Exercise 3.1 Solutions | Class 10 Maths

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: $x + (x + 4) = 10 \Rightarrow 2 x + 4 = 10 \Rightarrow 2 x = 6 \Rightarrow x = 3$
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 = $q$ .

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

Subtract the second from the first:

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

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

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

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 $a_{1} x + b_{1} y + c_{1} = 0$ and $a_{2} x + b_{2} y + c_{2} = 0$

If $\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}$ → intersecting (one unique solution).
If $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$ → parallel (no solution).
If $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c}$ → coincident (infinitely many solutions).

(i) $5 x - 4 y + 8 = 0$ and $7 x + 6 y - 9 = 0.$
$\frac{a_{1}}{a_{2}} = \frac{5}{7}, \frac{b_{1}}{b_{2}} = \frac{- 4}{6} = - \frac{2}{3}$ They are not equal ⇒ intersecting.

(ii) $9 x + 3 y + 12 = 0$ and $18 x + 6 y + 24 = 0.$
$\frac{a_{1}}{a_{2}} = \frac{9}{18} = \frac{1}{2}, \frac{b_{1}}{b_{2}} = \frac{3}{6} = \frac{1}{2}, \frac{c_{1}}{c_{2}} = \frac{12}{24} = \frac{1}{2}$ . All equal ⇒ coincident (infinitely many solutions).

(iii) $6 x - 3 y + 10 = 0$ and $2 x - y + 9 = 0.$
$\frac{a_{1}}{a_{2}} = \frac{6}{2} = 3, \frac{b_{1}}{b_{2}} = \frac{- 3}{- 1} = 3, \frac{c_{1}}{c_{2}} = \frac{10}{9} \neq 3$ . So $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$ ⇒ parallel (no solution).

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

(i) $3 x + 2 y = 5$ and $2 x - 3 y = 7.$
$\frac{a_{1}}{a_{2}} = \frac{3}{2}, \frac{b_{1}}{b_{2}} = \frac{2}{- 3}$ → not equal ⇒ consistent (unique solution).

(ii) $2 x - 3 y = 8$ and $4 x - 6 y = 9.$
$\frac{a_{1}}{a_{2}} = \frac{2}{4} = \frac{1}{2}, \frac{b_{1}}{b_{2}} = \frac{- 3}{- 6} = \frac{1}{2}, \frac{c_{1}}{c_{2}} = \frac{8}{9} \neq \frac{1}{2}$ ⇒ parallel ⇒ inconsistent (no solution).

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

Second: $9 x - 10 y = 14$ . 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) $5 x - 3 y = 11$ and $- 10 x + 6 y = - 22.$
Second is $- 2 \times$ (first) so all ratios equal ⇒ coincident, infinitely many solutions (consistent).

(v) $2 x + 3 y = 12$

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

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

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

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

$2 x + y = 6 and 4 x - 2 y = 4.$

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

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

Then $y = 6 - 4 = 2$

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

(iv) $2 x - 2 y - 2 = 0$ and $4 x - 4 y - 5 = 0.$

First ⇒ $2 x - 2 y = 2$ . Multiply first by 2 gives $4 x - 4 y = 4$ but second has $4 x - 4 y = 5$ ⇒ 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 $(w + 4) + w = 36 \Rightarrow 2 w + 4 = 36 \Rightarrow w = 16$

Then $l = 20$

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

Q6. Given the line $2 x + 3 y - 8 = 0$ , 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 $2 x + 3 y - 8 = 0$ and $x - y = 0$ intersect (unique solution).

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

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

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

Equations:

$x - y + 1 = 0 \Rightarrow y = x + 1.$ x-intercept: set $y = 0 \Rightarrow x = - 1$ → point $A (- 1, 0)$ .
$3 x + 2 y - 12 = 0 \Rightarrow y = - \frac{3}{2} x + 6.$ 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.

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

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

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

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

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

Q6. Given the line 2x+3y−8=0 2x+3y−8=0, write another linear equation so that the pair is:

Q7. Draw graphs of x−y+1=0x−y+1=0 and 3x+2y−12=03x+2y−12=0. Find vertices of the triangle formed by these lines and the x-axis; shade the triangular region.

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

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

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