Solution

(i) Let the number of girls be $x$ and the number of boys be $y$.

According to the question, the algebraic representation is

$x+y=10$

$x-y=4$

For $x+y=10$,

$x=10-y$

For $x-y=4$,

$x=4+y$

Hence, the graphic representation is as follows.

From the figure, it can be observed that these lines intersect each other at point $(7,3)$.

Therefore, the number of girls and boys in the class are 7 and 3 respectively.

(ii) Let the cost of 1 pencil be Rs $x$ and the cost of 1 pen be Rs $y$.

According to the question, the algebraic representation is

$5 x+7 y=50$

$7 x+5 y=46$

For $5 x+7 y=50$,

$x=\dfrac{50-7 y}{5}$

$7 x+5 y=46$

$x=\dfrac{46-5 y}{7}$

Hence, the graphic representation is as follows.

From the figure, it can be observed that these lines intersect each other at point $(3,5)$.

Therefore, the cost of a pencil and a pen are Rs 3 and Rs 5 respectively.

Solution

(i) $5 x-4 y+8=0$

$7 x+6 y-9=0$

Comparing these equations with $a_1 x+b_1 y+c_1=0$

and $a_2 x+b_2 y+c_2=0$, we obtain

$a_1=5, \quad b_1=-4, \quad c_1=8$

$a_2=7, \quad b_2=6, \quad c_2=-9$

$\dfrac{a_1}{a_2}=\dfrac{5}{7}$

$\dfrac{b_1}{b_2}=\dfrac{-4}{6}=\dfrac{-2}{3}$

Since $\dfrac{a_1}{a_2} \neq \dfrac{b_1}{b_2}$,

Hence, the lines representing the given pair of equations have a unique solution and the pair of lines intersects at exactly one point.

(ii) $9 x+3 y+12=0$

$18 x+6 y+24=0$

Comparing these equations with $a_1 x+b_1 y+c_1=0$

and $a_2 x+b_2 y+c_2=0$, we obtain

$a_1=9, \quad b_1=3, \quad c_1=12$

$a_2=18, \quad b_2=6, \quad c_2=24$

$\dfrac{a_1}{a_2}=\dfrac{9}{18}=\dfrac{1}{2}$

$\dfrac{b_1}{b_2}=\dfrac{3}{6}=\dfrac{1}{2}$

$\dfrac{c_1}{c_2}=\dfrac{12}{24}=\dfrac{1}{2}$

Since $\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=\dfrac{c_1}{c_2}$,

Hence, the lines representing the given pair of equations are coincident and there are infinite possible solutions for the given pair of equations.

(iii) $6 x-3 y+10=0$

$2 x-y+9=0$

Comparing these equations with $a_1 x+b_1 y+c_1=0$ and $a_2 x+b_2 y+c_2=0$, we obtain

$a_1=6, \quad b_1=-3, \quad c_1=10$

$a_2=2, \quad b_2=-1, \quad c_2=9$

$\dfrac{a_1}{a_2}=\dfrac{6}{2}=\dfrac{3}{1}$

$\dfrac{b_1}{b_2}=\dfrac{-3}{-1}=\dfrac{3}{1}$

$\dfrac{c_1}{c_2}=\dfrac{10}{9}$

Since $\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2}$

Hence, the lines representing the given pair of equations are parallel to each other and hence, these lines will never intersect each other at any point or there is no possible solution for the given pair of equations.

Solution

(i) $3 x+2 y=5$

$2 x-3 y=7$

$\dfrac{a_1}{a_2}=\dfrac{3}{2}, \quad \dfrac{b_1}{b_2}=\dfrac{-2}{3}, \quad \dfrac{c_1}{c_2}=\dfrac{5}{7}$

$\dfrac{a_1}{a_2} \neq \dfrac{b_1}{b_2}$

These linear equations are intersecting each other at one point and thus have only one possible solution. Hence, the pair of linear equations is consistent.

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

$\dfrac{a_1}{a_2}=\dfrac{2}{4}=\dfrac{1}{2}, \quad \dfrac{b_1}{b_2}=\dfrac{-3}{-6}=\dfrac{1}{2}, \quad \dfrac{c_1}{c_2}=\dfrac{8}{9}$

Since $\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2}$,

Therefore, these linear equations are parallel to each other and thus have no possible solution. Hence, the pair of linear equations is inconsistent.

(iii) $\dfrac{3}{2} x+\dfrac{5}{3} y=7$

$9 x-10 y=14$

$\dfrac{a_1}{a_2}=\dfrac{\dfrac{3}{2}}{9}=\dfrac{1}{6}, \quad \dfrac{b_1}{b_2}=\dfrac{\dfrac{5}{3}}{-10}=\dfrac{-1}{6}, \quad \dfrac{c_1}{c_2}=\dfrac{7}{14}=\dfrac{1}{2}$

Since $\dfrac{a_1}{a_2} \neq \dfrac{b_1}{b_2}$,

Therefore, these linear equations are intersecting each other at one point and thus have only one possible solution. Hence, the pair of linear equations is consistent.

(iv) $5 x-3 y=11$

$-10 x+6 y=-22$

$\dfrac{a_1}{a_2}=\dfrac{5}{-10}=\dfrac{-1}{2}, \quad \dfrac{b_1}{b_2}=\dfrac{-3}{6}=\dfrac{-1}{2}, \quad \dfrac{c_1}{c_2}=\dfrac{11}{-22}=\dfrac{-1}{2}$

Since $\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=\dfrac{c_1}{c_2}$

Therefore, these linear equations are coincident pair of lines and thus have infinite number of possible solutions. Hence, the pair of linear equations is consistent.

(v) $\dfrac{4}{3} x+2 y=8$

$2 x+3 y=12$

$\dfrac{a_1}{a_2}=\dfrac{\dfrac{4}{3}}{2}=\dfrac{2}{3}, \quad \dfrac{b_1}{b_2}=\dfrac{2}{3}, \quad \dfrac{c_1}{c_2}=\dfrac{8}{12}=\dfrac{2}{3}$

Since $\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=\dfrac{c_1}{c_2}$,

Therefore, these linear equations are coincident pair of lines and thus have infinite number of possible solutions. Hence, the pair of linear equations is consistent.

Solution

(i) $x+y=5$

$2 x+2 y=10$

$\dfrac{a_1}{a_2}=\dfrac{1}{2}, \quad \dfrac{b_1}{b_2}=\dfrac{1}{2}, \quad \dfrac{c_1}{c_2}=\dfrac{5}{10}=\dfrac{1}{2}$

Since $\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=\dfrac{c_1}{c_2}$

Therefore, these linear equations are coincident pair of lines and thus have infinite number of possible solutions. Hence, the pair of linear equations is consistent.

$x+y=5$

$x=5-y$

And, $2 x+2 y=10$

$x=\dfrac{10-2 y}{2}$

Hence, the graphic representation is as follows.

From the figure, it can be observed that these lines are overlapping each other. Therefore, infinite solutions are possible for the given pair of equations.

(ii) $x-y=8$

$3 x-3 y=16$

$\dfrac{a_1}{a_2}=\dfrac{1}{3}, \quad \dfrac{b_1}{b_2}=\dfrac{-1}{-3}=\dfrac{1}{3}, \quad \dfrac{c_1}{c_2}=\dfrac{8}{16}=\dfrac{1}{2}$

Since $\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2}$,

Therefore, these linear equations are parallel to each other and thus have no possible solution. Hence, the pair of linear equations is inconsistent.

(iii) $2 x+y-6=0$

$4 x-2 y-4=0$

$\dfrac{a_1}{a_2}=\dfrac{2}{4}=\dfrac{1}{2}, \quad \dfrac{b_1}{b_2}=\dfrac{-1}{2}, \quad \dfrac{c_1}{c_2}=\dfrac{-6}{-4}=\dfrac{3}{2}$

Since $\dfrac{a_1}{a_2} \neq \dfrac{b_1}{b_2}$,

Therefore, these linear equations are intersecting each other at one point and thus have only one possible solution. Hence, the pair of linear equations is consistent.

$2 x+y-6=0$

$y=6-2 x$

And $4 x-2 y-4=0$

$y=\dfrac{4 x-4}{2}$

Solution

Let the width of the garden be $x$ and length be $y$.

According to the question,

$y-x=4(1)$

$y+x=36(2)$

$y-x=4$

$y=x+4$

$y+x=36$

Hence, the graphic representation is as follows.

From the figure, it can be observed that these lines are intersecting each other at only point i.e., (16, 20). Therefore, the length and width of the given garden is $20 m$ and $16 m$ respectively.

Solution

(i)Intersecting lines:

For this condition,

$ \dfrac{a_1}{a_2} \neq \dfrac{b_1}{b_2} $

The second line such that it is intersecting the given line

is

$ 2 x+4 y-6=0 \quad \text{ as } \quad \dfrac{a_1}{a_2}=\dfrac{2}{2}=1, \quad \dfrac{b_1}{b_2}=\dfrac{3}{4} \text{ and } \dfrac{a_1}{a_2} \neq \dfrac{b_1}{b_2} \text{. } $

(ii) Parallel lines:

For this condition,

$\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2}$

Hence, the second line can be

$4 x+6 y-8=0$

as $\dfrac{a_1}{a_2}=\dfrac{2}{4}=\dfrac{1}{2}, \dfrac{b_1}{b_2}=\dfrac{3}{6}=\dfrac{1}{2}, \dfrac{c_1}{c_2}=\dfrac{-8}{-8}=1$

And clearly, $\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2}$

(iii)Coincident lines:

For coincident lines,

$\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=\dfrac{c_1}{c_2}$

Hence, the second line can be

$6 x+9 y-24=0$

as $\dfrac{a_1}{a_2}=\dfrac{2}{6}=\dfrac{1}{3}, \quad \dfrac{b_1}{b_2}=\dfrac{3}{9}=\dfrac{1}{3}, \dfrac{c_1}{c_2}=\dfrac{-8}{-24}=\dfrac{1}{3}$

And clearly, $\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=\dfrac{c_1}{c_2}$

Solution

$x-y+1=0$

$x=y-1$

$3 x+2 y-12=0$

$x=\dfrac{12-2 y}{3}$

Hence, the graphic representation is as follows.

From the figure, it can be observed that these lines are intersecting each other at point $(2,3)$ and $x$-axis at $(-1,0)$ and $(4,0)$. Therefore, the vertices of the triangle are $(2,3),(-1,0)$, and $(4,0)$.