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) Quiz Students Solution
Let the number of girls be $x$ and the number of boys be $y$.
According to the given conditions: $$\text{Total students} = 10 \implies x + y = 10 \quad \text{--- (1)}$$ $$\text{Girls are 4 more than boys} \implies x = y + 4 \implies x - y = 4 \quad \text{--- (2)}$$
Table of Values for Plotting:
For $x + y = 10$: when $x = 5 \implies y = 5$; when $x = 6 \implies y = 4$; when $x = 7 \implies y = 3$.
For $x - y = 4$: when $x = 4 \implies y = 0$; when $x = 5 \implies y = 1$; when $x = 7 \implies y = 3$.
Graph Intersection:
Both lines intersect at the point $(7, 3)$.
$$\therefore \text{Number of girls } (x) = 7, \quad \text{Number of boys } (y) = 3$$
According to the given conditions: $$\text{Total students} = 10 \implies x + y = 10 \quad \text{--- (1)}$$ $$\text{Girls are 4 more than boys} \implies x = y + 4 \implies x - y = 4 \quad \text{--- (2)}$$
Table of Values for Plotting:
For $x + y = 10$: when $x = 5 \implies y = 5$; when $x = 6 \implies y = 4$; when $x = 7 \implies y = 3$.
For $x - y = 4$: when $x = 4 \implies y = 0$; when $x = 5 \implies y = 1$; when $x = 7 \implies y = 3$.
Graph Intersection:
Both lines intersect at the point $(7, 3)$.
$$\therefore \text{Number of girls } (x) = 7, \quad \text{Number of boys } (y) = 3$$
(ii) Cost of Pencils and Pens
Let the cost of 1 pencil be ₹$x$ and 1 pen be ₹$y$.
According to the problem: $$5x + 7y = 50 \quad \text{--- (1)}$$ $$7x + 5y = 46 \quad \text{--- (2)}$$
Table of Values for Plotting:
For $5x + 7y = 50$: Points are $(3, 5), (10, 0), (-4, 10)$.
For $7x + 5y = 46$: Points are $(3, 5), (8, -2), (-2, 12)$.
Graph Intersection:
The lines intersect at point $(3, 5)$.
$$\therefore \text{Cost of 1 pencil } = ₹3, \quad \text{Cost of 1 pen } = ₹5$$
According to the problem: $$5x + 7y = 50 \quad \text{--- (1)}$$ $$7x + 5y = 46 \quad \text{--- (2)}$$
Table of Values for Plotting:
For $5x + 7y = 50$: Points are $(3, 5), (10, 0), (-4, 10)$.
For $7x + 5y = 46$: Points are $(3, 5), (8, -2), (-2, 12)$.
Graph Intersection:
The lines intersect at point $(3, 5)$.
$$\therefore \text{Cost of 1 pencil } = ₹3, \quad \text{Cost of 1 pen } = ₹5$$
Solutions: (i) Girls = 7, Boys = 3; (ii) Cost of 1 pencil = ₹3, Cost of 1 pen = ₹5.
Question 2
On comparing the ratios $\dfrac{a_1}{a_2}, \dfrac{b_1}{b_2}$ and $\dfrac{c_1}{c_2}$, 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) $6x - 3y + 10 = 0; \quad 2x - y + 9 = 0$
(ii) $9x + 3y + 12 = 0; \quad 18x + 6y + 24 = 0$
(iii) $6x - 3y + 10 = 0; \quad 2x - y + 9 = 0$
(i) $5x - 4y + 8 = 0$ and $7x + 6y - 9 = 0$
Here, $a_1 = 5, b_1 = -4, c_1 = 8$ and $a_2 = 7, b_2 = 6, c_2 = -9$.
$$\frac{a_1}{a_2} = \frac{5}{7}, \quad \frac{b_1}{b_2} = \frac{-4}{6} = -\frac{2}{3}$$ Since $\dfrac{a_1}{a_2} \neq \dfrac{b_1}{b_2}$, the lines intersect at exactly one point (unique solution).
$$\frac{a_1}{a_2} = \frac{5}{7}, \quad \frac{b_1}{b_2} = \frac{-4}{6} = -\frac{2}{3}$$ Since $\dfrac{a_1}{a_2} \neq \dfrac{b_1}{b_2}$, the lines intersect at exactly one point (unique solution).
(ii) $9x + 3y + 12 = 0$ and $18x + 6y + 24 = 0$
Here, $a_1 = 9, b_1 = 3, c_1 = 12$ and $a_2 = 18, b_2 = 6, c_2 = 24$.
$$\frac{a_1}{a_2} = \frac{9}{18} = \frac{1}{2}, \quad \frac{b_1}{b_2} = \frac{3}{6} = \frac{1}{2}, \quad \frac{c_1}{c_2} = \frac{12}{24} = \frac{1}{2}$$ Since $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}$, the lines are coincident (infinitely many solutions).
$$\frac{a_1}{a_2} = \frac{9}{18} = \frac{1}{2}, \quad \frac{b_1}{b_2} = \frac{3}{6} = \frac{1}{2}, \quad \frac{c_1}{c_2} = \frac{12}{24} = \frac{1}{2}$$ Since $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}$, the lines are coincident (infinitely many solutions).
(iii) $6x - 3y + 10 = 0$ and $2x - y + 9 = 0$
Here, $a_1 = 6, b_1 = -3, c_1 = 10$ and $a_2 = 2, b_2 = -1, c_2 = 9$.
$$\frac{a_1}{a_2} = \frac{6}{2} = 3, \quad \frac{b_1}{b_2} = \frac{-3}{-1} = 3, \quad \frac{c_1}{c_2} = \frac{10}{9}$$ Since $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2}$, the lines are parallel (no solution).
$$\frac{a_1}{a_2} = \frac{6}{2} = 3, \quad \frac{b_1}{b_2} = \frac{-3}{-1} = 3, \quad \frac{c_1}{c_2} = \frac{10}{9}$$ Since $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2}$, the lines are parallel (no solution).
Summary: (i) Intersecting lines; (ii) Coincident lines; (iii) Parallel lines.
Question 3
On comparing the ratios $\dfrac{a_1}{a_2}, \dfrac{b_1}{b_2}$ and $\dfrac{c_1}{c_2}$, 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) $\frac{4}{3}x + 2y = 8; \quad 2x + 3y = 12$
(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) $\frac{4}{3}x + 2y = 8; \quad 2x + 3y = 12$
Ratio Rules for Consistency
- Consistent: Has one or more solutions ($\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$ or $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$)
- Inconsistent: Has no solution ($\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$)
- Inconsistent: Has no solution ($\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$)
Step-by-Step Ratio Evaluation
(i) $\dfrac{a_1}{a_2} = \dfrac{3}{2}, \quad \dfrac{b_1}{b_2} = \dfrac{2}{-3} = -\dfrac{2}{3}$.
Since $\dfrac{a_1}{a_2} \neq \dfrac{b_1}{b_2} \implies$ Consistent.
(ii) $\dfrac{a_1}{a_2} = \dfrac{2}{4} = \dfrac{1}{2}, \quad \dfrac{b_1}{b_2} = \dfrac{-3}{-6} = \dfrac{1}{2}, \quad \dfrac{c_1}{c_2} = \dfrac{-8}{-9} = \dfrac{8}{9}$.
Since $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2} \implies$ Inconsistent.
(iii) $\dfrac{a_1}{a_2} = \dfrac{3/2}{9} = \dfrac{1}{6}, \quad \dfrac{b_1}{b_2} = \dfrac{5/3}{-10} = -\dfrac{1}{6}$.
Since $\dfrac{a_1}{a_2} \neq \dfrac{b_1}{b_2} \implies$ Consistent.
(iv) $\dfrac{a_1}{a_2} = \dfrac{5}{-10} = -\dfrac{1}{2}, \quad \dfrac{b_1}{b_2} = \dfrac{-3}{6} = -\dfrac{1}{2}, \quad \dfrac{c_1}{c_2} = \dfrac{-11}{22} = -\dfrac{1}{2}$.
Since $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2} \implies$ Consistent (Dependent).
(v) $\dfrac{a_1}{a_2} = \dfrac{4/3}{2} = \dfrac{2}{3}, \quad \dfrac{b_1}{b_2} = \dfrac{2}{3}, \quad \dfrac{c_1}{c_2} = \dfrac{-8}{-12} = \dfrac{2}{3}$.
Since $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2} \implies$ Consistent.
Since $\dfrac{a_1}{a_2} \neq \dfrac{b_1}{b_2} \implies$ Consistent.
(ii) $\dfrac{a_1}{a_2} = \dfrac{2}{4} = \dfrac{1}{2}, \quad \dfrac{b_1}{b_2} = \dfrac{-3}{-6} = \dfrac{1}{2}, \quad \dfrac{c_1}{c_2} = \dfrac{-8}{-9} = \dfrac{8}{9}$.
Since $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2} \implies$ Inconsistent.
(iii) $\dfrac{a_1}{a_2} = \dfrac{3/2}{9} = \dfrac{1}{6}, \quad \dfrac{b_1}{b_2} = \dfrac{5/3}{-10} = -\dfrac{1}{6}$.
Since $\dfrac{a_1}{a_2} \neq \dfrac{b_1}{b_2} \implies$ Consistent.
(iv) $\dfrac{a_1}{a_2} = \dfrac{5}{-10} = -\dfrac{1}{2}, \quad \dfrac{b_1}{b_2} = \dfrac{-3}{6} = -\dfrac{1}{2}, \quad \dfrac{c_1}{c_2} = \dfrac{-11}{22} = -\dfrac{1}{2}$.
Since $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2} \implies$ Consistent (Dependent).
(v) $\dfrac{a_1}{a_2} = \dfrac{4/3}{2} = \dfrac{2}{3}, \quad \dfrac{b_1}{b_2} = \dfrac{2}{3}, \quad \dfrac{c_1}{c_2} = \dfrac{-8}{-12} = \dfrac{2}{3}$.
Since $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2} \implies$ Consistent.
Results: (i) Consistent; (ii) Inconsistent; (iii) Consistent; (iv) Consistent; (v) 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) $2x - 2y - 2 = 0, \quad 4x - 4y - 5 = 0$
(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) $x + y = 5$ and $2x + 2y = 10$
$$\frac{a_1}{a_2} = \frac{1}{2}, \quad \frac{b_1}{b_2} = \frac{1}{2}, \quad \frac{c_1}{c_2} = \frac{5}{10} = \frac{1}{2}$$
Lines are coincident $\implies$ Consistent with infinitely many solutions lying on $x + y = 5$.
(ii) $x - y = 8$ and $3x - 3y = 16$
$$\frac{a_1}{a_2} = \frac{1}{3}, \quad \frac{b_1}{b_2} = \frac{-1}{-3} = \frac{1}{3}, \quad \frac{c_1}{c_2} = \frac{8}{16} = \frac{1}{2}$$
Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$, lines are parallel $\implies$ Inconsistent (no solution).
(iii) $2x + y - 6 = 0$ and $4x - 2y - 4 = 0$
$$\frac{a_1}{a_2} = \frac{2}{4} = \frac{1}{2}, \quad \frac{b_1}{b_2} = \frac{1}{-2} = -\frac{1}{2}$$
Since $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$, lines intersect $\implies$ Consistent.
Graphical Solution: Plotting the lines shows they intersect at $(2, 2)$.
$$\therefore x = 2, \quad y = 2$$
Graphical Solution: Plotting the lines shows they intersect at $(2, 2)$.
$$\therefore x = 2, \quad y = 2$$
(iv) $2x - 2y - 2 = 0$ and $4x - 4y - 5 = 0$
$$\frac{a_1}{a_2} = \frac{2}{4} = \frac{1}{2}, \quad \frac{b_1}{b_2} = \frac{-2}{-4} = \frac{1}{2}, \quad \frac{c_1}{c_2} = \frac{-2}{-5} = \frac{2}{5}$$
Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$, lines are parallel $\implies$ Inconsistent.
Results: (i) Consistent; (ii) Inconsistent; (iii) Consistent with $(2, 2)$; (iv) 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.
Formulate Equations
Let the length of the garden be $l$ metres and the width be $w$ metres.
Perimeter of rectangle $= 2(l + w)$.
$$\text{Half perimeter} = l + w = 36 \quad \text{--- (1)}$$ Given that length is 4 m more than width: $$l = w + 4 \implies l - w = 4 \quad \text{--- (2)}$$
Perimeter of rectangle $= 2(l + w)$.
$$\text{Half perimeter} = l + w = 36 \quad \text{--- (1)}$$ Given that length is 4 m more than width: $$l = w + 4 \implies l - w = 4 \quad \text{--- (2)}$$
Solve by Substitution
Substitute $l = w + 4$ into Equation (1):
$$(w + 4) + w = 36$$
$$2w + 4 = 36 \implies 2w = 32 \implies w = 16\text{ m}$$
Substitute $w = 16$ back:
$$l = 16 + 4 = 20\text{ m}$$
$$\therefore \text{Length} = 20\text{ m}, \quad \text{Width} = 16\text{ m}$$
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
(i) Intersecting Lines Condition
Condition: $\dfrac{a_1}{a_2} \neq \dfrac{b_1}{b_2}$.
Let second equation be: $\mathbf{3x + 2y - 7 = 0}$.
Check: $\dfrac{2}{3} \neq \dfrac{3}{2}$. Holds true.
Let second equation be: $\mathbf{3x + 2y - 7 = 0}$.
Check: $\dfrac{2}{3} \neq \dfrac{3}{2}$. Holds true.
(ii) Parallel Lines Condition
Condition: $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2}$.
Let second equation be: $\mathbf{4x + 6y - 12 = 0}$.
Check: $\dfrac{2}{4} = \dfrac{3}{6} = \dfrac{1}{2} \neq \dfrac{-8}{-12} = \dfrac{2}{3}$. Holds true.
Let second equation be: $\mathbf{4x + 6y - 12 = 0}$.
Check: $\dfrac{2}{4} = \dfrac{3}{6} = \dfrac{1}{2} \neq \dfrac{-8}{-12} = \dfrac{2}{3}$. Holds true.
(iii) Coincident Lines Condition
Condition: $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}$.
Multiply the given equation by 3: $\mathbf{6x + 9y - 24 = 0}$.
Check: $\dfrac{2}{6} = \dfrac{3}{9} = \dfrac{-8}{-24} = \dfrac{1}{3}$. Holds true.
Multiply the given equation by 3: $\mathbf{6x + 9y - 24 = 0}$.
Check: $\dfrac{2}{6} = \dfrac{3}{9} = \dfrac{-8}{-24} = \dfrac{1}{3}$. Holds true.
Possible equations: (i) $3x + 2y - 7 = 0$; (ii) $4x + 6y - 12 = 0$; (iii) $6x + 9y - 24 = 0$.
Question 7 (Board Favorite)
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.
Line 1: $x - y + 1 = 0 \implies y = x + 1$
- If $x = 0 \implies y = 1 \implies (0, 1)$
- If $x = -1 \implies y = 0 \implies (-1, 0)$ [Intersects $x$-axis]
- If $x = 2 \implies y = 3 \implies (2, 3)$
- If $x = -1 \implies y = 0 \implies (-1, 0)$ [Intersects $x$-axis]
- If $x = 2 \implies y = 3 \implies (2, 3)$
Line 2: $3x + 2y - 12 = 0 \implies 2y = 12 - 3x$
- If $x = 0 \implies y = 6 \implies (0, 6)$
- If $x = 4 \implies y = 0 \implies (4, 0)$ [Intersects $x$-axis]
- If $x = 2 \implies y = 3 \implies (2, 3)$
- If $x = 4 \implies y = 0 \implies (4, 0)$ [Intersects $x$-axis]
- If $x = 2 \implies y = 3 \implies (2, 3)$
Finding Vertices of Triangle
1. Point of intersection of the two lines is $A(2, 3)$.
2. Point of intersection of Line 1 with $x$-axis is $B(-1, 0)$.
3. Point of intersection of Line 2 with $x$-axis is $C(4, 0)$.
2. Point of intersection of Line 1 with $x$-axis is $B(-1, 0)$.
3. Point of intersection of Line 2 with $x$-axis is $C(4, 0)$.
$$\therefore \text{Vertices of the shaded triangle are: } A(2, 3), \quad B(-1, 0), \quad C(4, 0)$$