NCERT Exemplar Solutions Class 10 Maths Chapter 3

Slope-intercept cross-check. Reading the answer straight off the ratio test is fastest, but converting both lines to y=mx+c form gives a second, independent confirmation that the lines truly are parallel rather than coincident.

• Idea: Two lines are parallel when their slopes are equal but their y-intercepts differ.
• First line: Solve the first equation for y: 6x-3y+10=0 ⇒ 3y=6x+10 ⇒ y=2x+103.
• Second line: Solve the second equation for y: 2x-y+9=0 ⇒ y=2x+9.
• Slopes: Compare slopes: both have slope m=2, so the lines never tilt toward each other.
• Intercepts: Compare intercepts: 103≈ 3.33 while the other is 9. Equal slope but unequal intercept is the textbook signature of parallel lines, confirming there is no common point.

Equal slope 2, different intercepts ⇒ parallel; option (D).

Test by scaling one equation. Instead of three separate ratios, I scale the first equation and watch whether it can ever become the second; if the variable parts match but the constants clash, the pair is inconsistent.

• Principle: Two equations describe the same or parallel lines only if one is a constant multiple of the other in its variable terms.
• Scale up: Multiply the first equation x+2y+5=0 by -3: -3x-6y-15=0.
• Line up: Place it beside the given second equation -3x-6y+1=0.
• Variables: The x and y parts -3x-6y are now identical in both lines, so the lines are parallel.
• Constants: But the constants disagree: -15≠ 1. Identical variable parts with different constants means the two lines can never meet.

Variable parts match, constants differ ⇒ no solution; option (D).

Eliminate by counter-example. Rather than memorise, I knock out each wrong option by naming a consistent system it fails to describe; the only survivor is the answer.

• Logic: A definition that claims ``always'' fails the moment one valid example breaks it.
• Case one: Take x+y=2, x-y=0. It has the single solution (1,1): this is consistent and intersecting, so option (B) ``always coincident'' is false.
• Case two: Take x+y=2, 2x+2y=4. Every point of the line solves both, so it is consistent and coincident, so option (D) ``always intersecting'' is false.
• Parallel out: Parallel lines such as x+y=2, x+y=5 have no solution, so they are inconsistent: option (A) is false.
• Survivors: Both surviving cases are captured only by ``intersecting or coincident''.

Counter-examples kill (A), (B), (D); answer is (C).

Contradiction angle. A solution of a pair must satisfy both equations together; I test whether a single y can obey both, and a flat contradiction settles the count of solutions at zero.

• Idea: A value cannot equal two different numbers simultaneously, so the system is inconsistent.
• Assume: Suppose (x,y) solves both equations.
• First forces: The first equation forces y=0.
• Second forces: The second equation forces y=-7.
• Clash: Combining, 0=-7, which is false. No point can satisfy both, and x being free does not help because y is already trapped.

The demand 0=-7 is impossible ⇒ no solution; option (D).

Direct substitution check. I confirm the intersection by plugging the proposed point back into both equations; the correct point satisfies each one with no contradiction.

• Rule: The intersection point is the unique pair that satisfies both equations simultaneously.
• Test x=a: Test the candidate point (a,b) in x=a: the x-coordinate is a, so x=a holds.
• Test y=b: Test (a,b) in y=b: the y-coordinate is b, so y=b holds.
• Both hold: Both equations are satisfied, so (a,b) is the common point.
• Reject swap: Test the rival (b,a) in x=a: its x-coordinate is b, and b=a need not be true, so (b,a) generally fails.

(a,b) satisfies both lines; option (D).

Proportional-equation angle. For coincident lines, the second equation must be the first multiplied by a single constant; finding that constant pins down k in one move.

• Idea: Coincident lines are scalar multiples of each other.
• Find factor: Compare constants to find the multiplier: c₂/c₁=16/8=2, so the second equation should be twice the first.
• Double it: Multiply the first equation by 2: 2(3x-y+8)=0 gives 6x-2y+16=0.
• Match terms: Match this against 6x-ky+16=0 term by term: the x-terms (6x) and constants (+16) already agree.
• Read k: Equate the y-terms: -2y=-ky, so k=2.

The first equation doubled gives k=2; option (C).

Equal-slope angle. Parallel lines have identical slopes, so I write each in y=mx+c form and equate the slopes; this avoids handling the 2ky ratio with its variable inside.

• Rule: Slope of ax+by+c=0 is -ab, and parallel lines have equal slopes.
• First slope: Slope of the first line 3x+2ky-2=0 is -32k.
• Second slope: Slope of the second line 2x+5y+1=0 is -25.
• Equate: Set the slopes equal: -32k=-25, i.e. 32k=25.
• Solve: Cross-multiply: 15=4k, so k=154.

Equal slopes give k=154; option (C).

Check the fixed part first. The smart move is to test the ratios that do not contain the unknown before solving for it; if those clash, the answer is ``no value'' without any algebra on c.

• Idea: Infinitely many solutions require the full equality chain; a single broken link is fatal.
• Pick c-free: Isolate the c-free ratios b₁b₂ and c₁c₂.
• Evaluate: Evaluate them: -1-2=12 and -2-3=23.
• They clash: Since 12≠23, the equality b₁b₂=c₁c₂ is impossible.
• No fix: No adjustment of c touches these ratios, so the system can never be coincident.

No value of c works; option (D).

Ratio sweep across options. I compute the three ratios for the strongest-looking candidates and keep the one where the whole chain is equal, which guarantees coincident lines.

• Rule: Coincidence demands a₁a₂=b₁b₂=c₁c₂ all at once.
• x-ratio: For option (D) 10x-14y+4=0: a₁a₂=-510 =-12.
• y-ratio: Next, b₁b₂=7-14=-12.
• Constant: Finally, c₁c₂=-24=-12.
• All agree: All three equal -12, so the lines coincide; option (D) is the dependent equation.
• Exam habit: In a one-mark question stop the instant any two ratios disagree, because dependence needs every link equal and a single break already settles the answer without checking the rest.

All ratios =-12 for (D); option (D).

Reject-on-first-failure angle. I substitute the given point into the first equation of each option; the instant one equation fails I drop that option, which is faster than checking both lines everywhere.

• Idea: A solution must satisfy every equation, so one failure is enough to discard an option.
• Option B: Option (B) first equation 2x+5y=-11 at (2,-3): 4-15=-11 , second 4x+10y=-22: 8-30=-22 ; but 24=510=-11-22, so (B) is dependent (infinitely many, not unique) and is rejected.
• Option C: Option (C) first equation 2x-y=1 at (2,-3): 4+3=7≠ 1, reject.
• Option A: Option (A) fails its second equation as shown, reject.
• Option D: Option (D) passes both equations and is not dependent, so it is the unique-solution pair.

(D) is the only pair with a genuine unique solution at (2,-3); option (D).

Subtract-then-add angle. Subtracting the equations isolates y just as quickly; I show this route so the answer is confirmed by a second, independent elimination.

• Method: Subtracting equations cancels the common x-term, giving y directly.
• Subtract: Subtract (1) from (2): (x+y)-(x-y)=4-2, so 2y=2.
• Get y: Hence y=1, which matches b=1.
• Get x: Put y=1 into (1) x-y=2: x-1=2, so x=3, matching a=3.
• Agree: Both routes agree on (a,b)=(3,1).

a=3, b=1; option (C).

Verify by value, not just count. I confirm the solution by checking that (25,25) matches the rupee total, since several wrong options satisfy only the coin count.

• Idea: The correct split must satisfy both the count equation and the value equation at once.
• Take values: Take the solution x=25, y=25.
• Count check: Count check: 25+25=50 , matching equation (1).
• Value check: Value check: 1(25)+2(25)=25+50=75 , matching equation (2).
• Contrast: For contrast, option (A) (35,15) gives the value 35+2(15)=65≠ 75, so it fails the rupee total and is rejected.
• Watch out: Notice that option (A) survives the count test (35+15=50), which is exactly why a quick count check alone would mislead a hurried student.
• Verdict: The rupee total is therefore the real discriminator here, and only (25,25) clears it, so the answer is firmly option (D).

Re 1 coins =25, Rs 2 coins =25; option (D).

Back-check the option angle. With clean options available, I take the keyed pair and confirm it obeys both conditions, which both verifies the answer and models how to eliminate wrong options under time pressure.

• Method: The correct ages must satisfy the present ratio and the future ratio simultaneously.
• Pick option: Take option (C): son =6, father =36.
• Now check: Present check: 36=6× 6 , so the ``six times'' rule holds.
• Later check: Four years later: son =10, father =40. Is 40=4× 10? Yes .
• Verdict: Both conditions pass for (C); for contrast option (A) son 4, father 24 gives future 28 vs 4× 8=32, which fails.

Son 6, father 36; option (C).

Decimal-ratio angle. I convert each ratio to a decimal so the ``equal-equal-different'' pattern jumps out at a glance, which is handy when fractions like 23 make the comparison fiddly.

• Trick: The same parallel-line test, read as decimals.
• Part i: (i) 0.33, 0.33, 0.5: first two match, last differs, so parallel, no solution (Yes).
• Part ii: (ii) 0.5 versus 2: unequal at the first comparison, so they intersect (No).
• Part iii: (iii) 1.5, 1.5, 1.5: all equal, so coincident, infinitely many solutions (No).
• Confirm: The decimals confirm the fraction work in every part.
• Why it helps: Reading the ratios as decimals removes the awkward 23 in part (iii), where 12/3=1.5 matches the other two decimals cleanly.
• Verdict: Because all three decimals agree only in part (iii), it is the lone coincident pair, while (i) is parallel and (ii) intersecting.

(i) Yes, (ii) No, (iii) No.

Scale-to-match angle. Coincident means one equation is a clean multiple of the other; I scale the simpler equation and see whether it reproduces the second exactly, including the constant.

• Method: Coincident lines are exact scalar multiples.
• Part i: (i) Doubling or scaling 3x+17y-3 can never give 7x+3y-7 because 37≠1/73: No.
• Part ii: (ii) Multiply -2x-3y-1=0 by -2: 4x+6y+2=0, which is exactly the second equation: Yes, coincident.
• Part iii: (iii) Multiply x2+y+25=0 by 8: 4x+8y+165=0. The variable parts match the second equation but the constant 165≠516: No.
• Only one: Only (ii) reproduces fully.
• Decisive step: The scale-to-match test is decisive in part (iii): the variable parts line up after multiplying by 8, but the constant 165 refuses to become 516.
• Verdict: So only part (ii) is a true scalar copy of its partner, confirming it is the single coincident pair among the three.

(i) No, (ii) Yes, (iii) No.

Spot the disguise angle. Parts (i) and (iv) hide the structure: (i) is the same line negated, (iv) has a bracket multiplier. I simplify first, then read consistency from a single ratio comparison.

• Rule: Simplify each equation to lowest standard form before comparing ratios.
• Part i: (i) The two equations are 3x+4y=-12 and 3x+4y=12: same left side, different right side, so parallel and inconsistent (No).
• Part ii: (ii) Slopes 3 and 13 differ, so a unique solution exists: consistent (Yes).
• Part iii: (iii) Doubling the first equation gives the second exactly, so the lines coincide: consistent (Yes).
• Part iv: (iv) Reducing 2(2x+6y)=22 to 2x+6y=11 shows it is the first line x+3y=11 doubled on the left but not the right (11≠ 22), so parallel: inconsistent (No).
• The disguises: The disguises matter: part (i) is the same line with both sides negated on one copy, and part (iv) hides a factor of 2 inside a bracket.
• Reduce first: Once each pair is reduced to lowest standard form, a single ratio comparison settles consistency, giving No, Yes, Yes, No across the four parts.

(i) No, (ii) Yes, (iii) Yes, (iv) No.

Substitute the claim angle. I take the claimed λ=1 at face value, build the actual pair, and run the full ratio test to expose the contradiction directly.

• Test: A claimed solution either survives the equality chain or it does not.
• Set lambda: Put λ=1: the pair becomes x+3y+7=0 and 2x+6y-14=0.
• Variables: Variable ratios: 12 and 36=12, which match.
• Constant: Constant ratio: 7-14=-12, which does not match 12.
• Result: So at λ=1 the lines are parallel (no solution), the exact opposite of infinitely many. The statement is false.

False; λ=1 actually gives parallel lines, not infinitely many solutions.

Slope-equality angle. Both lines turn out to have the same slope, so they can never cross once; I show this and note how the value of c only slides the second line, never tilts it.

• Idea: Equal slopes rule out a single intersection point.
• First slope: Slope of x-2y=8 is 12 (from y=x-82).
• Second slope: Slope of 5x-10y=c is also 510=12.
• Equal slopes: Equal slopes mean the lines stay parallel; changing c just shifts the second line up or down without changing its direction.
• Verdict: Therefore the lines are parallel for c≠ 40 and coincident for c=40, but never uniquely intersecting. The statement is false.

False; equal slopes (12) forbid a unique solution for any c.

Point-sampling angle. I list a few points on x=7 and watch which coordinate stays constant; the constant coordinate reveals the line's direction and hence which axis it parallels.

• Method: The coordinate that stays fixed across all points of a line tells you its orientation.
• Sample: Sample points on x=7: (7,0), (7,1), (7,5), (7,-3).
• Pattern: In every point the x-value is locked at 7 while y changes freely.
• Vertical: A constant x with varying y traces a vertical line.
• Verdict: Vertical lines are parallel to the y-axis, so the claim that it parallels the x-axis is wrong.

Not true; x=7 is vertical and parallel to the y-axis.

Determinant angle. The pair fails to have a unique solution exactly when the coefficient determinant a₁b₂-a₂b₁ is zero; I use that to find the special λ values, then split them into the two degenerate cases.

• Tool: Unique solution ⇔ a₁b₂-a₂b₁≠ 0.
• Determinant: Determinant =λλ-1· 1=λ²-1.
• Zero case: Set it to zero for the degenerate cases: λ²-1=0, so λ=1; for all other λ the solution is unique (part iii).
• At +1: At λ=1 both equations become x+y=1: identical, so infinitely many (part ii).
• At -1: At λ=-1 they become -x+y=1 and x-y=1, i.e. x-y=-1 and x-y=1: parallel, so no solution (part i).
• Why it works: The determinant λ²-1 vanishing pinpoints the two boundary values λ=1 without testing the constant ratio first.
• Split roots: Splitting those two roots by inspecting the actual equations cleanly separates the coincident case λ=1 from the parallel case λ=-1.

(i) λ=-1; (ii) λ=1; (iii) λ≠1.

Substitute each root angle. I plug each candidate k back into the original pair and read off the geometry, which removes any doubt about which root is the no-solution case.

• Method: Parallel lines (no solution) have equal variable ratios but a different constant ratio.
• k=6: For k=6: equations are 6x+3y=3 and 12x+6y=6. The second is exactly twice the first, so coincident: infinitely many, rejected.
• k=-6: For k=-6: equations are -6x+3y=-9 and 12x-6y=-6, i.e. -6x+3y=-9 and -6x+3y=3 after dividing the second by -2.
• Same side: Same left side -6x+3y but right sides -9 and 3 differ, so the lines are parallel.
• No solution: Parallel means no solution, confirming k=-6.

k=-6.

Back-verify the values angle. After solving, I substitute a=3, b=1 into the second equation to confirm it is genuinely a multiple of the first, which guarantees coincident lines.

• Check: The found (a,b) must turn the second equation into a scalar multiple of the first.
• Plug a,b: With a=3, b=1: a-b=2, a+b=4, a+b-2=2.
• Rewrite: The second equation becomes 2x+4y=2.
• Divide: Divide by 2: x+2y=1, which is exactly the first equation.
• Coincident: Identical equations mean coincident lines and infinitely many solutions, so (a,b)=(3,1) is correct.

a=3, b=1.

Find the forbidden value angle. For the ``unique'' parts I locate the single value of p that makes the lines parallel (slopes equal); the answer is then everything except that value, which is faster than checking a range.

• Method: The boundary value is where a₁a₂=b₁b₂; uniqueness fails only there.
• Part iii: (iii) Parallel boundary: -32p=5-3 gives 9=-10p, so p=-910; exclude it.
• Part iv: (iv) Parallel boundary: 2p=-12 gives p=-4; exclude it.
• Parts i,ii: (i) and (ii) are the equal-slope (parallel / no solution) cases, solved by the constant-ratio test above (p10 and p=1).
• Part v: (v) needs all three ratios equal; matching against 14 gives p=4, q=8.
• Range parts: For the ``intersecting / unique'' parts (iii) and (iv) the answer is a whole range, so I find only the single forbidden slope-matching value and exclude it.
• Part v again: Part (v) is the opposite extreme, demanding all three ratios equal, which the match against 14 resolves to p=4, q=8.

(i) p10, (ii) p=1, (iii) p≠-910, (iv) p≠-4, (v) p=4,q=8.

Try-to-solve angle. I attempt to solve the pair; if the algebra collapses into a false statement, the system is inconsistent and the paths cannot meet.

• Method: An inconsistent system reduces to a contradiction like 0= (nonzero).
• Scale up: Multiply the first equation by 2: 2x-6y=4.
• Add them: Add it to the second -2x+6y=5: the x and y terms cancel, leaving 0=9.
• False line: 0=9 is false, so no (x,y) satisfies both equations.
• Parallel: No common point means the paths are parallel and never cross.

No common solution (0=9), so the paths do not cross.

Geometric-count angle. I explain why the count is infinite: through a single point pass infinitely many lines, and almost any two of them with different directions form a valid unique-solution pair.

• Idea: Infinitely many distinct lines pass through one fixed point.
• Pencil: Through (-1,3) there is a whole pencil of lines, one for every slope.
• Pick two: Pick any two of them with different slopes; they cross only at (-1,3).
• Each pair: Each such pair is a distinct ``pair of equations'' with the unique solution (-1,3).
• Infinitely many: Since the choice of two slopes is unlimited, there are infinitely many valid pairs; my sample is x-y=-4, 2x+3y=7.
• Exam takeaway: When a question asks you to write a pair with a given solution, just invent any two simple lines that both pass through that point, and you are free to pick easy whole-number coefficients to keep the marking quick.

Sample pair x-y=-4, 2x+3y=7; infinitely many pairs possible.

Subtraction route. I isolate x first by subtracting to confirm the same (x,y), then evaluate, giving an independent check on the elimination.

• Method: Subtracting the equations cancels y just as adding does, since y appears as +y and -y.
• Subtract: Subtract: (4x-y)-(2x+y)=19-23, so 2x-2y=-4, i.e. x-y=-2.
• Add back: Combine x-y=-2 with 2x+y=23 by adding: 3x=21, so x=7, and then y=9.
• First value: Evaluate 5y-2x=45-14=31.
• Second value: Evaluate yx-2=97-2=-57.

5y-2x=31, yx-2=-57.

Substitution route. I solve equation (2) for y and substitute into (1), which avoids scaling and keeps the arithmetic small.

• Method: Substitution replaces one variable using the simpler equation.
• Express y: From (2) 3x+y=7, express y=7-3x.
• Substitute: Substitute into (1) x+3y=13: x+3(7-3x)=13.
• Expand: Expand: x+21-9x=13, so -8x=-8, giving x=1.
• Get y: Then y=7-3(1)=4.
• No fractions: Substituting y=7-3x keeps every coefficient an integer, so there is no fraction handling at any stage of this rectangle problem.
• One equation: The single equation -8x=-8 that results gives x=1 at once, and back-substitution then fixes y=4, matching the elimination route.

x=1, y=4.

Cross-check by back-substitution. I verify each solution in the original (fractional) equations, since clearing denominators is where sign and LCM slips usually happen.

• Check: A correct solution satisfies the original equations, not just the cleared ones.
• Part i: (i) Check x+y=1.2+2.1=3.3 , and 3x-2y=3.6-4.2=-0.6, so 0.6-0.6=-1 .
• Part ii: (ii) Check 63+84=2+2=4 .
• More of ii: And 566-88=5-1=4 .
• Confirm: Both solutions satisfy the originals, confirming the answers.
• Why this matters: Students lose marks here by checking the simplified equation they wrote, not the printed one, so a hidden arithmetic slip while clearing fractions slips through unnoticed until the final answer is already wrong.

(i) x=1.2, y=2.1; (ii) x=6, y=8.

Verify the flip-back angle. The most common error is forgetting to invert p,q at the end, so I substitute the final x,y into the original reciprocal equations as a check.

• Check: After solving for the reciprocals, the answer is their inverse, which must satisfy the original equations.
• Part iii: (iii) Check 4(3)+62=12+3=15 and 6(3)-82=18-4=14 .
• Part iv: (iv) With x=16, y=14: 12x=3 and 1y=4, so 3-4=-1 .
• More of iv: Also 1x=6 and 12y=2, so 6+2=8 .
• Confirm: Both checks pass, confirming the flip-back was done correctly.

(iii) x=3, y=2; (iv) x=16, y=14.

Elimination route for (vi). Rather than guessing, I eliminate properly: clear denominators and combine, which derives x=a², y=b² with no luck involved.

• Method: Multiplying to align one variable's coefficient lets us eliminate it.
• Scale eq 2: Multiply the second equation by a: xa+yb²· a=2a, i.e. xa+ayb²=2a.
• Subtract: Subtract this from the first xa+yb=a+b: yb-ayb²=a+b-2a=b-a.
• Factor: Factor: yb(1-ab)=yb·b-ab=b-a, so y(b-a)b²=b-a, giving y=b² (for a≠ b).
• Back-sub: Substitute back into the first equation to get x=a².

(v) x=1, y=-1; (vi) x=a², y=b².

Verify in the original angle. Because reciprocating is error-prone, I substitute x=12, y=-32 straight into the two original equations to confirm the values.

• Check: The found values must satisfy the original (non-reciprocal) equations.
• Sum,product: Compute x+y=12-32=-1 and xy=12·(-32)=-34.
• First eq: First equation: 2xyx+y=2(-3/4)-1=-3/2-1=32 .
• Get 2x-y: Compute 2x-y=1-(-32)=52.
• Second eq: Second equation: xy2x-y=-3/45/2=-34·25=-310 .
• Why check: Both original equations are ratios of products, so checking them directly is the surest guard against a slip while taking reciprocals.
• Verdict: Since 2xyx+y=32 and xy2x-y=-310 both hold for (12,-32), the reciprocal work was done correctly.

x=12, y=-32.

Substitution route. I solve (1) for x and feed it into (2), keeping the big number 340 until the end as a check on the elimination.

• Method: Substitution reduces the pair to one equation in y.
• Express x: From (1) x=10-2y. Substitute into (2): 3(10-2y)+4y=360.
• Expand: Expand: 30-6y+4y=360, so -2y=330, giving y=-165.
• Get x: Then x=10-2(-165)=10+330=340.
• Find lambda: Finally λ=y-5x=-165-5340=-170340=-12.
• Why substitution: Substitution is the cleaner choice here because one variable already has a small coefficient, so expressing it first avoids the large numbers that elimination would carry through every line.

x=340, y=-165, λ=-12.

Two-point plotting angle. Graphically, I find two points per line; crossing lines mean a unique solution, identical lines mean infinitely many, and never-meeting lines mean inconsistent.

• Method: The graph of each consistency type matches its ratio verdict.
• Part i: (i) Plotting 3x+y=-4 and 6x-2y=-4 gives two lines crossing at (-1,-1): consistent, unique.
• Part ii: (ii) Plotting x-2y=6 and 3x-6y=0 gives two parallel lines (same slope 12, different intercepts): inconsistent.
• Part iii: (iii) Plotting x+y=3 and 3x+3y=9 gives the same line twice: consistent, infinitely many.
• Plots agree: The plotted pictures agree with the ratio test in all three parts.
• Two points: Plotting two points per line makes the three pictures unmistakable: a crossing in (i), separated parallels in (ii), and one line drawn twice in (iii).
• Verdict: The visual verdict matches the ratio test exactly, so (i) is uniquely solvable at (-1,-1), (ii) is inconsistent, and (iii) has the whole line y=3-x.

(i) (-1,-1); (ii) inconsistent; (iii) infinitely many, y=3-x.

Coordinate-area-formula angle. I plug the three vertices into the determinant area formula, which is a fully algebraic check that does not rely on identifying base and height.

• Tool: Area =12 |x₁(y₂-y₃)+x₂(y₃-y₁) +x₃(y₁-y₂)|.
• Vertices: Take (x₁,y₁)=(2,0), (x₂,y₂)=(0,4), (x₃,y₃)=(0,-4).
• Substitute: Substitute: 12|2(4-(-4))+0((-4)-0)+0(0-4)|.
• Simplify: Simplify inside: 2(8)+0+0=16.
• Area: Area =12× 16=8 square units, matching the base-height result.
• Why it works: The determinant area formula needs only the three vertices, so it sidesteps any argument about which side is the base and which is the height.
• Verdict: Plugging (2,0),(0,4),(0,-4) into it gives 12|16|=8, confirming the base-height answer of 8 square units independently.
• Sign safety: Keep the modulus bars in place at every step, because the bracket inside can come out negative depending on the order in which you list the vertices, and the area is always the positive value.
• Exam takeaway: This formula is the safer choice in the board exam whenever the triangle is tilted, since you never have to draw a perpendicular or guess which side serves as the base.

Vertices (2,0),(0,4),(0,-4); area =8 sq units.

Family-of-lines angle. I write the general line through (1,1) explicitly, which both gives a sample answer and shows why the count is infinite.

• Idea: Every line through (1,1) has the form y-1=m(x-1) for some slope m.
• Solution: The solution point is (1,1) from the elimination above.
• Family: The point-slope family is y-1=m(x-1).
• Vary slope: Each real value of m gives a different line, e.g. m=1 gives y=x, m=0 gives y=1, m=2 gives y=2x-1.
• Infinitely many: Infinitely many choices of m mean infinitely many lines.

Sample line y=x; infinitely many lines through (1,1).

Verification angle. After solving, I plug a=5, b=2 back into the cubic and confirm x=-1 is genuinely a root, closing the loop.

• Check: A correct (a,b) must make x=-1 a zero of the cubic.
• Build cubic: With a=5, b=2 the cubic is 2x³+5x²+4x+1.
• Plug x=-1: Evaluate at x=-1: 2(-1)+5(1)+4(-1)+1=-2+5-4+1=0 .
• Relation: Check the given relation: 2a-3b=2(5)-3(2)=10-6=4 .
• Verdict: Both conditions hold, so a=5, b=2 is correct.
• Exam takeaway: Whenever a value is claimed to be a root, the fastest confirmation is to substitute it and watch the whole expression collapse to zero, which catches any sign error from the earlier algebra in one line.

a=5, b=2.

Substitution route. I express x from the difference equation and substitute, which also makes it clear the answer is the same if we had taken y>x (the two angles just swap).

• Method: Substitution from the simpler difference equation.
• Express x: From (2) x=y+30^∘.
• Substitute: Substitute into (1): (y+30^∘)+y=140^∘, so 2y=110^∘, giving y=55^∘.
• Get x: Then x=55^∘+30^∘=85^∘.
• Verify: Check: 85^∘+55^∘+40^∘=180^∘ .

x=85^∘, y=55^∘.

Verify both timelines angle. I check the present ages against both the past and future statements, since age problems can satisfy one timeline and fail the other.

• Check: The present ages must obey every time-shifted condition.
• Present: Present: Salim 38, daughter 14.
• Two years ago: Two years ago: 36 and 12. Is 36=312? Yes .
• Six years on: Six years later: 44 and 20. Is 44=220+4=44? Yes .
• Verdict: Both timelines check out, confirming the ages.

Salim 38, daughter 14.

Single-variable angle. Because the father's age is tied to the children's sum by (1), I reduce everything to one variable y and solve in a single line.

• Method: Substitute x=2y immediately to work with one unknown.
• Future link: Future condition: x+20=(y+40).
• Replace x: Replace x by 2y: 2y+20=y+40.
• Simplify: Simplify: y=20, the children's combined present age.
• Father: Father's age x=2y=40 years.
• Why it helps: Because the father's age is locked to twice the children's sum, collapsing to the single unknown y removes all chance of a stray second variable.
• One line: The one-line equation 2y+20=y+40 then yields y=20, so the father is 2y=40 years old, agreeing with the two-variable setup.

Father's age =40 years.

Two-variable angle. I instead use two unknowns a,b and two equations to show the same answer arises without the k-trick.

• Setup: Each ratio becomes a cross-multiplied linear equation.
• First ratio: Let the numbers be a and b. First ratio 5b=6a, i.e. 6a-5b=0   (1).
• Second ratio: Second ratio 5(a-8)=4(b-8), so 5a-4b=8   (2).
• Substitute: From (1) b=6a5. Substitute into (2): 5a-4·6a5=8, so 25a-24a5=8, giving a=40.
• Get b: Then b=6(40)5=48.

The numbers are 40 and 48.

Plug-back verification. I confirm x=100, y=80 reproduces both transfer conditions exactly, since transfer problems are easy to set up with a sign flipped.

• Check: The found counts must satisfy both transfer statements.
• First move: First transfer: 100-10=90 and 80+10=90, equal .
• Second move: Second transfer: A becomes 100+20=120, B becomes 80-20=60.
• Twice as much: Is 120=260? Yes .
• Verdict: Both conditions hold, so the counts are correct.

Hall A =100, hall B =80.

Verify both bills angle. I plug x=10, y=3 back into both rentals to confirm the totals Rs 22 and Rs 16 come out right.

• Check: The charges must reproduce both customers' bills.
• Latika: Latika: fixed 10 plus 4 extra days at 3: 10+4(3)=10+12=22 .
• Anand: Anand: fixed 10 plus 2 extra days at 3: 10+2(3)=10+6=16 .
• Both match: Both bills match the given amounts.
• Charges: So the fixed charge is Rs 10 and each extra day costs Rs 3.

Fixed charge Rs 10, extra-day charge Rs 3.

Check the score angle. I confirm x=100, y=20 produces exactly 90 marks, since negative marking problems are easy to set up with the wrong sign.

• Check: Score = (correct) -12(wrong) must equal 90.
• Totals: Correct =100, wrong =20, total =120 .
• From correct: Marks from correct: 1001=100.
• From wrong: Marks lost to wrong: 20×12=10.
• Net score: Net score =100-10=90 , matching the data.

100 correct answers.

Total-angle cross-check. Besides the two supplementary pairs, all four angles of any quadrilateral sum to 360^∘; I use this as an independent check on the final angles.

• Check: The four interior angles of a quadrilateral sum to 360^∘.
• Angles: From the solution, the angles are 130^∘,100^∘,50^∘,80^∘.
• Opposite pairs: Check the opposite pairs: ∠ A+∠ C=130+50=180^∘ and ∠ B+∠ D=100+80=180^∘ .
• Total: Check the total: 130+100+50+80=360^∘ .
• Verdict: All consistency checks pass, confirming x=20, y=30.

x=20, y=30; angles 130^∘,100^∘,50^∘,80^∘.

Determinant-area angle. I compute each triangle's area from its three vertices using the coordinate formula, an algebraic route that needs no base-height identification.

• Tool: Area =12|x₁(y₂-y₃)+x₂(y₃-y₁) +x₃(y₁-y₂)|.
• x-axis: x-axis triangle (3,0),(-1,0),(1,4): 12|3(0-4)+(-1)(4-0)+1(0-0)|=12|-12-4|=8.
• y-axis: y-axis triangle (0,6),(0,2),(1,4): 12|0(2-4)+0(4-6)+1(6-2)|=12|4|=2.
• Ratio: Ratio =8:2=4:1.
• Agree: Both methods agree on 4:1.
• Why uniform: The coordinate area formula handles both triangles uniformly, so the x-axis triangle and the y-axis triangle are computed by the same single expression.
• Quotient: It returns 8 and 2 square units respectively, and their quotient 82=4 gives the required ratio 4:1 with no base-height bookkeeping.
• Vertex care: Find the third vertex of each triangle as the point where the two slanting lines meet, because the two points on the axis are read straight off the intercepts and only the apex needs solving.
• Ratio shortcut: Since both triangles share the same apex, their areas scale with the lengths cut on the two axes, so once you trust the formula the final ratio drops out as a single clean division.
• Exam takeaway: Questions that ask for a ratio of areas rarely need the actual units, so cancel common factors early and quote the answer as a tidy whole-number ratio like 4:1.

Areas 8 and 2; ratio 4:1.

Substitution-only angle. All three lines are easy to write as x in terms of y (or vice versa), so I solve every pairing by plain substitution and list the results.

• Method: Substituting one line into another gives their common point.
• First vertex: y=x with 3y=x forces x=y=0, vertex (0,0).
• Second vertex: y=x with x+y=8 forces x=y=4, vertex (4,4).
• Third vertex: x=3y with x+y=8 forces y=2, x=6, vertex (6,2).
• List: Vertices: (0,0),(4,4),(6,2).
• Why this works: Each vertex of the triangle is simply where two of the three lines meet, so solving the three pairings one at a time delivers all three corners without any graph paper.
• Exam takeaway: For vertices of a triangle made by three lines, always solve the lines two at a time, and list the corners clearly so the marker can see each intersection was found separately.

Vertices (0,0),(4,4),(6,2).

Split-into-shapes angle. I split the trapezium into a rectangle plus a right triangle, an arithmetic route that avoids the trapezium formula.

• Method: A trapezium is a rectangle plus a triangle.
• Rectangle: Rectangle: width 2 (from x=3 to x=5), height 2 (the shorter side), area 22=4.
• Triangle: Triangle on top: base 2 (same width), height 6-2=4, area 1224=4.
• Total: Total area =4+4=8 square units.
• Matches: This matches the trapezium-formula result.
• Why split: Splitting the trapezium into a 22 rectangle plus a triangle of base 2 and height 4 turns the area into two easy products.
• Add up: Adding the rectangle's 4 and the triangle's 4 gives 8 square units, matching the trapezium formula and confirming the figure's area.
• Why this helps: Breaking an awkward shape into a rectangle and a triangle means you only ever multiply length by width and halve a product, so there is nothing to memorise beyond two areas you already know.
• Exam takeaway: When a figure does not match a standard formula, slice it along a horizontal or vertical line into familiar pieces and add the parts, which is far safer under exam pressure than recalling a special formula.

Area =8 sq units.

Substitution route. I express y from the simplified first equation and substitute, then verify both original sentences.

• Method: Substitution and a money-sense check.
• Express y: From (1) y=25-x. Substitute into (2): 3x-(25-x)=15.
• Simplify: Simplify: 4x-25=15, so 4x=40, giving x=10; then y=15.
• Check one: Check sentence 1: 4(10)+4(15)=40+60=100 .
• Check two: Check sentence 2: 3(10)=30 and 15+15=30 .
• One move: Substituting y=25-x into 3x-y=15 collapses the pair to 4x-25=15 in a single move, so x=10 and y=15 follow at once.
• Both check: The two original sentences both check out (4(10)+4(15)=100 and 3(10)=15+15), so a pen costs Rs 10 and a pencil box Rs 15.

Pen Rs 10, pencil box Rs 15.

Elimination route. I solve one pairing by elimination to give an independent confirmation of a vertex, then list all three.

• Method: Elimination aligns a variable's coefficient and subtracts.
• Make pair: Pair 3x-y=3 and x+2y=8. Multiply the first by 2: 6x-2y=6.
• Add up: Add to x+2y=8: 7x=14, so x=2; then y=3, vertex (2,3), as before.
• Other two: The other two vertices follow by the substitutions above: (1,0) and (4,2).
• Agree: All three vertices agree across methods.

Vertices (1,0),(2,3),(4,2).

Plug-back time check. I verify x=10, y=40 reproduce both stated travel times exactly, which catches any reciprocal slip.

• Check: Total time = sum of (distance/speed) over the two legs.
• Trip 1: Trip 1: 210+1240=0.2+0.3=0.5 hour =12 hour .
• Trip 2: Trip 2: 410+1040=0.4+0.25=0.65 hour.
• Convert: Convert: 0.65 hour =39 minutes =30+9 minutes, i.e. 9 minutes longer than trip 1 .
• Verdict: Both times match the data, so the speeds are correct.
• Why convert: Converting minutes to hours first (9 minutes =0.15 hour) is what makes the time check come out cleanly to 0.65 hour for trip 2.
• Both trips: Trip 1 takes exactly 0.5 hour and trip 2 takes 0.65 hour, a gap of 9 minutes, so the speeds 10 and 40 km/h are confirmed.

Rickshaw 10 km/h, bus 40 km/h.

Time-ratio check. I confirm s=2.5 makes the upstream time exactly three times the downstream time, verifying the answer.

• Check: The ratio of times equals the inverse ratio of speeds for a fixed distance.
• Speeds: With s=2.5: downstream speed =7.5 km/h, upstream speed =2.5 km/h.
• Downstream: Downstream time =407.5=163 hour.
• Upstream: Upstream time =402.5=16 hour.
• Ratio: Ratio 1616/3=3, so upstream is thrice downstream .

Stream speed =2.5 km/h.

Time-verification angle. I substitute the recovered speeds into both trips to confirm the 7-hour and 5-hour totals.

• Check: Total time is the sum of distance/speed over each leg.
• Speeds: Boat 10, stream 4 give upstream 6 km/h, downstream 14 km/h.
• Trip 1: Trip 1: 306+2814=5+2=7 hours .
• Trip 2: Trip 2: 216+2114=3.5+1.5=5 hours .
• Verdict: Both totals match, confirming boat 10 km/h and stream 4 km/h.
• Sign care: Converting the still-water and stream speeds back into upstream 6 and downstream 14 km/h is the step where a sign error would surface, so I test both trips.
• Both trips: Trip 1 totals 5+2=7 hours and trip 2 totals 3.5+1.5=5 hours, both matching the data, so boat 10 km/h and stream 4 km/h are correct.

Boat 10 km/h, stream 4 km/h.