Exercise-3.1, Class 10th, Maths, Chapter 3, NCERT

Q1. Form the pair of linear equations and solve graphically.

(i) 10 students; number of girls is 4 more than number of boys.
Let number of boys = $x$, girls = $y$.
Equations:

$$x+y=10,y=x+4.$$

Substitute: $x+(x+4)=10\Rightarrow 2x+4=10\Rightarrow 2x=6\Rightarrow x=3$
So $y=7$.

Answer: 3 boys and 7 girls.

(ii) 5 pencils + 7 pens cost ₹50; 7 pencils + 5 pens cost ₹46.
Let pencil = $p$, pen = $q$.

$$5p+7q=50,7p+5q=46.$$

Subtract the second from the first:

$$(5p+7q)-(7p+5q)=50-46\Rightarrow -2p+2q=4\Rightarrow -p+q=2\Rightarrow q=p+2.$$

Substitute into $5p+7q=50$:

$$5p+7(p+2)=50\Rightarrow 5p+7p+14=50\Rightarrow 12p=36\Rightarrow p=3.$$

So $q=5$

Answer: pencil = ₹3, pen = ₹5.

---

Q2. Determine whether the pairs represent intersecting, parallel or coincident lines (compare ratios).

General rule for two linear equations $a_{1}x+b_{1}y+c_1=\mathrm{0 }$and $a_{2}x+b_{2}y+c_2=0$

• If $\frac{a_1}{a_2}\mathrm{\ne }\frac{b_1}{b_2}$→ intersecting (one unique solution).

• If $\frac{a_1}{a_2}=\frac{b_1}{b_2}\mathrm{\ne }\frac{c_1}{c_2}$→ parallel (no solution).

• If $\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_{}}$ → coincident (infinitely many solutions).

(i) $5x-4y+8=\mathrm{0 }$and $7x+6y-9=0.$
$\frac{a_1}{a_2}=\frac{5}{7},\frac{b_1}{b_2}=\frac{-4}{6}=-\frac{2}{3}$ They are not equal ⇒ intersecting.

(ii) $9x+3y+12=0$ and $18x+6y+24=0.$
$\frac{a_1}{a_2}=\frac{9}{18}=\frac{1}{2},\frac{b_1}{b_2}=\frac{3}{6}=\frac{1}{2},\frac{c_1}{c_2}=\frac{12}{24}=\frac{1}{2}$. All equal ⇒ coincident (infinitely many solutions).

(iii) $6x-3y+10=0$ and $2x-y+9=0.$
$\frac{a_1}{a_2}=\frac{6}{2}=3,\frac{b_1}{b_2}=\frac{-3}{-1}=3,\frac{c_1}{c_2}=\frac{10}{9}\mathrm{\ne }3$. So $\frac{a_1}{a_2}=\frac{b_1}{b_2}\mathrm{\ne }\frac{c_1}{c_2}$⇒ parallel (no solution).

---

Q3. Using ratio comparisons, say whether pairs are consistent or inconsistent.

(i) $3x+2y=5$ and $2x-3y=7.$
$\frac{a_1}{a_2}=\frac{3}{2},\frac{b_1}{b_2}=\frac{2}{-3}$→ not equal ⇒ consistent (unique solution).

(ii) $2x-3y=8$ and $4x-6y=9.$
$\frac{a_1}{a_2}=\frac{2}{4}=\frac{1}{2},\frac{b_1}{b_2}=\frac{-3}{-6}=\frac{1}{2},\frac{c_1}{c_2}=\frac{8}{9}\mathrm{\ne }\frac{1}{2}$ ⇒ parallel ⇒ inconsistent (no solution).

(iii)  $\frac{3}{5}x+\frac{7}{2}y=3$, and the second is $9x-10y=14$.
Multiply first by 10 to clear denominators: $6x+35y=30$

Second: $9x-10y=14$. These two have $\frac{a_1}{a_2}=\frac{6}{9}=\frac{2}{3}$ and $\frac{b_1}{b_2}=\frac{35}{-10}=-\frac{7}{2}$ — not equal ⇒ consistent (unique solution).

(iv) $5x-3y=11$ and $-10x+6y=-22.$
Second is $-2\times$(first) so all ratios equal ⇒ coincident, infinitely many solutions (consistent).

(v)  $2x+3y=12$

---

Q4. Which of the following are consistent/inconsistent? If consistent, solve graphically.

(i) $x+y=5,2x+2y=10.$Second is $2\times$ first ⇒ coincident (infinitely many solutions).

(ii) $x-y=8,3x-3y=16$ If second were $3\times$ first it would be 24 on RHS, but it’s 16 ⇒ contradiction ⇒ inconsistent (no solution).

(iii) $2x+y-6=0$ and $4x-2y-4=0.$ Rewrite:

$$2x+y=6\text{and}4x-2y=4.$$

Solve: from first $y=6-2x$. Substitute into second:

$$4x-2(6-2x)=4\Rightarrow 4x-12+4x=4\Rightarrow 8x=16\Rightarrow x=2.$$

Then $y=6-4=2$

Answer: unique solution $(2,2)$ — consistent.

(iv) $2x-2y-2=0$ and $4x-4y-5=0.$

First ⇒ $2x-2y=2$. Multiply first by 2 gives $4x-4y=4$ but second has $4x-4y=5$ ⇒ contradiction ⇒ inconsistent (no solution).

---

Q5. Rectangle: half-perimeter = 36 m; length = width + 4. Find dimensions.

Half-perimeter = $l+w=36$. Given $l=w+4$. So $(w+4)+w=36\Rightarrow 2w+4=36\Rightarrow w=16$

 Then $l=20$

Answer: Width = 16 m, Length = 20 m.

---

Q6. Given the line $2x+3y-8=\mathrm{0 }$, write another linear equation so that the pair is:

(i) intersecting lines — pick any line not proportional in $a,b$.
Example: $x-y=0$. The pair $2x+3y-8=0$ and $x-y=0$ intersect (unique solution).

(ii) parallel lines — choose $a,b$ proportional but $c$ not proportional.
Example: $4x+6y-10=0$. Here $\frac{2}{4}=\frac{3}{6}=\frac{1}{2}$ but $\frac{-8}{-10}=0.8\mathrm{\ne }\frac{1}{2}$ ⇒ parallel, no solution.

(iii) coincident lines — exact scalar multiple, e.g. $4x+6y-16=0$ (= $2\times$(first)). ⇒ coincident, infinitely many solutions.

---

Q7. Draw graphs of $x-y+1=0$ and $3x+2y-12=0$. Find vertices of the triangle formed by these lines and the x-axis; shade the triangular region.

Equations:

• $x-y+1=0\Rightarrow y=x+\mathrm{1. }$ x-intercept: set $y=0\Rightarrow x=-1$ → point $A(-1,0)$.

• $3x+2y-12=0\Rightarrow y=-\frac{3}{2}x+6.$ x-intercept: set $y=0\Rightarrow x=4$ → point $B(4,0)$.

• Intersection of the two lines: solve $x+1=-\frac{3}{2}x+6\Rightarrow \frac{5}{2}x=5\Rightarrow x=2$. Then $y=3$ → point $C(2,3)$.

Vertices of triangle: $(-1,0),(4,0),(2,3)$. Shade the triangular region bounded by the two lines and the x-axis.