Chapter 4.3 — Exercise 4.2 — Elimination and Substitution

Substitution, Step by Step

For 2x + y − 4 = 0 and x + 2y − 5 = 0, substitution isolates one variable in one equation, then feeds it into the other:

• Step 1: from the first equation, y = 4 − 2x.
• Step 2: substitute into the second: x + 2(4 − 2x) − 5 = 0.
• Step 3: simplify: x + 8 − 4x − 5 = 0 → −3x + 3 = 0 → x = 1.
• Step 4: substitute x = 1 back into y = 4 − 2x: y = 2.
• Step 5: check both originals: 2(1) + 2 − 4 = 0 ✓ and 1 + 2(2) − 5 = 0 ✓.

Elimination, Step by Step

For 3x + 2y = 11 and 2x + 3y = 4, elimination instead scales both equations so one variable cancels when they're combined:

• Step 1–2: multiply the first by 2 (6x + 4y = 22) and the second by 3 (6x + 9y = 12), matching the x-coefficient in both.
• Step 3: subtract: (6x + 4y) − (6x + 9y) = 22 − 12, giving −5y = 10, so y = −2.
• Step 4: substitute into the first original: 3x + 2(−2) = 11 → 3x = 15.
• Step 5: solve: x = 5. So x = 5, y = −2.

Substitution suits equations where one variable is already isolated or easy to isolate; elimination suits equations where matching a coefficient by multiplication is quicker than rearranging. Both always reach the same answer — the choice is purely about which path involves less arithmetic for a given pair. A useful habit before starting either method: glance at both equations first and check whether any variable already has a coefficient of 1 or −1 somewhere — that's usually the fastest signal that substitution will be less work than elimination for that particular pair.

Incomes From a Ratio

Two people have incomes in ratio 9:7 and expenditures in ratio 4:3, each saving ₹2000 a month. Writing incomes as 9x and 7x, expenditures as 4y and 3y:

Eliminating y (multiply the first by 7, the second by 9, then subtract) gives x = 2000, so the incomes are 9(2000) = ₹18,000 and 7(2000) = ₹14,000. Notice the setup itself: naming the incomes 9x and 7x — rather than two unrelated letters — bakes the given 9:7 ratio directly into the algebra, so the final answer is automatically in that ratio without needing a separate check.

A Two-Digit Number From Its Reverse

A two-digit number plus its digit-reversal equals 66, and the digits differ by 2. Writing the units digit as x and tens digit as y, the number is 10y + x and its reverse is 10x + y:

Solving gives x = 4, y = 2 — so the number is 10(2) + 4 = 24, and its reverse, 42, is an equally valid answer, since the problem never specifies which digit comes first. Both satisfy the original conditions: 24 + 42 = 66, and the digits 2 and 4 differ by exactly 2. The step from 11x + 11y = 66 to x + y = 6 is worth noticing on its own — dividing an entire equation by a common factor before combining it with the second equation often keeps the arithmetic that follows noticeably smaller.

Recovering a Fraction From Two Clues

A fraction becomes 4/5 when 1 is added to both its numerator and denominator, and becomes 1/2 when 5 is subtracted from both. Writing the fraction as x/y and translating each clue into an equation:

From the second equation, y = 2x − 5. Substituting into the first: 5x − 4(2x − 5) = −1 → 5x − 8x + 20 = −1 → −3x = −21 → x = 7, then y = 2(7) − 5 = 9. The original fraction is 7/9 — and checking both clues confirms it: (7+1)/(9+1) = 8/10 = 4/5 ✓, and (7−5)/(9−5) = 2/4 = 1/2 ✓.

Six More Problems, Solved

The same two methods handle every situation below — only the setup changes, never the underlying algebra:

The acid-mixture and investment problems share the same underlying structure: a total-quantity equation paired with a weighted-average equation, where the weights are percentages instead of plain counts. Recognising that shared shape makes the setup automatic once you've solved one of the two — ₹12,000 split at 10% and 15% averaging 12% is really the same skeleton as 100 ml of acid split at 50% and 80% averaging 68%, just with money and interest rates standing in for volume and concentration.

The two cars problem is worth a second look at its setup, since "same direction" and "opposite direction" translate into surprisingly different equations from the same 100 km gap. Travelling the same direction, the faster car has to cover the full 100 km gap relative to the slower one before catching up, over 5 hours — giving 5x − 5y = 100, or x − y = 20. Travelling toward each other, their combined distance covered has to add up to the 100 km gap in just 1 hour — giving x + y = 100. The same 100 km appears in both equations, but as a difference in one and a sum in the other, purely because of which direction the cars are moving.

Where Elimination Falls Short

Every equation solved in this exercise, no matter how the word problem was dressed up, was genuinely linear once translated. Exercise 4.3 shows what to do when that's not quite true — equations built from 1/x and 1/y terms, which look nonlinear until a clever substitution turns them linear again, at which point the exact same substitution and elimination methods from this exercise take over. For the graphical picture behind why a unique solution exists at all, see Exercise 4.1, which plots several of these same kinds of pairs by hand.