Pair of Linear Equations – Exercise 3.1

Question 1

Form the pair of linear equations in the following problems, and find their solutions graphically.

(i) 10 students of Class X took part in a Mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls who took part in the quiz.

(ii) 5 pencils and 7 pens together cost ₹ 50, whereas 7 pencils and 5 pens together cost ₹ 46. Find the cost of one pencil and that of one pen.

(i)
Let the number of girls be $x$ and the number of boys be $y$.
Total students = 10 $\Rightarrow x + y = 10$ ...(1)
Girls are 4 more than boys $\Rightarrow x = y + 4 \Rightarrow x - y = 4$ ...(2)

Plotting the graph for these equations:
For $x + y = 10$: Points (5,5), (6,4), (7,3)
For $x - y = 4$: Points (4,0), (5,1), (6,2)

The lines intersect at point $(7, 3)$.
So, $x = 7$ and $y = 3$.
✅ Number of girls = 7, Number of boys = 3.

(ii)
Let cost of 1 pencil be ₹ $x$ and 1 pen be ₹ $y$.
Case 1: $5x + 7y = 50$ ...(1)
Case 2: $7x + 5y = 46$ ...(2)

Plotting the graph:
For $5x + 7y = 50$: Points (3, 5), (10, 0), (-4, 10)
For $7x + 5y = 46$: Points (3, 5), (8, -2), (-2, 12)

The lines intersect at point $(3, 5)$.
✅ Cost of one pencil = ₹ 3, Cost of one pen = ₹ 5.

Question 2

On comparing the ratios $\frac{a_1}{a_2}, \frac{b_1}{b_2}$ and $\frac{c_1}{c_2}$, find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincident:

(i) $5x - 4y + 8 = 0; \quad 7x + 6y - 9 = 0$

(ii) $9x + 3y + 12 = 0; \quad 18x + 6y + 24 = 0$

(iii) $6x - 3y + 10 = 0; \quad 2x - y + 9 = 0$

(i)
$\frac{a_1}{a_2} = \frac{5}{7}, \quad \frac{b_1}{b_2} = \frac{-4}{6} = \frac{-2}{3}$
Since $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$, the lines intersect at a point.

(ii)
$\frac{a_1}{a_2} = \frac{9}{18} = \frac{1}{2}, \quad \frac{b_1}{b_2} = \frac{3}{6} = \frac{1}{2}, \quad \frac{c_1}{c_2} = \frac{12}{24} = \frac{1}{2}$
Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$, the lines are coincident.

(iii)
$\frac{a_1}{a_2} = \frac{6}{2} = 3, \quad \frac{b_1}{b_2} = \frac{-3}{-1} = 3, \quad \frac{c_1}{c_2} = \frac{10}{9}$
Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$, the lines are parallel.

Question 3

On comparing the ratios $\frac{a_1}{a_2}, \frac{b_1}{b_2}$ and $\frac{c_1}{c_2}$, find out whether the following pair of linear equations are consistent, or inconsistent.

(i) $3x + 2y = 5; \quad 2x - 3y = 7$

(ii) $2x - 3y = 8; \quad 4x - 6y = 9$

(iii) $\frac{3}{2}x + \frac{5}{3}y = 7; \quad 9x - 10y = 14$

(iv) $5x - 3y = 11; \quad -10x + 6y = -22$

(v) $\frac{4}{3}x + 2y = 8; \quad 2x + 3y = 12$

(i) $\frac{3}{2} \neq \frac{2}{-3}$ (Intersecting) $\Rightarrow$ Consistent
(ii) $\frac{2}{4} = \frac{-3}{-6} \neq \frac{8}{9}$ (Parallel) $\Rightarrow$ Inconsistent
(iii) $\frac{3/2}{9} = \frac{1}{6}, \frac{5/3}{-10} = \frac{-1}{6}$. Not equal (Intersecting) $\Rightarrow$ Consistent
(iv) $\frac{5}{-10} = \frac{-3}{6} = \frac{11}{-22} = -\frac{1}{2}$ (Coincident) $\Rightarrow$ Consistent (Dependent)
(v) $\frac{4/3}{2} = \frac{2}{3}, \frac{2}{3} = \frac{2}{3}, \frac{8}{12} = \frac{2}{3}$ (Coincident) $\Rightarrow$ Consistent

Question 4

Which of the following pairs of linear equations are consistent/inconsistent? If consistent, obtain the solution graphically:

(i) $x + y = 5, \quad 2x + 2y = 10$

(ii) $x - y = 8, \quad 3x - 3y = 16$

(iii) $2x + y - 6 = 0, \quad 4x - 2y - 4 = 0$

(iv) $2x - 2y - 2 = 0, \quad 4x - 4y - 5 = 0$

(i) $\frac{1}{2} = \frac{1}{2} = \frac{5}{10}$. Coincident lines. Consistent. Infinitely many solutions.
(ii) $\frac{1}{3} = \frac{-1}{-3} \neq \frac{8}{16}$. Parallel lines. Inconsistent.
(iii) $\frac{2}{4} \neq \frac{1}{-2}$. Intersecting lines. Consistent.
Graph solution: Lines intersect at $(2, 2)$.
(iv) $\frac{2}{4} = \frac{-2}{-4} \neq \frac{-2}{-5}$. Parallel lines. Inconsistent.

Question 5

Half the perimeter of a rectangular garden, whose length is 4 m more than its width, is 36 m. Find the dimensions of the garden.

Let length be $l$ and width be $w$.
Perimeter $P = 2(l + w)$. Half perimeter = $l + w = 36$.
Given: $l = w + 4$.

Substitute $l$ in the first equation:
$(w + 4) + w = 36$
$2w = 32 \Rightarrow w = 16$ m.
$l = 16 + 4 = 20$ m.

✅ Length = 20 m, Width = 16 m.

Question 6

Given the linear equation $2x + 3y - 8 = 0$, write another linear equation in two variables such that the geometrical representation of the pair so formed is:

(i) intersecting lines

(ii) parallel lines

(iii) coincident lines

Given: $2x + 3y - 8 = 0$

(i) Intersecting lines ($\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$):
Example: $3x + 2y - 7 = 0$ (Since $\frac{2}{3} \neq \frac{3}{2}$)

(ii) Parallel lines ($\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$):
Example: $4x + 6y - 12 = 0$ (Since $\frac{2}{4} = \frac{3}{6} \neq \frac{-8}{-12}$)

(iii) Coincident lines ($\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$):
Example: $6x + 9y - 24 = 0$ (Multiplying given equation by 3)

Question 7

Draw the graphs of the equations $x - y + 1 = 0$ and $3x + 2y - 12 = 0$. Determine the coordinates of the vertices of the triangle formed by these lines and the x-axis, and shade the triangular region.

Equation 1: $x - y + 1 = 0 \Rightarrow y = x + 1$
Points: $(0, 1), (-1, 0), (2, 3)$

Equation 2: $3x + 2y - 12 = 0 \Rightarrow 2y = 12 - 3x$
Points: $(0, 6), (4, 0), (2, 3)$

Intersection: The lines intersect at $(2, 3)$.
X-axis Intercepts:
Line 1 intersects x-axis at $(-1, 0)$.
Line 2 intersects x-axis at $(4, 0)$.

✅ Vertices of the triangle are $(-1, 0)$, $(4, 0)$, and $(2, 3)$.

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