Chapter 4.2 — Exercise 4.1 — Graphical Method

Three Pairs, Classified by Ratio Alone

Comparing a₁/a₂, b₁/b₂, and c₁/c₂ for three given pairs settles each one without drawing anything — a direct application of the exact test introduced in the chapter's opening lesson, now run on three fresh pairs of coefficients:

Two Pairs, Solved and Verified by Graph

For 3x + 2y = 5 and 2x − 3y = 7, the ratios a₁/a₂ = 3/2 and b₁/b₂ = 2/(−3) are unequal, so the pair is consistent with a unique solution. Plotting both lines confirms it: they cross at (2.2, 0.8).

For 2x − 3y = 8 and 4x − 6y = 9, the ratios a₁/a₂ = 2/4 = 1/2 and b₁/b₂ = −3/(−6) = 1/2 are equal, but c₁/c₂ = −8/−9 = 8/9 doesn't match — so the pair is inconsistent. The graph confirms it: two parallel lines that never meet, with no solution. Notice how close this pair is to being coincident: the second equation is almost exactly double the first, and only the constant term — 9 instead of the 16 that true doubling would require — keeps the two lines from landing on top of each other.

Seven More Pairs, Classified and Solved

The same consistent-or-inconsistent check, run across the exercise's remaining pairs, produces every possible outcome from the ratio test:

Three separate pairs turn out coincident here, not just one — a reminder that "infinitely many solutions" isn't a rare edge case in this exercise, it's a genuinely common outcome whenever one equation is just a scaled copy of the other. The two "no solution" rows are worth comparing directly too: x−y=8 vs 3x−3y=16 and 2x−2y=2 vs 4x−4y=5 both follow the identical pattern — the second equation's coefficients are exactly a whole-number multiple of the first, but the constant term breaks the pattern by just enough to keep the lines from ever actually meeting or coinciding.

Real Situations, Turned Into Line Pairs

Word problems in this exercise translate into exactly the same graphing process, once the right two variables are named — the graphing itself never changes, only the naming step at the very start.

• Neha's shopping: skirts are "two less than twice the pants" and "four less than four times the pants" — y = 2x − 2 and y = 4x − 4. The lines cross at (1, 0): 1 pant, 0 skirts.
• A maths quiz: 10 students total, with 4 more girls than boys — x + y = 10 and y = x + 4. The lines cross at (3, 7): 3 boys, 7 girls, and 3+7=10 checks out against the total given.
• Pencils and pens: 5 pencils + 7 pens cost ₹50, and 7 pencils + 5 pens cost ₹46 — 5x + 7y = 50 and 7x + 5y = 46. The lines cross at (3, 5): pencils cost ₹3, pens cost ₹5. Checking: 5(3)+7(5) = 15+35 = 50 ✓, and 7(3)+5(5) = 21+25 = 46 ✓ — both original conditions hold at once.
• A rectangular garden: half the perimeter is 36 m, and the length is 4 m more than the width — x + y = 36 and x = y + 4. The lines cross at (20, 16): length 20 m, width 16 m, and indeed 20+16=36 confirms half the perimeter matches exactly.

Building Your Own Companion Equation

Given a single equation, 2x + 3y − 8 = 0, the ratio test also works to construct a second equation with a required relationship, rather than only classify one that's already given. There's no single correct answer for any of the three cases below — infinitely many equations would work equally well — so what matters is understanding why the chosen example satisfies the required ratio pattern.

• Intersecting: any equation with a different a₁/a₂ ratio, such as 4x + 9y − 7 = 0 — here 4/2 ≠ 9/3, so the lines are guaranteed to cross somewhere.
• Parallel: keep a₁/a₂ = b₁/b₂ but change the constant, such as 8x + 12y + 13 = 0 — the coefficients scale by 4 throughout (8/2 = 12/3 = 4), but the constant doesn't match that same scaling, so the lines run side by side forever.
• Coincident: scale every term by the same factor, such as 6x + 9y − 24 = 0 (exactly 3 times the original, constant included) — since every term scales identically, this "different-looking" equation describes the exact same line.

Two Larger Applications

The same technique scales up to more involved word problems without changing the method at all. A rectangle whose area shrinks by 80 sq units when the length is reduced by 5 and the width is increased by 2, but grows by 50 sq units under a different length/width change, translates to 2x − 5y + 70 = 0 and x − 2y + 20 = 0 — solved graphically, the lines cross at (40, 30): length 40 units, width 30 units. And a classroom where 3-per-bench leaves one student standing but 4-per-bench leaves one bench empty translates to y = 3x + 1 and y = 4x − 4 — the lines cross at (5, 16): 5 benches, 16 students, with both seating arrangements checking out against that count. Both problems needed one extra step before any graphing could begin: translating a sentence about "reduced area" or "one bench empty" into an actual algebraic equation, which is consistently the harder half of any word problem — once the two equations are written down correctly, solving them is identical to every other pair in this exercise.

Faster Than Plotting

Every solution in this exercise came from drawing two lines and reading off where they cross — reliable, and genuinely useful for building intuition about what "no solution" or "infinitely many" actually look like, but slow, and imprecise whenever the intersection lands on a fraction like (2.2, 0.8) or (3.1, 1.4). Exercise 4.2 replaces plotting with two purely algebraic methods that reach the exact same answers without ever drawing an axis. For the ratio-test classification this exercise leans on throughout, revisit the chapter introduction, where all three outcomes were first established.