Pair of Linear Equations in Two Variables

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CLASS X Mathematics ~5 marks/year Ch 3 of 14
Pair of Linear Equations in Two Variables

Class 10 · Mathematics · NCERT chapter notes · Akanksha Classes

Snapshot
  • A linear equation in two variables $ax+by+c=0$ is a straight line; a pair of them is two lines on the same graph.
  • Two lines can do only three things: intersect (one solution), be parallel (no solution) or be coincident (infinitely many).
  • Without drawing, the ratios $\dfrac{a_1}{a_2},\dfrac{b_1}{b_2},\dfrac{c_1}{c_2}$ tell you which case you are in — the single most-tested idea of the chapter.
  • Three ways to solve: graphical, substitution and elimination. Algebraic methods give exact answers even when the solution is a fraction.
  • Most board questions are word problems (ages, fractions, money, speed, geometry) turned into two equations and then solved.
  • Board weightage: ~5 marks/year — typically one consistency/ratio question (1-2 marks) and one word problem solved algebraically (3 marks).
Detailed notes

1. What this chapter is about

In Class 9 you met a linear equation in two variables — any equation of the form

$$a x + b y + c = 0,\qquad a,b\ \text{not both zero}.$$

Its graph is a straight line, and it has infinitely many solutions (every point on that line). This chapter studies two such equations together — a pair of linear equations in two variables. The big question is: do the two lines have a common point, and if so, how do we find it?

The fair example (NCERT): Akhila plays at a fair. Rides cost ₹3 each, Hoopla costs ₹4 per game, she plays Hoopla half as many times as she rides, and spends ₹20. Let $x$ = rides, $y$ = Hoopla games. Then

$$y=\tfrac12 x \quad(1),\qquad 3x+4y=20\quad(2).$$

Finding the $x,y$ that satisfy both equations is exactly what the whole chapter teaches.

2. The general form and key words

A pair of linear equations is written in the general form

$$a_1 x + b_1 y + c_1 = 0,\qquad a_2 x + b_2 y + c_2 = 0.$$

Learn these words — board questions use them directly:

  • Consistent: the pair has at least one solution (lines intersect or coincide).
  • Inconsistent: the pair has no solution (lines are parallel).
  • Dependent: the two equations are really the same line (coincident) — infinitely many solutions. A dependent pair is always consistent.

3. Graphical method — three possible pictures

Plot both lines on one graph. Only three things can happen:

  • (i) Lines intersect at a single point → exactly one solution (consistent).
  • (ii) Lines are parallel (never meet) → no solution (inconsistent).
  • (iii) Lines coincide (lie on top of each other) → infinitely many solutions (dependent, consistent).

To draw a line, find two solutions (two points), plot and join. Usually take $x=0$ to get the $y$-intercept and $y=0$ to get the $x$-intercept.

4. The ratio test — the heart of the chapter

You do not need a graph to know which case you are in. Compare the coefficient ratios:

$$\begin{aligned} \text{(i) Intersecting (unique):}\quad &\frac{a_1}{a_2}\neq\frac{b_1}{b_2}\\[4pt] \text{(ii) Coincident (infinite):}\quad &\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\\[4pt] \text{(iii) Parallel (no solution):}\quad &\frac{a_1}{a_2}=\frac{b_1}{b_2}\neq\frac{c_1}{c_2} \end{aligned}$$

Memory trick: if the first two ratios are unequal → one solution. If all three are equal → infinite. If first two equal but the third differs → none. Always write both equations in the form $ax+by+c=0$ first (move everything to one side) before reading off $a,b,c$.

NCERT Table 3.1 examples:

  • $x-2y=0$ and $3x+4y-20=0$: $\dfrac{1}{3}\neq\dfrac{-2}{4}$ → intersecting, unique.
  • $2x+3y-9=0$ and $4x+6y-18=0$: $\dfrac24=\dfrac36=\dfrac{-9}{-18}$ → coincident, infinite.
  • $x+2y-4=0$ and $2x+4y-12=0$: $\dfrac12=\dfrac24\neq\dfrac{-4}{-12}$ → parallel, none.

5. Worked graphical Examples

NCERT Example 1 — is $x+3y=6,\ 2x-3y=12$ consistent? Solve graphically.

From line 1: when $x=0,\ y=2$; when $x=6,\ y=0$. Points $A(0,2),B(6,0)$.

From line 2: when $x=0,\ y=-4$; when $x=3,\ y=-2$. Points $P(0,-4),Q(3,-2)$.

The two lines meet at $B(6,0)$. So the pair is consistent with solution $x=6,\ y=0$.

NCERT Example 2 — no solution, unique, or infinite? $5x-8y+1=0$ and $3x-\tfrac{24}{5}y+\tfrac35=0$.

Multiply the second equation by $\dfrac53$: $5x-8y+1=0$ — identical to the first. So the lines are coincident and the pair has infinitely many solutions.

NCERT Example 3 — Champa's pants and skirts (graphical).

Let $x$ = pants, $y$ = skirts. "Skirts are two less than twice the pants" → $y=2x-2$. "Skirts are four less than four times the pants" → $y=4x-4$.

Line 1 points: $(2,2),(0,-2)$. Line 2 points: $(0,-4),(1,0)$. The lines intersect at $(1,0)$. So Champa bought 1 pant and 0 skirts.

6. Algebraic methods — why we need them

Graphs are clumsy when the answer is a fraction like $\left(\dfrac{49}{29},\dfrac{19}{29}\right)$ or involves surds like $\big(\sqrt3,\,2\sqrt7\big)$ — you cannot read such a point off graph paper accurately. So we use exact algebraic methods: substitution and elimination.

7. Substitution method

Steps:

  • Step 1: from either equation, express one variable in terms of the other (pick the easier one).
  • Step 2: substitute that into the other equation → a single equation in one variable; solve it. (A true statement with no variable means infinite solutions; a false one means no solution.)
  • Step 3: put the value back into Step 1's expression to get the second variable.
NCERT Example 4 — solve $7x-15y=2,\ x+2y=3$ by substitution.

From eq (2): $x=3-2y$. Substitute in (1): $7(3-2y)-15y=2\Rightarrow 21-14y-15y=2\Rightarrow -29y=-19\Rightarrow y=\dfrac{19}{29}$.

Then $x=3-2\left(\dfrac{19}{29}\right)=\dfrac{49}{29}$. Solution: $x=\dfrac{49}{29},\ y=\dfrac{19}{29}$.

NCERT Example 5 — Aftab and his daughter's ages.

Let $s,t$ be present ages of Aftab and daughter. "Seven years ago I was 7 times you": $s-7=7(t-7)\Rightarrow s-7t+42=0$. "Three years from now I shall be 3 times you": $s+3=3(t+3)\Rightarrow s-3t=6$.

From the second: $s=3t+6$. Substitute: $(3t+6)-7t+42=0\Rightarrow -4t+48=0\Rightarrow t=12$, and $s=3(12)+6=42$. So Aftab is 42 and his daughter is 12.

NCERT Example 6 — 2 pencils + 3 erasers = ₹9; 4 pencils + 6 erasers = ₹18.

$2x+3y=9,\ 4x+6y=18$. From (1): $x=\dfrac{9-3y}{2}$. Substitute in (2): $\dfrac{4(9-3y)}{2}+6y=18\Rightarrow 18-6y+6y=18\Rightarrow 18=18$ — true for all $y$. So there are infinitely many solutions (the equations are the same line); cost cannot be uniquely found.

NCERT Example 7 — do rails $x+2y-4=0$ and $2x+4y-12=0$ cross?

From (1): $x=4-2y$. Substitute in (2): $2(4-2y)+4y-12=0\Rightarrow 8-4y+4y-12=0\Rightarrow -4=0$ — false. So no common solution: the rails are parallel and never cross.

8. Elimination method

Steps:

  • Step 1: multiply the equations by suitable numbers so the coefficient of one variable becomes numerically equal.
  • Step 2: add or subtract to eliminate that variable. (True statement means infinite; false means none.)
  • Step 3: solve the resulting single-variable equation.
  • Step 4: substitute back to get the other variable.
NCERT Example 8 — incomes 9:7, expenditures 4:3, each saves ₹2000.

Let incomes be $9x,7x$ and expenditures $4y,3y$. Savings: $9x-4y=2000$ and $7x-3y=2000$.

Multiply (1) by 3 and (2) by 4: $27x-12y=6000$ and $28x-12y=8000$. Subtract: $x=2000$. Then $9(2000)-4y=2000\Rightarrow y=4000$. Incomes $=9(2000)=$ ₹18000 and $7(2000)=$ ₹14000.

NCERT Example 9 — eliminate to solve $2x+3y=8,\ 4x+6y=7$.

Multiply (1) by 2: $4x+6y=16$. Subtract (2): $0=9$ — false. So the pair has no solution (parallel lines).

NCERT Example 10 — two-digit number; reversed sum 66; digits differ by 2.

Let tens digit $x$, units digit $y$. Number $=10x+y$, reversed $=10y+x$. Sum: $(10x+y)+(10y+x)=66\Rightarrow 11(x+y)=66\Rightarrow x+y=6$. Digits differ by 2: $x-y=2$ or $y-x=2$.

With $x-y=2$: adding to $x+y=6$ gives $x=4,y=2$ → number 42. With $y-x=2$: gives $x=2,y=4$ → number 24. So there are two such numbers: 42 and 24.

9. NCERT Exercise 3.1 — fully solved

Q1. Form the equations and solve graphically.

  • (i) 10 students; girls are 4 more than boys. $x+y=10,\ x-y=4$ (girls $x$, boys $y$). Solving: $x=7,\ y=3$ → 7 girls, 3 boys.
  • (ii) 5 pencils + 7 pens = ₹50; 7 pencils + 5 pens = ₹46. $5x+7y=50,\ 7x+5y=46$. Solving: $x=3,\ y=5$ → pencil ₹3, pen ₹5.

Q2. Compare ratios — intersect, parallel or coincident?

  • (i) $5x-4y+8=0,\ 7x+6y-9=0$: $\dfrac57\neq\dfrac{-4}{6}$ → intersect at a point.
  • (ii) $9x+3y+12=0,\ 18x+6y+24=0$: $\dfrac{9}{18}=\dfrac36=\dfrac{12}{24}=\dfrac12$ → coincident.
  • (iii) $6x-3y+10=0,\ 2x-y+9=0$: $\dfrac62=\dfrac{-3}{-1}=3$ but $\dfrac{10}{9}\neq3$ → parallel.

Q3. Consistent or inconsistent?

  • (i) $3x+2y=5,\ 2x-3y=7$: $\dfrac32\neq\dfrac{2}{-3}$ → consistent (unique).
  • (ii) $2x-3y=8,\ 4x-6y=9$: $\dfrac24=\dfrac{-3}{-6}\neq\dfrac89$ → inconsistent.
  • (iii) $\tfrac32x+\tfrac53y=7,\ 9x-10y=14$: $\dfrac{3/2}{9}=\dfrac16,\ \dfrac{5/3}{-10}=-\dfrac16$, unequal → consistent.
  • (iv) $5x-3y=11,\ -10x+6y=-22$: $\dfrac{5}{-10}=\dfrac{-3}{6}=\dfrac{11}{-22}=-\dfrac12$ → consistent (dependent, infinite).
  • (v) $\tfrac43x+2y=8,\ 2x+3y=12$: $\dfrac{4/3}{2}=\dfrac23,\ \dfrac23,\ \dfrac{8}{12}=\dfrac23$ — all equal → consistent (dependent).

Q4. Consistent/inconsistent; if consistent solve graphically.

  • (i) $x+y=5,\ 2x+2y=10$: same line → consistent, infinitely many solutions.
  • (ii) $x-y=8,\ 3x-3y=16$: $\dfrac13=\dfrac{-1}{-3}\neq\dfrac{8}{16}$ → inconsistent.
  • (iii) $2x+y-6=0,\ 4x-2y-4=0$: $\dfrac24\neq\dfrac{1}{-2}$ → consistent; solving gives $x=2,\ y=2$.
  • (iv) $2x-2y-2=0,\ 4x-4y-5=0$: $\dfrac24=\dfrac{-2}{-4}\neq\dfrac{-2}{-5}$ → inconsistent.

Q5. Rectangular garden: half-perimeter $=36$ m, length $4$ m more than width. $x+y=36,\ x-y=4$ → $x=20,\ y=16$. Dimensions: length 20 m, width 16 m.

Q6. Given $2x+3y-8=0$, write a second equation that gives:

  • (i) intersecting lines: e.g. $3x-2y-6=0$ (ratios of $a,b$ unequal).
  • (ii) parallel lines: e.g. $4x+6y-10=0$ ($\dfrac24=\dfrac36\neq\dfrac{-8}{-10}$).
  • (iii) coincident lines: e.g. $4x+6y-16=0$ (just double the given equation).

Q7. Triangle from $x-y+1=0,\ 3x+2y-12=0$ and the $x$-axis. The two lines meet at $(2,3)$. Line 1 meets $x$-axis ($y=0$) at $(-1,0)$; line 2 meets $x$-axis at $(4,0)$. Vertices of the triangle: $(-1,0),\ (4,0),\ (2,3)$; shade that region.

10. NCERT Exercise 3.2 — fully solved (substitution)

Q1. Solve by substitution.

  • (i) $x+y=14,\ x-y=4$: $x=14-y$ → $14-2y=4\Rightarrow y=5,\ x=9$.
  • (ii) $s-t=3,\ \dfrac{s}{3}+\dfrac{t}{2}=6$: $s=t+3$ → $\dfrac{t+3}{3}+\dfrac t2=6\Rightarrow 2(t+3)+3t=36\Rightarrow 5t=30,\ t=6,\ s=9$.
  • (iii) $3x-y=3,\ 9x-3y=9$: same line → infinitely many solutions (any $x$ with $y=3x-3$).
  • (iv) $0.2x+0.3y=1.3,\ 0.4x+0.5y=2.3$: clear decimals → $2x+3y=13,\ 4x+5y=23$. $x=\dfrac{13-3y}{2}$ → $2(13-3y)+5y=23\Rightarrow 26-y=23\Rightarrow y=3,\ x=2$.
  • (v) $\sqrt2\,x+\sqrt3\,y=0,\ \sqrt3\,x-\sqrt8\,y=0$: both pass through origin and meet only there → $x=0,\ y=0$.
  • (vi) $\dfrac{3x}{2}-\dfrac{5y}{3}=-2,\ \dfrac x3+\dfrac y2=\dfrac{13}{6}$: multiply through → $9x-10y=-12,\ 2x+3y=13$. Solving: $x=2,\ y=3$.

Q2. Solve $2x+3y=11,\ 2x-4y=-24$; then find $m$ with $y=mx+3$. Subtract: $7y=35\Rightarrow y=5$, then $2x+15=11\Rightarrow x=-2$. So $5=m(-2)+3\Rightarrow -2m=2\Rightarrow m=-1$.

Q3. Form equations and solve by substitution.

  • (i) Difference is 26, one is three times the other: $x-y=26,\ x=3y\Rightarrow 3y-y=26\Rightarrow y=13,\ x=39$.
  • (ii) Supplementary angles, larger exceeds smaller by 18°: $x+y=180,\ x-y=18\Rightarrow x=99^\circ,\ y=81^\circ$.
  • (iii) 7 bats + 6 balls = ₹3800; 3 bats + 5 balls = ₹1750. $7x+6y=3800,\ 3x+5y=1750$. Solving: bat $x=$ ₹500, ball $y=$ ₹50.
  • (iv) Taxi: fixed charge $x$ + per-km $y$. $x+10y=105,\ x+15y=155\Rightarrow 5y=50,\ y=10,\ x=5$. For 25 km: $5+25(10)=$ ₹255. (Fixed ₹5, ₹10/km.)
  • (v) Fraction $\dfrac{x}{y}$: $\dfrac{x+2}{y+2}=\dfrac{9}{11}$ and $\dfrac{x+3}{y+3}=\dfrac56$. These give $11x-9y=-4$ and $6x-5y=-3$; solving, $x=7,\ y=9$. Fraction $=\dfrac79$.
  • (vi) Jacob: in 5 years he is 3× son; 5 years ago he was 7× son. $(x+5)=3(y+5),\ (x-5)=7(y-5)$ → $x-3y=10,\ x-7y=-30$. Subtract: $4y=40,\ y=10,\ x=40$. Jacob 40, son 10.

11. NCERT Exercise 3.3 — fully solved (elimination)

Q1. Solve by elimination (and check by substitution).

  • (i) $x+y=5,\ 2x-3y=4$: multiply (1) by 2 → $2x+2y=10$; subtract (2): $5y=6\Rightarrow y=\dfrac65,\ x=5-\dfrac65=\dfrac{19}{5}$.
  • (ii) $3x+4y=10,\ 2x-2y=2$: multiply (2) by 2 → $4x-4y=4$; add (1): $7x=14\Rightarrow x=2,\ y=1$.
  • (iii) $3x-5y-4=0,\ 9x=2y+7$ i.e. $9x-2y-7=0$: multiply (1) by 3 → $9x-15y-12=0$; subtract: $-13y-5=0\Rightarrow y=-\dfrac{5}{13},\ x=\dfrac{9}{13}$.
  • (iv) $\dfrac x2+\dfrac{2y}{3}=-1,\ x-\dfrac y3=3$: clear → $3x+4y=-6,\ 3x-y=9$. Subtract: $5y=-15\Rightarrow y=-3,\ x=2$.

Q2. Form equations and solve by elimination.

  • (i) Fraction $\dfrac xy$: $\dfrac{x+1}{y-1}=1$ and $\dfrac{x}{y+1}=\dfrac12$. So $x-y=-2$ and $2x-y=1$. Subtract: $x=3,\ y=5$. Fraction $=\dfrac35$.
  • (ii) Five years ago Nuri was thrice Sonu; in ten years twice. $(x-5)=3(y-5),\ (x+10)=2(y+10)$ → $x-3y=-10,\ x-2y=10$. Subtract: $-y=-20\Rightarrow y=20,\ x=50$. Nuri 50, Sonu 20.
  • (iii) Digit sum 9; nine times the number = twice reversed. $x+y=9,\ 9(10x+y)=2(10y+x)\Rightarrow 88x-11y=0\Rightarrow 8x=y$. With $x+y=9$: $9x=9\Rightarrow x=1,\ y=8$. Number 18.
  • (iv) Meena's ₹2000 in 25 notes of ₹50 and ₹100. $x+y=25,\ 50x+100y=2000\Rightarrow x+2y=40$. Subtract: $y=15,\ x=10$. Ten ₹50 notes, fifteen ₹100 notes.
  • (v) Library: fixed charge for first 3 days $+$ extra per day. Saritha (7 days) ₹27: $x+4y=27$. Susy (5 days) ₹21: $x+2y=21$. Subtract: $2y=6\Rightarrow y=3,\ x=15$. Fixed ₹15, extra ₹3/day.

12. Equations reducible to linear form

Some equations are not linear but become linear with a clever substitution. Whenever you see $\dfrac1x$ or $\dfrac1y$, let $u=\dfrac1x,\ v=\dfrac1y$ — the pair turns into ordinary linear equations in $u,v$. Solve, then take reciprocals to recover $x,y$. The same idea handles fraction word-problems and the decimal/fraction-clearing seen in Exercises 3.2 and 3.3.

13. Summary of conditions (must memorise)

$$\frac{a_1}{a_2}\neq\frac{b_1}{b_2}\ \Rightarrow\ \text{unique solution (consistent, intersecting)}$$ $$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\ \Rightarrow\ \text{infinite solutions (consistent, dependent, coincident)}$$ $$\frac{a_1}{a_2}=\frac{b_1}{b_2}\neq\frac{c_1}{c_2}\ \Rightarrow\ \text{no solution (inconsistent, parallel)}$$

To find a missing constant $k$ that makes a pair have, say, infinitely many solutions, set the relevant ratios equal and solve for $k$.

14. Common mistakes to avoid

  • Reading off $a,b,c$ before writing both equations in $ax+by+c=0$ form — signs of $c$ go wrong otherwise.
  • Mixing up the conditions: "all three equal" is infinite, "first two equal, third different" is none.
  • Calling a dependent (coincident) pair "inconsistent" — it is consistent (it has solutions, just infinitely many).
  • In word problems, defining variables loosely. Always write "let $x=\dots,\ y=\dots$" clearly.
  • Forgetting the second case in problems like "digits differ by 2" — there can be two valid answers.
  • Not clearing fractions/decimals first — leads to arithmetic slips.
  • Choosing the graphical method when the answer is a fraction — use substitution/elimination for exactness.

15. Quick revision checklist

  • One linear equation = one line; a pair = two lines.
  • Three cases: intersect (unique), parallel (none), coincident (infinite).
  • Ratio test: $\dfrac{a_1}{a_2}$ vs $\dfrac{b_1}{b_2}$ vs $\dfrac{c_1}{c_2}$.
  • Solve exactly by substitution (express one variable, plug in) or elimination (match coefficients, add/subtract).
  • True statement with no variable means infinite; false means none.
  • For $\dfrac1x,\dfrac1y$ type, substitute $u=\dfrac1x,\ v=\dfrac1y$.
  • Always define variables, then verify the answer in both original equations.
Practice MCQs
1. The pair $x+2y=4,\ 2x+4y=12$ has:
  1. a unique solution
  2. exactly two solutions
  3. infinitely many solutions
  4. no solution
Answer: (D) $\dfrac12=\dfrac24\neq\dfrac{4}{12}$ → parallel, no solution.
2. For what value of $k$ does $2x+3y=5,\ 4x+ky=10$ have infinitely many solutions?
  1. $3$
  2. $6$
  3. $2$
  4. $9$
Answer: (B) Need $\dfrac24=\dfrac3k=\dfrac{5}{10}$ → $\dfrac3k=\dfrac12\Rightarrow k=6$.
3. A pair of linear equations is inconsistent when the lines are:
  1. intersecting
  2. coincident
  3. parallel
  4. perpendicular
Answer: (C) parallel lines never meet → no solution.
4. The solution of $x+y=14,\ x-y=4$ is:
  1. $x=9,y=5$
  2. $x=5,y=9$
  3. $x=10,y=4$
  4. $x=4,y=10$
Answer: (A) adding gives $2x=18,\ x=9,\ y=5$.
5. If $\dfrac{a_1}{a_2}\neq\dfrac{b_1}{b_2}$, the pair has:
  1. no solution
  2. a unique solution
  3. infinitely many
  4. cannot say
Answer: (B) intersecting lines → exactly one solution.
6. The lines $3x-y=3$ and $9x-3y=9$ are:
  1. intersecting
  2. parallel
  3. coincident
  4. perpendicular
Answer: (C) the second is just 3× the first → coincident, infinite solutions.
7. Two numbers differ by 26 and one is three times the other. The numbers are:
  1. $39,13$
  2. $26,52$
  3. $30,4$
  4. $36,12$
Answer: (A) $x-y=26,\ x=3y\Rightarrow y=13,\ x=39$.
8. A two-digit number with tens digit $x$ and units digit $y$ equals:
  1. $x+y$
  2. $xy$
  3. $10x+y$
  4. $10y+x$
Answer: (C) place value gives $10x+y$.
9. The pair $5x-8y+1=0,\ 3x-\tfrac{24}{5}y+\tfrac35=0$ has:
  1. a unique solution
  2. no solution
  3. infinitely many solutions
  4. exactly one integer solution
Answer: (C) multiplying the 2nd by $\tfrac53$ gives the 1st → coincident.
10. The point where $x+3y=6$ meets the $x$-axis is:
  1. $(0,2)$
  2. $(6,0)$
  3. $(0,6)$
  4. $(2,0)$
Answer: (B) on the $x$-axis $y=0\Rightarrow x=6$, i.e. $(6,0)$.
Assertion–Reason
A: The pair $2x-3y=8,\ 4x-6y=9$ has no solution.   R: If $\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}\neq\dfrac{c_1}{c_2}$, the lines are parallel.
Answer: Both A and R are true, and R correctly explains A — here $\dfrac24=\dfrac{-3}{-6}\neq\dfrac89$.
A: A dependent pair of linear equations is inconsistent.   R: A dependent pair represents two coincident lines.
Answer: A is false, R is true — a dependent (coincident) pair is consistent with infinitely many solutions.
Previous-year questions
Q1. Find the value of $k$ for which $kx-y=2$ and $6x-2y=3$ have no solution. (CBSE, 2 marks)
Answer: No solution needs $\dfrac k6=\dfrac{-1}{-2}\neq\dfrac{2}{3}$. So $\dfrac k6=\dfrac12\Rightarrow k=3$ (and $\dfrac12\neq\dfrac23$ ✓). Hence $k=3$.
Q2. The coach buys 7 bats and 6 balls for ₹3800 and 3 bats and 5 balls for ₹1750. Find the cost of each. (CBSE, 3 marks)
Answer: $7x+6y=3800,\ 3x+5y=1750$. Solving: bat = ₹500, ball = ₹50.
Q3. A fraction becomes $\dfrac9{11}$ if 2 is added to numerator and denominator, and $\dfrac56$ if 3 is added to both. Find it. (CBSE, 3 marks)
Answer: $11(x+2)=9(y+2),\ 6(x+3)=5(y+3)$ → $11x-9y=-4,\ 6x-5y=-3$. Solving: $x=7,\ y=9$, fraction $=\dfrac79$.
Q4. Solve $\dfrac x2+\dfrac{2y}{3}=-1$ and $x-\dfrac y3=3$. (CBSE, 3 marks)
Answer: Clear fractions → $3x+4y=-6,\ 3x-y=9$. Subtract: $5y=-15\Rightarrow y=-3,\ x=2$.
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