Solution

(i) $x+y=14(1)$

$x-y=4(2)$

From (1), we obtain

$x=14-y(3)$

Substituting this value in equation (2), we obtain $(14-y)-y=4$

$14-2 y=4$

$10=2 y$

$y=5$

Substituting this in equation (3), we obtain

$x=9$

$\therefore x=9, y=5$

(ii) $s-t=3$

$\dfrac{s}{3}+\dfrac{t}{2}=6$

From (1), we obtain

$s=t+3$

Substituting this value in equation (2), we obtain

$\dfrac{t+3}{3}+\dfrac{t}{2}=6$

$2 t+6+3 t=36$

$5 t=30$

$t=6$

Substituting in equation (3), we obtain

$s=9$

$\therefore s=9, t=6$

(iii) $3 x-y=3(1)$

$9 x-3 y=9(2)$

From (1), we obtain

$y=3 x-3(3)$

Substituting this value in equation (2), we obtain

$9 x-3(3 x-3)=9$

$9 x-9 x+9=9$

$9=9$

This is always true.

Hence, the given pair of equations has infinite possible solutions and the relation between these variables can be given by

$y=3 x-3$

Therefore, one of its possible solutions is $x=1, y=0$.

(iv) $0.2 x+0.3 y=1.3$

$0.4 x+0.5 y=2.3$

From equation (1), we obtain

$x=\dfrac{1.3-0.3 y}{0.2}$

Substituting this value in equation (2), we obtain

$0.4(\dfrac{1.3-0.3 y}{0.2})+0.5 y=2.3$

$2.6-0.6 y+0.5 y=2.3$

$2.6-2.3=0.1 y$

$0.3=0.1 y$

$y=3$

Substituting this value in equation (3), we obtain

$x=\dfrac{1.3-0.3 \times 3}{0.2}$

$=\dfrac{1.3-0.9}{0.2}=\dfrac{0.4}{0.2}=2$

$\therefore x=2, y=3$

(v) $\sqrt{2} x+\sqrt{3} y=0$

$\sqrt{3} x-\sqrt{8} y=0$

From equation (1), we obtain

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

Substituting this value in equation (2), we obtain

$$ \begin{align*} & \sqrt{3}(-\dfrac{\sqrt{3} y}{\sqrt{2}})-\sqrt{8} y=0 \\ & -\dfrac{3 y}{\sqrt{2}}-2 \sqrt{2} y=0 \\ & y(-\dfrac{3}{\sqrt{2}}-2 \sqrt{2})=0 \\ & y=0 \tag{4} \end{align*} $$

Substituting this value in equation (3), we obtain

$x$

Solution

$2 x+3 y=11$

$2 x-4 y=-24$

From equation (1), we obtain

$x=\dfrac{11-3 y}{2}$

Substituting this value in equation (2), we obtain

$2(\dfrac{11-3 y}{2})-4 y=-24$

$11-3 y-4 y=-24$

$-7 y=-35$

$y=5$

Putting this value in equation (3), we obtain

$x=\dfrac{11-3 \times 5}{2}=-\dfrac{4}{2}=-2$

Hence, $x=-2, y=5$

Also,

$y=m x+3$

$5=-2 m+3$

$-2 m=2$

$m=-1$

Solution

(i) Let the first number be $x$ and the other number be $y$ such that $y>x$.

According to the given information,

$$ \begin{align*} & y=3 x \tag{1}\\ & y-x=26 \tag{2}\\ & \text{ On substituting } \\ & 3 x-x=26 \tag{3}\\ & x=13 \end{align*} $$

On substituting the value of $y$ from equation (1) into equation (2), we obtain

Substituting this in equation (1), we obtain

$y=39$

Hence, the numbers are 13 and 39.

(ii) Let the larger angle be $x$ and smaller angle be $y$.

We know that the sum of the measures of angles of a supplementary pair is always $180^{\circ}$.

According to the given information,

$x+y=180^{\circ}$

$x-y=18^{\circ}$

From (1), we obtain

$x=180^{\circ}-y(3)$

Substituting this in equation (2), we obtain

$180^{\circ}-y-y=18^{\circ}$

$162^{\circ}=2 y$

$81^{\circ}=y$

Putting this in equation (3), we obtain

$x=180^{\circ}-81^{\circ}$

$=99^{\circ}$

Hence, the angles are $99^{\circ}$ and $81^{\circ}$.

(iii) Let the cost of a bat and a ball be $x$ and $y$ respectively.

According to the given information,

$7 x+6 y=3800$

$3 x+5 y=1750$

From (1), we obtain

$y=\dfrac{3800-7 x}{6}$

Substituting this value in equation (2), we obtain

$3 x+5(\dfrac{3800-7 x}{6})=1750$

$3 x+\dfrac{9500}{3}-\dfrac{35 x}{6}=1750$

$3 x-\dfrac{35 x}{6}=1750-\dfrac{9500}{3}$

$$ \begin{align*} & \dfrac{18 x-35 x}{6}=\dfrac{5250-9500}{3} \\ & -\dfrac{17 x}{6}=\dfrac{-4250}{3} \\ & -17 x=-8500 \\ & x=500 \tag{4} \end{align*} $$

Substituting this in equation (3), we obtain

$ \begin{aligned} y & =\dfrac{3800-7 \times 500}{6} \\ & =\dfrac{300}{6}=50 \end{aligned} $

Hence, the cost of a bat is Rs 500 and that of a ball is Rs 50 .

(iv) Let the fixed charge be Rs $x$ and per km charge be Rs $y$.

According to the given information,

$x+10 y=105$

$x+15 y=155$

From (1), we obtain

$x=105-10 y$

Substituting this in equation (2), we obtain

$105-10 y+15 y=155$

$5 y=50$

$y=10$

Putting this in equation (3), we obtain

$x=105-10 \times 10$

$x=5$

Hence, fixed charge $=$ Rs 5

And per km charge $=$ Rs 10

Charge for $25 km=x+25 y$

$=5+250=$ Rs 255

(v) Let the fraction be $\dfrac{x}{y}$.

According to the given information,

$ \dfrac{x+2}{y+2}=\dfrac{9}{11} $

$11 x+22=9 y+18$

$11 x-9 y=-4$

$\dfrac{x+3}{y+3}=\dfrac{5}{6}$

$6 x+18=5 y+15$

$6 x-5 y=-3$

From equation (1), we obtain