.

[Board Important] Elimination is the go-to method when both equations have awkward coefficients (e.g. $2x + 3y = 8$ and $3x - 2y = 1$).

The Elimination Steps

Let's solve $2x + 3y = 8$ and $3x + 2y = 7$.

1. Make one variable's coefficients equal. To eliminate $x$, multiply the first by 3 and the second by 2:

• $6x + 9y = 24$
• $6x + 4y = 14$

1. Subtract (same sign on $x$): $(6x + 9y) - (6x + 4y) = 24 - 14 \Rightarrow 5y = 10 \Rightarrow y = 2$.
2. Back-substitute: $2x + 3(2) = 8 \Rightarrow 2x = 2 \Rightarrow x = 1$.

So the solution is $x = 1$, $y = 2$.

Key Point: If the matching coefficients have the same sign, subtract; if opposite signs, add. Either way the targeted variable disappears.

[Board Important] Choose to eliminate whichever variable needs the smaller multipliers — it keeps the arithmetic clean.

Add or Subtract? A Quick Guide

The single most common slip is adding when you should subtract (or vice versa). Here's the rule:

• Coefficients like $+6x$ and $+6x$ (same sign): subtract the equations.
• Coefficients like $+6x$ and $-6x$ (opposite signs): add the equations.

Tip

Sometimes it's easier to arrange for opposite signs and add, since adding avoids sign errors. For $3x + 2y = 7$ and $2x - 3y = 4$, to eliminate $y$ multiply the first by 3 and the second by 2: $9x + 6y = 21$ and $4x - 6y = 8$. Now the $y$-terms are opposite, so add: $13x = 29$.

Key Point: Always check the signs of the matched coefficients before deciding to add or subtract.

[Board Important] Write the two aligned equations one above the other before combining — it makes the add/subtract decision obvious and reduces errors.

Special Cases and Word Problems

Like substitution, elimination reveals the special cases:

• If both variables cancel and you get a false statement (e.g. $0 = 6$) → no solution.
• If both cancel to a true statement (e.g. $0 = 0$) → infinitely many solutions.

Word-problem recipe

1. Define variables, form two equations.
2. Eliminate one variable.
3. Solve, back-substitute, and interpret with units.

Key Point: Elimination shines in word problems where the two equations naturally have similar structure, such as 'cost of $a$ of one item and $b$ of another' types.

[Board Important] Many 3-mark word problems (cost, speed, fractions, ages) can be solved by either substitution or elimination — pick whichever gives cleaner numbers.

Solved Examples

Example 1: Basic elimination (subtract)

Solve: $2x + 3y = 8$ and $3x + 2y = 7$.

Solution:

1. Multiply first by 3, second by 2: $6x + 9y = 24$ and $6x + 4y = 14$.
2. Subtract: $5y = 10 \Rightarrow y = 2$.
3. Back-substitute: $2x + 6 = 8 \Rightarrow x = 1$.

Final Answer: $x = 1$, $y = 2$.

Takeaway: Match coefficients, subtract, then back-substitute.

Example 2: Elimination (add)

Solve: $3x + 2y = 7$ and $2x - 3y = 4$, eliminating $y$.

Solution:

1. Multiply first by 3, second by 2: $9x + 6y = 21$ and $4x - 6y = 8$.
2. Add (opposite $y$ signs): $13x = 29 \Rightarrow x = \dfrac{29}{13}$.
3. Back-substitute into $3x + 2y = 7$: $2y = 7 - 3\cdot\dfrac{29}{13} = 7 - \dfrac{87}{13} = \dfrac{4}{13} \Rightarrow y = \dfrac{2}{13}$.

Final Answer: $x = \dfrac{29}{13}$, $y = \dfrac{2}{13}$.

Takeaway: Opposite signs ⇒ add to eliminate.

Example 3: Simple add (ready-made)

Solve: $x + y = 7$ and $x - y = 3$.

Solution:

1. Add the equations: $2x = 10 \Rightarrow x = 5$.
2. Subtract: $2y = 4 \Rightarrow y = 2$.

Final Answer: $x = 5$, $y = 2$.

Takeaway: When coefficients already match, just add and subtract directly.

Example 4: Eliminate after scaling

Solve: $4x + 3y = 24$ and $3x - 2y = 8$.

Solution:

1. To eliminate $y$: multiply first by 2, second by 3: $8x + 6y = 48$ and $9x - 6y = 24$.
2. Add: $17x = 72$… let's recheck: actually $8x + 9x = 17x$ and $48 + 24 = 72$, so $x = \dfrac{72}{17}$. Hmm — let's instead verify with cleaner elimination of $x$.
3. Eliminate $x$: multiply first by 3, second by 4: $12x + 9y = 72$ and $12x - 8y = 32$. Subtract: $17y = 40 \Rightarrow y = \dfrac{40}{17}$, and $x = \dfrac{72}{17}$.

Final Answer: $x = \dfrac{72}{17}$, $y = \dfrac{40}{17}$.

Takeaway: Not all systems give whole numbers; carry fractions carefully.

Example 5: No solution by elimination

Solve: $2x + 3y = 5$ and $4x + 6y = 15$.

Solution:

1. Multiply first by 2: $4x + 6y = 10$.
2. Subtract from the second: $(4x + 6y) - (4x + 6y) = 15 - 10 \Rightarrow 0 = 5$ (false).

Final Answer: No solution (inconsistent).

Takeaway: Both variables cancel to a false statement ⇒ no solution.

Example 6: Infinitely many by elimination

Solve: $x + 2y = 4$ and $3x + 6y = 12$.

Solution:

1. Multiply first by 3: $3x + 6y = 12$.
2. Subtract from the second: $0 = 0$ (true).

Final Answer: Infinitely many solutions.

Takeaway: $0 = 0$ ⇒ same line ⇒ infinitely many solutions.

Example 7: Cost word problem

2 chairs and 3 tables cost ₹1300; 3 chairs and 2 tables cost ₹1200. Find the cost of each.

Solution:

1. Let chair $= x$, table $= y$: $2x + 3y = 1300$, $3x + 2y = 1200$.
2. Multiply first by 3, second by 2: $6x + 9y = 3900$, $6x + 4y = 2400$. Subtract: $5y = 1500 \Rightarrow y = 300$.
3. $2x + 900 = 1300 \Rightarrow x = 200$.

Final Answer: Chair ₹200, table ₹300.

Takeaway: Cost problems with two items per equation suit elimination well.

Example 8: Incomes and expenditures

Two persons' incomes are in ratio 9:7 and expenditures in ratio 4:3. Each saves ₹2000. If incomes are $9k$ and $7k$ and expenditures $4m$ and $3m$, then $9k - 4m = 2000$ and $7k - 3m = 2000$. Find $k$.

Solution:

1. Multiply first by 3, second by 4: $27k - 12m = 6000$, $28k - 12m = 8000$.
2. Subtract: $k = 2000$.

Final Answer: $k = 2000$ (so incomes are ₹18000 and ₹14000).

Takeaway: Elimination handles ratio-based savings problems neatly.

Example 9: Choose the easier variable to eliminate

Solve: $x + 4y = 14$ and $7x - 3y = 5$.

Solution:

1. Eliminate $x$: multiply first by 7: $7x + 28y = 98$.
2. Subtract the second: $(7x + 28y) - (7x - 3y) = 98 - 5 \Rightarrow 31y = 93 \Rightarrow y = 3$.
3. $x + 12 = 14 \Rightarrow x = 2$.

Final Answer: $x = 2$, $y = 3$.

Takeaway: Multiplying just one equation can be enough when a coefficient is 1.

Example 10: Digits problem by elimination

The sum of the digits of a two-digit number is 12. The number is 18 more than the number formed by reversing the digits. Find the number.

Solution:

1. Tens $= x$, units $= y$: $x + y = 12$ and $(10x + y) - (10y + x) = 18 \Rightarrow 9x - 9y = 18 \Rightarrow x - y = 2$.
2. Add $x + y = 12$ and $x - y = 2$: $2x = 14 \Rightarrow x = 7$, $y = 5$.
3. Number $= 75$.

Final Answer: The number is 75.

Takeaway: Adding the digit-sum and digit-difference equations eliminates $y$ instantly.