Solution

(d) The given equations are

and table for $2 x-y+9=0$,

Hence, the pair of equations represents two parallel lines.

Solution

(d) Given, equations are $x+2 y+5=0$ and $-3 x-6 y+1=0$

Here, $\quad a_1=1, b_1=2, c_1=5$ and $\quad a_2=-3, b_2=-6, c_2=1$

$ \begin{aligned} & \dfrac{c_1}{c_2}=\dfrac{5}{1} \\ \therefore \quad & \dfrac{a_1}{a_2}=\dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2} \end{aligned} $

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

Hence, the pair of equations has no Solution.

Solution

(c) Condition for a consistent pair of linear equations

and

$ \begin{aligned} & \dfrac{a_1}{a_2} \neq \dfrac{b_1}{b_2} \quad \text{ [intersecting lines having unique Solution] } \\ & \dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=\dfrac{c_1}{c_2} \quad \text{ [coincident or dependent] } \end{aligned} $

Solution

(d) The given pair of equations are $y=0$ and $y=-7$.

By graphically, both lines are parallel and having no Solution.

Solution

(d) By graphically in every condition, if $a, b>0 ; a, b<0, a>0, b<0 ; a<0, b>0$ but $a=b \neq 0$.

The pair of equations $x=a$ and $y=b$ graphically represents lines which are intersecting at $(a, b)$. If $a, b>0$

Similarly, in all cases two lines intersect at $(a, b)$.

Solution

(c) Condition for coincident lines is

Given lines,

and

$$ \begin{align*} \dfrac{a_1}{a_2}=\dfrac{b_1}{b_2} & =\dfrac{c_1}{c_2} \tag{i}\\ 3 x-y+8 & =0 \\ 6 x-k y+16 & =0 \end{align*} $$

Here,

and

From Eq. (i),

$ \begin{matrix} \Rightarrow & 6 \\ \dfrac{1}{k} & =\dfrac{1}{2} \end{matrix} $

$ \begin{aligned} a_1=3, b_1 & =-1, c_1=8 \\ a_2=6, b_2 & =-k, c_2=16 \\ \dfrac{3}{6} & =\dfrac{-1}{-k}=\dfrac{8}{16} \\ \dfrac{1}{k} & =\dfrac{1}{2} \\ k & =2 \end{aligned} $

Solution

(c) Condition for parallel lines is

Solution

(d) Condition for infinitely many Solutions

$$ \begin{equation*} \dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=\dfrac{c_1}{c_2} \tag{i} \end{equation*} $$

The given lines are $c x-y=2$ and $6 x-2 y=3$

Here,

$a_1=c, b_1=-1, c_1=-2$

and

$a_2=6, b_2=-2, c_2=-3$

From Eq. (i),

$ \dfrac{c}{6}=\dfrac{-1}{-2}=\dfrac{-2}{-3} $

Here,

$ \dfrac{c}{6}=\dfrac{1}{2} \quad \text{ and } \quad \dfrac{c}{6}=\dfrac{2}{3} $

$ \Rightarrow \quad c=3 \text{ and } \quad c=4 $

Since, $c$ has different values.

Hence, for no value of $c$ the pair of equations will have infinitely many Solutions.

Solution

(d) Condition for dependent linear equations

$$ \begin{equation*} \dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=\dfrac{c_1}{c_2}=\dfrac{1}{k} \tag{i} \end{equation*} $$

Given equation of line is, $-5 x+7 y-2=0$

Here,

$$ \begin{align*} a_1 & =-5, b_1=7, c_1=-2 \\ -\dfrac{5}{a_2} & =\dfrac{7}{b_2}=-\dfrac{2}{c_2}=\dfrac{1}{k} \tag{say}\\ a_2 & =-5 k, b_2=7 k, c_2=-2 k \end{align*} $$

From Eq. (i),

$\Rightarrow$

where, $k$ is any arbitrary constant. Putting $k=2$, then and $a_2=-10, b_2=14$ $c_2=-4$

$\therefore$ The required equation of line becomes

$ \begin{array}{lrl} & \quad \quad\quad\quad\quad& a_2x+b_2y+c_2=0 \\ \Rightarrow & & -10x+14y-4=0 \\ \Rightarrow & & 10x-14y+4=0 \end{array} $

Solution

(b) If $x=2, y=-3$ is a unique Solution of any pair of equation, then these values must satisfy that pair of equations.

From option (b),

$ \begin{aligned} & LHS=2 x+5 y=2(2)+5(-3)=4-15=-11=RHS \\ & LHS=4 x+10 y=4(2)+10(-3)=8-30=-22=RHS \end{aligned} $

$ \text{ and } $

Solution

(c) Since, $x=a$ and $y=b$ is the Solution of the equations $x-y=2$ and $x+y=4$, then these values will satisfy that equations

and

$$ \begin{align*} & a-b=2 \tag{i}\\ & a+b=4 \tag{ii} \end{align*} $$

On adding Eqs. (i) and (ii), we get

$ 2 a=6 $

$ \therefore \quad a=3 \text{ and } b=1 $

Solution

(d) Let number of $\text {₹} 1$ coins $=x$

and number of $\text {₹} 2$ coins $=y$

Now, by given conditions $\quad x+y=50$

Also, $\quad x \times 1+y \times 2=75$

$\Rightarrow \quad x+2 y=75$

On subtracting Eq. (i) from Eq. (ii), we get

$ \Rightarrow \quad y=25 $

When $y=25$, then $x=25$

$ (x+2 y)-(x+y)=75-50 $

Solution

(c) Let $x$ yr be the present age of father and $y$ yr be the present age of son.

Four years hence, it has relation by given condition,

On putting the value of $x$ from Eq. (ii) in Eq. (i), we get

$ \begin{matrix} \Rightarrow & 2 y=12 \\ \Rightarrow & y=6 \end{matrix} $

When $y=6$, then $x=36$

Hence, present age of father is $36 yr$ and age of son is $6 yr$.

Solution

Condition for no Solution $\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2}$

(i) Yes, given pair of equations,

$ \begin{aligned} & 2 x+4 y=3 \text{ and } 12 y+6 x=6 \\ & \text{ Here, } \quad a_1=2, b_1=4, c_1=-3 \text{, } \\ & a_2=6, b_2=12, c_2=-6 \\ & \therefore \quad \dfrac{a_1}{a_2}=\dfrac{2}{6}=\dfrac{1}{3}, \dfrac{b_1}{b_2}=\dfrac{4}{12}=\dfrac{1}{3} \\ & \dfrac{c_1}{c_2}=\dfrac{-3}{-6}=\dfrac{1}{2} \\ & \therefore \quad \dfrac{a_1}{a_2}=\dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2} \end{aligned} $

Hence, the given pair of linear equations has no Solution.

(ii) No, given pair of equations,

or

$ x=2 y \text{ and } y=2 x $

$ x-2 y=0 \text{ and } 2 x-y=0 $

Here,

$ \begin{aligned} & a_1=1, b_1=-2, c_1=0 \\ & a_2=2, b_2=-1, c_2=0 \end{aligned} $

$ \begin{matrix} \therefore & \dfrac{a_1}{a_2}=\dfrac{1}{2} \text{ and } \dfrac{b_1}{b_2}=\dfrac{2}{1} \\ \because & \dfrac{a_1}{a_2} \neq \dfrac{b_1}{b_2} \end{matrix} $

Hence, the given pair of linear equations has unique Solution.

(iii) No, given pair of equations,

$ 3 x+y-3=0 \text{ and } 2 x+\dfrac{2}{3} y-2=0 $

Here,

$ \begin{aligned} & a_1=3, b_1=1, c_1=-3, \\ & a_2=2, b_2=\dfrac{2}{3}, c_2=-2 \\ & \dfrac{a_1}{a_2}=\dfrac{3}{2}, \dfrac{b_1}{b_2}=\dfrac{1}{2 / 3}=\dfrac{3}{2} \\ & \dfrac{c_1}{c_2}=\dfrac{-3}{-2}=\dfrac{3}{2} \end{aligned} $

$ \begin{matrix} \therefore & \dfrac{a_1}{a_2}=\dfrac{3}{2}, \dfrac{b_1}{b_2}=\dfrac{1}{2 / 3}=\dfrac{3}{2} \\ \because \quad & \dfrac{c_1}{c_2}=\dfrac{-3}{-2}=\dfrac{3}{2} \\ & \dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=\dfrac{c_1}{c_2}=\dfrac{3}{2} \end{matrix} $

Hence, the given pair of linear equations is coincident and having infinitely many Solutions.

Solution

Condition for coincident lines,

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

(i) No, given pair of linear equations

and

$ 3 x+\dfrac{y}{7}-3=0 $

where,

$ \begin{gathered} 7 x+3 y-7=0, \\ a_1=3, b_1=\dfrac{1}{7}, c_1=-3 ; \\ a_2=7, b_2=3, c_2=-7 \end{gathered} $

$ \text{ Now, } \quad \dfrac{a_1}{a_2}=\dfrac{3}{7}, \dfrac{b_1}{b_2}=\dfrac{1}{21}, \dfrac{c_1}{c_2}=\dfrac{3}{7} $

Hence, the given pair of linear equations has unique Solution. (ii) Yes, given pair of linear equations

$ \begin{aligned} & -2 x-3 y-1=0 \text{ and } 6 y+4 x+2=0 \\ & \text{ where, } \quad a_1=-2, b_1=-3, c_1=-1 \text{; } \\ & a_2=4, b_2=6, c_2=2 \\ & \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{1}{2} \\ & \because \quad \dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=\dfrac{c_1}{c_2}=-\dfrac{1}{2} \end{aligned} $

Hence, the given pair of linear equations is coincident.

(iii) No, the given pair of linear equations are

Here,

$ \dfrac{x}{2}+y+\dfrac{2}{5}=0 \text{ and } 4 x+8 y+\dfrac{5}{16}=0 $

.

$a_1=\dfrac{1}{2}, b_1=1, c_1=\dfrac{2}{5}$

$a_2=4, b_2=8, \quad c_2=\dfrac{5}{16}$

Now,

$ \begin{aligned} & \dfrac{a_1}{a_2}=\dfrac{1}{8}, \dfrac{b_1}{b_2}=\dfrac{1}{8}, \dfrac{c_1}{c_2}=\dfrac{32}{25} \\ & \dfrac{a_1}{a_2}=\dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2} \end{aligned} $

Hence, the given pair of linear equations has no Solution.

Solution

Conditions for pair of linear equations are consistent

and

$ \begin{aligned} & \dfrac{a_1}{a_2} \neq \dfrac{b_1}{b_2} & \text{[unique Solution]}\\ & \dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=\dfrac{c_1}{c_2} & \text{[infinitely many Solutions]} \end{aligned} $

(i) No, the given pair of linear equations

$$ \begin{align*} & -3 x-4 y=12 \text{ and } 3 x+4 y=12 \\ \text{ Here, } & a_1=-3, b_1=-4, c_1=-12 \\ & a_2=3, b_2=4, c_2=-12 \\ \text{ Now, } & \dfrac{a_1}{a_2}=-\dfrac{3}{3}=-1, \dfrac{b_1}{b_2}=-\dfrac{4}{4}=-1, \dfrac{c_1}{c_2}=\dfrac{-12}{-12}=1 \\ \because & \dfrac{a_1}{a_2}=\dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2} \end{align*} $$

Hence, the pair of linear equations has no Solution, i.e., inconsistent. (ii) Yes, the given pair of linear equations

$ \begin{aligned} & \quad\quad\quad\quad\quad\quad\quad \dfrac{3}{5} x-y=\dfrac{1}{2} \text{ and } \dfrac{1}{5} x-3 y=\dfrac{1}{6} \\ & \text{ Here, } \quad \begin{aligned} a_1 & =\dfrac{3}{5}, b_1=-1, c_1=-\dfrac{1}{2} \\ a_2 & =\dfrac{1}{5}, b_2=-3, c_2=-\dfrac{1}{6} \\ \text{ and } \quad \dfrac{a_1}{a_2} & =\dfrac{3}{1}, \dfrac{b_1}{b_2}=\dfrac{-1}{-3}=\dfrac{1}{3}, \dfrac{c_1}{c_2}=\dfrac{3}{1} & \quad \because \dfrac{a_1}{a_2} \neq \dfrac{b_1}{b_2} \end{aligned} \end{aligned} $

Hence, the given pair of linear equations has unique Solution, i.e., consistent.

(iii) Yes, the given pair of linear equations

$ \begin{aligned} & \quad\quad\quad 2 a x+b y-a=0 \\ & \text{ and } \quad 4 a x+2 b y-2 a=0 ; a, b \neq 0 \\ & \quad \begin{aligned} \text{ Here, } \quad\quad a_1 & =2 a, b_1=b, c_1=-a ; \\ a_2 & =4 a, b_2=2 b, c_2=-2 a \\ \text{ Now, } \quad \dfrac{a_1}{a_2} & =\dfrac{2 a}{4 a}=\dfrac{1}{2}, \dfrac{b_1}{b_2}=\dfrac{b}{2 b}=\dfrac{1}{2}, \dfrac{c_1}{c_2}=\dfrac{-a}{-2 a}=\dfrac{1}{2} \\ \because \quad \dfrac{a_1}{a_2} & =\dfrac{b_1}{b_2}=\dfrac{c_1}{c_2}=\dfrac{1}{2} \end{aligned} \end{aligned} $

Hence, the given pair of linear equations has infinitely many Solutions, i.e., consistent or dependent.

(iv) No, the given pair of linear equations

$ \begin{aligned} & x+3 y=11 \text{ and } 2 x+6 y=11 \\ & a_1 =1, b_1=3, c_1=-11 \\ & a_2 =2, b_2=6, c_2=-11 \\ \text{ Now, } \dfrac{a_1}{a_2} & =\dfrac{1}{2}, \dfrac{b_1}{b_2}=\dfrac{3}{6}=\dfrac{1}{2}, \dfrac{c_1}{c_2}=\dfrac{-11}{-11}=1 \\ \therefore & \dfrac{a_1}{a_2}=\dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2} \end{aligned} $

Hence, the pair of linear equation have no Solution i.e., inconsistent.

Solution

No, the given pair of linear equations

$ \lambda x+3 y+7=0 \text{ and } 2 x+6 y-14=0 $

$ \text{ Here, } \quad a_1=\lambda, b_1=3, c_1=7 ; a_2=2, b_2=6, c_2=-14 $

If $\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=\dfrac{c_1}{c_2}$, then system has infinitely many Solutions.

$ \begin{matrix} \Rightarrow & \dfrac{\lambda}{2}=\dfrac{3}{6}=-\dfrac{7}{14} \\ \because & \dfrac{\lambda}{2}=\dfrac{3}{6} \Rightarrow \lambda=1 \\ \text{ and } & \dfrac{\lambda}{2}=-\dfrac{7}{14} \Rightarrow \lambda=-1 \end{matrix} $

Hence, $\lambda=-1$ does not have a unique value.

So, for no value of $\lambda$ the given pair of linear equations has infinitely many Solutions.

Solution

False, the given pair of linear equations

and

$ \begin{aligned} x-2 y-8 & =0 \\ 5 x-10 y-c & =0 \\ a_1 & =1, b_1=-2, c_1=-8 \\ a_2 & =5, b_2=-10, c_2=-c \\ \dfrac{a_1}{a_2} & =\dfrac{1}{5}, \dfrac{b_1}{b_2}=\dfrac{-2}{-10}=\dfrac{1}{5} \\ \dfrac{c_1}{c_2} & =\dfrac{-8}{-c}=\dfrac{8}{c} \end{aligned} $

Here,

Now,

But if $c=40$ (real value), then the ratio $\dfrac{c_1}{c_2}$ becomes $\dfrac{1}{5}$ and then the system of linear equations has an infinitely many Solutions.

Hence, at $c=40$, the system of linear equations does not have a unique Solution.

Solution

Not true, by graphically, we observe that $x=7$ line is parallel to $Y$-axis and perpendicular to $X$-axis.

Solution

The given pair of linear equations is

$ \lambda x+y=\lambda^{2} \text{ and } x+\lambda y=1 $

Here,

$ \begin{aligned} & a_1=\lambda, b_1=1, c_1=-\lambda^{2} \\ & a_2=1, b_2=\lambda, c_2=-1 \end{aligned} $

(i) For no Solution,

$ \begin{aligned} & \quad\quad\quad \dfrac{a_1}{a_2}=\dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2} \\ \Rightarrow & \quad\quad\quad \dfrac{\lambda}{1}=\dfrac{1}{\lambda} \neq \dfrac{-\lambda ^2}{-1} \end{aligned} $

$ \begin{aligned} \Rightarrow &\quad\quad\quad \dfrac{\lambda}{1}=\dfrac{\lambda ^2}{1} \\ \Rightarrow & \quad\quad\quad \lambda(\lambda - 1)=0 \end{aligned} $

When $\lambda \neq 0$, then $\lambda =1 $

Here, we take only $\lambda=-1$ because at $\lambda=1$ the system of linear equations has infinitely many Solutions.

(ii) For infinitely many Solutions,

$ \begin{aligned}& \dfrac{a_1}{a_2} & =\dfrac{b_1}{b_2}=\dfrac{c_1}{c_2} \\ \Rightarrow & \dfrac{\lambda}{1} & =\dfrac{1}{\lambda}=\dfrac{\lambda^{2}}{1} \\ \Rightarrow & \dfrac{\lambda}{1} & =\dfrac{\lambda^{2}}{1} \\ \Rightarrow & \lambda(\lambda-1) & =0 \end{aligned} $

When $\lambda \neq 0$, then $\lambda=1$

(iii) For a unique Solution,

$ \begin{aligned}& \dfrac{a_1}{a_2} \neq \dfrac{b_1}{b_2} \\ \Rightarrow \dfrac{\lambda}{1} \neq \dfrac{1}{\lambda} \\ \Rightarrow & \quad \lambda ^{2} \neq 1 \\ \Rightarrow & \lambda \neq \pm 1 \end{aligned} $

$ \begin{array}{lll} & & &\dfrac{a_1}{a_2} \neq \dfrac{b_1}{b_2} \Rightarrow \dfrac{\lambda}{1} \neq \dfrac{1}{\lambda} \\ \Rightarrow & & &\lambda ^2 \neq 1 \\ \Rightarrow & & &\lambda \neq \pm 1 \end{array} $

So, all real values of $\lambda$ except $\pm 1$

Solution

Given pair of linear equations is

and

$$ \begin{equation*} k x+3 y=k-3 \tag{i} \end{equation*} $$

$$ \begin{equation*} 12 x+k y=k \tag{ii} \end{equation*} $$

On comparing with $a x+b y+c=0$, we get

$ \begin{aligned} & a_1=k, b_1=3 \text{ and } c_1=-(k-3) \\ & a_2=12, b_2=k \text{ and } c_2=-k \end{aligned} $

For no Solution of the pair of linear equations,

[from Eq. (i)] [from Eq. (ii)]

$ \begin{matrix} \text{ Taking first two parts, we get } & \\ \Rightarrow & \dfrac{k}{12}=\dfrac{3}{k} \\ \Rightarrow & k^{2}=36 \\ \Rightarrow & k= \pm 6 \end{matrix} $

$ \begin{aligned} & \dfrac{a_1}{a_2}=\dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2} \\ & \dfrac{k}{12}=\dfrac{3}{k} \neq \dfrac{-(k-3)}{-k} \end{aligned} $

Taking last two parts, we get

$\begin{array}{lr} & \dfrac{3}{k} \neq \dfrac{k-3}{k} \\ \Rightarrow & 3 k \neq k(k-3) \\ \Rightarrow & 3 k-k(k-3) \neq 0 \\ \Rightarrow & k(3-k+3) \neq 0 \\ \Rightarrow & k(6-k) \neq 0 \\ \Rightarrow & k \neq 0 \text { and } k \neq 6\end{array}$

Hence, required value of $k$ for which the given pair of linear equations has no Solution is -6 .

Solution

Given pair of linear equations are

For infinitely many Solutions of the the pairs of linear equations,

$ \begin{aligned} \dfrac{a_1}{a_2} & =\dfrac{b_1}{b_2}=\dfrac{c_1}{c_2} \\ \Rightarrow \quad \dfrac{1}{a-b} & =\dfrac{2}{a+b}=\dfrac{-1}{-(a+b-2)} \end{aligned} $

Taking first two parts,

$$ \begin{array}{lr}& \dfrac{1}{a-b} & =\dfrac{2}{a+b} \\ \Rightarrow & a+b =2 a-2 b \\ \Rightarrow & 2 a-a =2 b+b \\ \Rightarrow & a =3 b \tag{iii} \end{array} $$

$ \begin{array}{lll} & \quad \quad \quad\dfrac{1}{a-b} & =\dfrac{2}{a+b} \\ \Rightarrow & \quad\quad\quad a+b & =2a-2b \\ \Rightarrow & \quad\quad\quad 2a-a & =2b+b \\ \Rightarrow & \quad\quad\quad a & =3b \end{array} $

Taking last two parts,

$$ \begin{align*} & \dfrac{2}{a+b} & =\dfrac{1}{(a+b-2)} \\ \Rightarrow & 2 a+2 b-4 & =a+b \\ \Rightarrow & a+b & =4 \tag{iv} \end{align*} $$

Now, put the value of a from Eq. (iii) in Eq. (iv), we get

$ 3 b+b=4 $

$ \begin{matrix} \Rightarrow & 4 b & =4 \\ \Rightarrow & b & =1 \end{matrix} $

Put the value of $b$ in Eq. (iii), we get

$ \Rightarrow \quad a=3 $

So, the values $(a, b)=(3,1)$ satisfies all the parts. Hence, required values of $a$ and $b$ are 3 and 1 respectively for which the given pair of linear equations has infinitely many Solutions.

Solution

(i) Given pair of linear equations is $3 x-y-5=0$ $6 x-2 y-p=0$ and On comparing with $a x+b y+c=0$, we get and

$ \begin{aligned} & a_1=3, b_1=-1 \\ & c_1=-5 \\ & a_2=6, b_2=-2 \end{aligned} $

$ c_2=-p $

and [from Eq. (ii)]

Since, the lines represented by these equations are parallel, then

$ \begin{matrix} \dfrac{a_1}{a_2} & =\dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2} \\ \dfrac{3}{6} & =\dfrac{-1}{-2} \neq \dfrac{-5}{-p} \\ \text{ Taking last two parts, we get } & \dfrac{-1}{-2} \neq \dfrac{-5}{-p} \\ \Rightarrow \quad & \dfrac{1}{2} \neq \dfrac{5}{p} \\ \Rightarrow \quad & p \neq 10 \end{matrix} $

Hence, the given pair of linear equations are parallel for all real values of $p$ except 10 i.e., $p \in R-{10}$.

(ii) Given pair of linear equations is

$ \begin{aligned} & -x+p y-1=0 \\ & \text{ and } \quad p x-y-1=0 \\ & \text{ On comparing with } a x+b y+c=0 \text{, we get } \\ & \begin{matrix} a_1=-1, b_1=p \text{ and } c_1=-1 \\ a_2=p, b_2=-1 \text{ and } c_2=-1 \end{matrix} \end{aligned} $

Since, the pair of linear equations has no Solution i.e., both lines are parallel to each other.

$ \therefore \quad \dfrac{a_1}{a_2}=\dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2} \Rightarrow \dfrac{-1}{p}=\dfrac{p}{-1} \neq \dfrac{-1}{-1} $

Taking last two parts, we get

$ \begin{aligned} & \dfrac{p}{-1} \neq \dfrac{-1}{-1} \\ & \Rightarrow \quad p \neq-1 \end{aligned} $

Taking first two parts, we get

$\begin{array}{ll} & \dfrac{-1}{p}=\dfrac{p}{-1} \\ \Rightarrow & p^2=1 \\ \Rightarrow & p= \pm 1 \\ \text { but } & p \neq-1 \\ \therefore & p=1\end{array}$

Hence, the given pair of linear equations has no Solution for $p=1$.

(iii) Given, pair of linear equations is

$\quad\quad\quad -3 x+5 y-7=0$ $\ldots$ (i)
and $\quad 2 p x-3 y-1=0$ $\ldots$ (ii)

On comparing with $a x+b y+c=0$, we get

$\quad\quad\quad\quad a_1=-3, b_1=5$

and $\quad\quad\quad c_1=-7 \quad\quad\quad\quad$ [from Eq. (i)]

$ \quad\quad\quad\quad a_2=2p,b_2=-3 $

and $\quad\quad\quad c_1=-1 \quad\quad\quad\quad$ [from Eq. (ii)]

Since, the lines are intersecting at a unique point i.e., it has a unique Solution.

$ \begin{matrix} \therefore & \dfrac{a_1}{a_2} \neq \dfrac{b_1}{b_2} \\ \Rightarrow & \dfrac{-3}{2 p} \neq \dfrac{5}{-3} \\ \Rightarrow & 9 \neq 10 p \\ \Rightarrow & p \neq \dfrac{9}{10} . \end{matrix} $

Hence, the lines represented by these equations are intersecting at a unique point for all real values of $p$ except $\dfrac{9}{10}$

(iv) Given pair of linear equations is

$ \begin{aligned} & 2 x+3 y-5=0 \\ & p x-6 y-8=0 \\ & a_1=2, b_1=3 \\ & c_1=-5 \\ & a_2=p, b_2=-6 \\ & c_2=-8 \end{aligned} $

Since, the pair of linear equations has a unique Solution.

$ \begin{matrix} \therefore & \dfrac{a_1}{a_2} \neq \dfrac{b_1}{b_2} \\ \Rightarrow & \dfrac{2}{p} \neq \dfrac{3}{-6} \\ \Rightarrow & p \neq-4 \end{matrix} $

Hence, the pair of linear equations has a unique Solution for all values of $p$ except -4 i.e.,

$ p \in R-{-4,} . $

(v) Given pair of linear equations is

$$ \begin{align*} & 2 x+3 y & =7 \tag{i}\\ \text{ and } & 2 p x+p y & =28-q y \\ \Rightarrow & 2 p x+(p+q) y & =28 \tag{ii} \end{align*} $$

On comparing with $a x+b y+c=0$, we get

$ \begin{aligned} & a_1=2, b_1=3 \\ & c_1=-7 \\ & a_2=2 p, b_2=(p+q) \end{aligned} $

$ \begin{matrix} \text{ and } & c_1=-7 \\ a_2=2 p, b \\ \text{ and } & c_2=-28 \end{matrix} $

Since, the pair of equations has infinitely many Solutions i.e., both lines are coincident.

$ \begin{matrix} \therefore & \dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=\dfrac{c_1}{c_2} \\ \Rightarrow & \dfrac{2}{2 p}=\dfrac{3}{(p+q)}=\dfrac{-7}{-28} \end{matrix} $

Taking first and third parts, we get

$ \begin{aligned} & \dfrac{2}{2 p} & =\dfrac{-7}{-28} \\ \Rightarrow & \dfrac{1}{p} & =\dfrac{1}{4} \\ \Rightarrow & p & =4 \end{aligned} $

Again, taking last two parts, we get

$ \begin{aligned} & \dfrac{3}{p+q}=\dfrac{-7}{-28} \Rightarrow \dfrac{3}{p+q}=\dfrac{1}{4} \\ & \Rightarrow \quad p+q=12 \\ & \Rightarrow \quad 4+q=12 \quad[\because p=4] \\ & \therefore \quad q=8 \end{aligned} $

Here, we see that the values of $p=4$ and $q=8$ satisfies all three parts.

Hence, the pair of equations has infinitely many Solutions for the values of $p=4$ and $q=8$

Solution

Given linear equations are

$x-3 y-2=0$

$$ \begin{equation*} -2 x+6 y-5=0 \tag{i} \end{equation*} $$

On comparing both the equations with $a x+b y+c=0$, we get

and $$ a_1=1, b_1=-3 $$

$$ c_1=-2 \quad\quad\quad [\text{from Eq.} (i)] $$

and $a_2=-2, b_2=6$

$c_2=-5 \quad\quad$ [from Eq. (ii)]

Here, $\dfrac{a_1}{a_2}=\dfrac{1}{-2}$

$\dfrac{b_1}{b_2}=\dfrac{-3}{6}=-\dfrac{1}{2}$ and $\dfrac{c_1}{c_2}=\dfrac{-2}{-5}=\dfrac{2}{5}$

i.e., $\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2}$ [parallel lines]

i.e.,

Hence, two straight paths represented by the given equations never cross each other, because they are parallel to each other.

Solution

Condition for the pair of system to have unique Solution

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

Let the equations are,

and

$ \begin{matrix} a_1 x+b_1 y+c_1=0 \\ a_2 x+b_2 y+c_2=0 \end{matrix} $

Since, $x=-1$ and $y=3$ is the unique Solution of these two equations, then

$$ \begin{matrix} \Rightarrow & a_1(-1)+b_1(3)+c_1=0 \\ \text{ and } & -a_1+3 b_1+c_1=0 \\ \Rightarrow & a_2(-1)+b_2(3)+c_2=0 \\ & -a_2+3 b_2+c_2=0 \tag{ii} \end{matrix} $$

So, the different values of $a_1, a_2, b_1, b_2, c_1$ and $c_2$ satisfy the Eqs. (i) and (ii).

Hence, infinitely many pairs of linear equations are possible.

Solution

Given equations are

and $\quad$| $2 x+y=23$ | | —: | | $4 x-y=19$ |

On adding both equations, we get

$$ \begin{equation*} 6 x=42 \Rightarrow x=7 \tag{ii} \end{equation*} $$

Put the value of $x$ in Eq. (i), we get

$$ \begin{array}{lrlrl} \Rightarrow & 14+y & =23 \\ \Rightarrow & y & =23-14 \\ \Rightarrow & y & =9 \\ \text { We have, } & 5 y-2 x & =5 \times 9-2 \times 7 \\ & & =45-14=31 \\ \text { and } & \dfrac{y}{x}-2=\dfrac{4}{x}-2 & =\dfrac{9}{7}-2=\dfrac{9-14}{7}=-\dfrac{5}{7} \end{array} $$

We have,

Hence, the values of $(5 y-2 x)$ and $\left(\dfrac{y}{x}-2\right)$ are 31 and $\dfrac{-5}{7}$, respectively.

Solution

By property of rectangle,

Lengths are equal, i.e.,

$\Rightarrow$

$$ \begin{align*} C D & =A B \\ x+3 y & =13 \tag{i}\\ A D & =B C \\ 3 x+y & =7 \tag{ii} \end{align*} $$

On multiplying Eq. (ii) by 3 and then subtracting Eq. (i), we get

$ \begin{matrix} 9 x+3 y=21 \\ x+3 y=13 \\ \hline 8 x \quad=8 \\ \hline x=1 \end{matrix} $

On putting $x=1$ in Eq. (i), we get

$ 3 y=12 \Rightarrow y=4 $

Hence, the required values of $x$ and $y$ are 1 and 4, respectively.

Solution

(i) Given pair of linear equations are is

$$ \begin{array} x+y & =3.3 \tag {i} \\ \dfrac{0.6}{3 x-2 y} & =-1 \end{array} $$

$$ \begin{array}{rrr} \Rightarrow & 0.6=-3 x+2 y \\ \Rightarrow & 3 x-2 y=-0.6 \tag{ii} \end{array} $$

Now, multiplying Eq. (i) by 2 and then adding with Eq. (ii), we get

$ \begin{aligned} \Rightarrow & 2 x+2 y & =6.6 \\ \Rightarrow & 3 x-2 y & =-0.6 \\ 5 x & =6 \quad \Rightarrow \quad x=\dfrac{6}{5}=1.2 \end{aligned} $

Now, put the value of $x$ in Eq. (i), we get

$ \begin{aligned} & 1.2+y =3.3 \\ \Rightarrow & \quad\quad y =3.3-1.2 \\ \Rightarrow & \quad\quad y =2.1 \end{aligned} $

Hence, the required values of $x$ and $y$ are 1.2 and 2.1, respectively.

(ii) Given, pair of linear equations is $$ \dfrac{x}{3}+\dfrac{y}{4}=4 $$

On multiplying both sides by $\operatorname{LCM}(3,4)=12$, we get $$ 4 x+3 y=48 $$

$$ \dfrac{5 x}{6}-\dfrac{y}{8}=4 $$

On multiplying both sides by $LCM(6,8)=24$, we get

$$ \begin{equation*} 20 x-3 y=96 \tag{ii} \end{equation*} $$

Now, adding Eqs. (i) and (ii), we get

$ 24 x=144 $

$ \Rightarrow \quad x=6 $

Now, put the value of $x$ in Eq. (i), we get

$ \begin{aligned} & & 4 \times 6+3 y & =48 \\ \Rightarrow & & 3 y & =48-24 \\ \Rightarrow & & 3 y & =24 \Rightarrow \quad y=8 \end{aligned} $

Hence, the required values of $x$ and $y$ are 6 and 8 , respectively.

(iii) Given pair of linear equations are

$$ \begin{align*} & 4 x+\dfrac{6}{y}=15 \tag{i}\\ & 6 x-\dfrac{8}{y}=14, y \neq 0 \tag{ii} \end{align*} $$

and

Let $u=\dfrac{1}{y}$, then above equation becomes

$$ \begin{align*} & 4 x+6 u=15 \tag{iii}\\ & 6 x-8 u=14 \tag{iv} \end{align*} $$

On multiplying Eq. (iii) by 8 and Eq. (iv) by 6 and then adding both of them, we get

$$ \begin{aligned} 32 x+48 u & =120 \\ 36 x-48 u & =84 \quad \Rightarrow \quad 68 x=204 \\ x & =3 \end{aligned} $$

$\Rightarrow$

Now, put the value of $x$ in Eq. (iii), we get

$ \begin{aligned} \Rightarrow & & 4 \times 3+6 u & =15 \\ \Rightarrow & & 6 u & =15-12 \Rightarrow 6 u=3 \\ \Rightarrow & & u & =\dfrac{1}{2} \Rightarrow \dfrac{1}{y}=\dfrac{1}{2} \\ \Rightarrow & & y & =2 \end{aligned} $

$ \Rightarrow \quad u=\dfrac{1}{2} \Rightarrow \dfrac{1}{y}=\dfrac{1}{2} \quad \because u=\dfrac{1}{y} $

Hence, the required values of $x$ and $y$ are 3 and 2, respectively.

(iv) Given pair of linear equations is

and

$$ \begin{align*} & \dfrac{1}{2 x}-\dfrac{1}{y}=-1 \tag{i}\\ & \dfrac{1}{x}+\dfrac{1}{2 y}=8, x, y \neq 0 \tag{ii} \end{align*} $$

Let $u=\dfrac{1}{x}$ and $v=\dfrac{1}{y}$, then the above equations becomes

$\begin{array}{lc} & \dfrac{u}{2}-v=-1 \\ \Rightarrow & u-2 v=-2 \\ \text { and } & u+\dfrac{v}{2}=8 \\ \Rightarrow & 2 u+v=16\end{array}$

On, multiplying Eq. (iv) by 2 and then adding with Eq. (iii), we get

$ \begin{aligned} 4 u+2 v & =32 \\ u-2 v & =-2 \\ \hline 5 u & =30 \\ \Rightarrow \quad u & =6 \end{aligned} $

Now, put the value of $u$ in Eq. (iv), we get

$ 2 \times 6+v=16 $

$ \begin{matrix} \Rightarrow & v=16-12=4 \\ \Rightarrow & v=4 \\ \therefore & x=\dfrac{1}{u}=\dfrac{1}{6} \text{ and } y=\dfrac{1}{v}=\dfrac{1}{4} \end{matrix} $

Hence, the required values of $x$ and $y$ are $\dfrac{1}{6}$ and $\dfrac{1}{4}$, respectively.

(v) Given pair of linear equations is

$$ \begin{align*} & 43 x+67 y=-24 \tag{i}\\ & 67 x+43 y=24 \tag{ii} \end{align*} $$

and

On multiplying Eq. (i) by 43 and Eq. (ii) by 67 and then subtracting both of them, we get

$ \begin{aligned} & (67)^{2} x+43 \times 67 y=24 \times 67 \\ & (43)^{2} x+43 \times 67 y=-24 \times 43 \\ & {(67)^{2}-(43)^{2}} x=24(67+43) \end{aligned} $

$ \begin{aligned} & \Rightarrow \quad(67+43)(67-43) x=24 \times 110 \quad[\because(a^{2}-b^{2})=(a-b)(a+b)] \\ & \Rightarrow \quad 110 \times 24 x=24 \times 110 \\ & \Rightarrow \quad x=1 \end{aligned} $

Now, put the value of $x$ in Eq. (i), we get

$ \begin{aligned} 43 \times 1+67 y & =-24 \\ 67 y & =-24-43 \end{aligned} $

$ \begin{matrix} \Rightarrow & 67 y=-24 \\ \Rightarrow & 67 y=-67 \\ \Rightarrow & y=-1 \end{matrix} $

Hence, the required values of $x$ and $y$ are 1 and -1 , respectively. (vi) Given pair of linear equations is

and

$$ \begin{align*} \dfrac{x}{a}+\dfrac{y}{b} & =a+b \tag{i}\\ \dfrac{x}{a^{2}}+\dfrac{y}{b^{2}} & =2, a, b \neq 0 \tag{ii} \end{align*} $$

On multiplying Eq. (i) by $\dfrac{1}{a}$ and then subtracting from Eq. (ii), we get

$ \begin{aligned} \dfrac{x}{a^{2}}+\dfrac{y}{b^{2}} & =2 \\ \dfrac{x}{a^{2}}+\dfrac{y}{a b} & =1+\dfrac{b}{a} \\ \dfrac{-1}{y \dfrac{1}{b^{2}}-\dfrac{1}{a b}} & =2-1-\dfrac{b}{a} \\ \Rightarrow \quad y \dfrac{a-b}{a b^{2}} & =1-\dfrac{b}{a}=\dfrac{a-b}{a} \\ \Rightarrow \quad y & =\dfrac{a b^{2}}{a} \Rightarrow y=b^{2} \end{aligned} $

Now, put the value of $y$ in Eq. (ii), we get

$ \begin{matrix} \Rightarrow & \dfrac{x}{a^{2}}+\dfrac{b^{2}}{b^{2}} & =2 \\ \Rightarrow & & \dfrac{x}{a^{2}} & =2-1=1 \\ & x & =a^{2} \end{matrix} $

Hence, the required values of $x$ and $y$ are $a^{2}$ and $b^{2}$, respectively.

(vii) Given pair of equations is

$ \begin{matrix} \Rightarrow & \dfrac{2 x y}{x+y}=\dfrac{3}{2}, \text{ where } x+y \neq 0 \\ \Rightarrow & \dfrac{x+y}{2 x y}=\dfrac{2}{3} \\ \Rightarrow & \dfrac{x}{x y}+\dfrac{y}{x y}=\dfrac{4}{3} \\ \text{ and } & \dfrac{1}{y}+\dfrac{1}{x}=\dfrac{4}{3} \\ \Rightarrow & \dfrac{x y}{2 x-y}=\dfrac{-3}{10}, \text{ where } 2 x-y \neq 0 \\ \Rightarrow & \dfrac{2 x-y}{x y}=\dfrac{-10}{3} \\ \Rightarrow & \dfrac{2 x}{x y}-\dfrac{y}{x y}=\dfrac{-10}{3} \\ \Rightarrow & \dfrac{2}{y}-\dfrac{1}{x}=\dfrac{-10}{3} \end{matrix} $

Now, put $\dfrac{1}{x}=u$ and $\dfrac{1}{y}=v$, then the pair of equations becomes

and

$$ \begin{align*} v+u & =\dfrac{4}{3} \tag{iii}\\ 2 v-u & =\dfrac{-10}{3} \tag{iv} \end{align*} $$

On adding both equations, we get

$ \begin{aligned} \Rightarrow & & 3 v & =\dfrac{4}{3}-\dfrac{10}{3}=\dfrac{-6}{3} \\ \Rightarrow & & 3 v & =-2 \\ & & v & =\dfrac{-2}{3} \end{aligned} $

Now, put the value of $v$ in Eq. (iii), we get

$ \begin{aligned} & & \dfrac{-2}{3}+u & =\dfrac{4}{3} \\ & & u & =\dfrac{4}{3}+\dfrac{2}{3}=\dfrac{6}{3}=2 \\ & & x & =\dfrac{1}{u}=\dfrac{1}{2} \\ & \text{ and } & y & =\dfrac{1}{v}=\dfrac{1}{(-2 / 3)}=\dfrac{-3}{2} \end{aligned} $

Hence, the required values of $x$ and $y$ are $\dfrac{1}{2}$ and $\dfrac{-3}{2}$, respectively.

Solution

Given pair of equations is

and

$$ \begin{align*} \dfrac{x}{10}+\dfrac{y}{5}-1 & =0 \tag{i}\\ \dfrac{x}{8}+\dfrac{y}{6} & =15 \tag{ii} \end{align*} $$

Now, multiplying both sides of Eq. (i) by $LCM(10,5)=10$, we get

$ x+2 y-10=0 $

$$ \begin{equation*} \Rightarrow \quad x+2 y=10 \tag{iii} \end{equation*} $$

Again, multiplying both sides of Eq. (iv) by $LCM(8,6)=24$, we get

$$ \begin{equation*} 3 x+4 y=360 \tag{iv} \end{equation*} $$

On, multiplying Eq. (iii) by 2 and then subtracting from Eq. (iv), we get

$ \begin{gathered} 3 x+4 y=360 \\ 2 x+4 y=20 \\ -\quad-\quad \\ \hline x=340 \end{gathered} $

Put the value of $x$ in Eq. (iii), we get

$ \begin{aligned} & & 340+2 y & =10 \\ \Rightarrow & & 2 y & =10-340=-330 \\ \Rightarrow & & y & =-165 \end{aligned} $

Given that, the linear relation between $x, y$ and $\lambda$ is

$ y=\lambda x+5 $

Now, put the values of $x$ and $y$ in above relation, we get

$ \begin{aligned} \Rightarrow & 340 \lambda & =-170 \\ \Rightarrow & \lambda & =-\dfrac{1}{2} \end{aligned} $

Hence, the Solution of the pair of equations is $x=340, y=-165$ and the required value of $\lambda$ is $-\dfrac{1}{2}$.

Solution

(i) Given pair of equations is

$ \begin{aligned} & 3 x+y+4=0 \\ & \begin{matrix} \text{ and } \quad 6 x-2 y+4=0 \\ \text{ On comparing with } a x+b y+c=0 \text{, we get } \end{matrix} \\ & a_1=3, b_1=1 \\ & c_1=4 \\ & a_2=6, b_2=-2 \\ & \text{ and } \\ & c_2=4 \\ & \dfrac{a_1}{a_2}=\dfrac{3}{6}=\dfrac{1}{2} ; \dfrac{b_1}{b_2}=\dfrac{1}{-2} \\ & \dfrac{c_1}{c_2}=\dfrac{4}{4}=\dfrac{1}{1} \\ & \dfrac{a_1}{a_2} \neq \dfrac{b_1}{b_2} \end{aligned} $

So, the given pair of linear equations are intersecting at one point, therefore these lines have unique Solution.

Hence, given pair of linear equations is consistent.

We have,

$ \begin{gathered} 3 x+y+4=0 \\ y=-4-3 x \end{gathered} $

$\Rightarrow$

When $x=0$, then $y=-4$

When $x=-1$, then $y=-1$

When $x=-2$, then $y=2$

and

$6 x-2 y+4=0$

$\Rightarrow$

$2 y=6 x+4$

$\Rightarrow$

$ y=3 x+2 $

When $x=0$, then $y=2$

When $x=-1$, then $y=-1$

When $x=1$, then $y=5$

Plotting the points $B(0,-4)$ and $A(-2,2)$, we get the straight tine $A B$. Plotting the points $Q(0,2)$ and $P(1,5)$, we get the straight line $P Q$. The lines $A B$ and $P Q$ intersect at $C$ $(-1,-1)$.

(ii) Given pair of equations is

$$ \begin{matrix} x-2 y=6 \\ 3 x-6 y=0 \tag{ii} \end{matrix} $$

and

On comparing with $a x+b y+c=0$, we get

$$ \begin{align*} & a_1=1, b_1=-2 \text{ and } c_1=-6 \tag{i}\\ & a_2=3, b_2=-6 \text{ and } c_2=0 \end{align*} $$

Here,

$ \dfrac{a_1}{a_2}=\dfrac{1}{3}, \dfrac{b_1}{b_2} \dfrac{-2}{-6}=\dfrac{1}{3} \text{ and } \dfrac{c_1}{c_2}=\dfrac{-6}{0} $

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

Hence, the lines represented by the given equations are parallel. Therefore, it has no Solution. So, the given pair of lines is inconsistent.

(iii) Given pair of equations is $x+y=3$

and

$$ \begin{equation*} 3 x+3 y=9 \tag{i} \end{equation*} $$

On comparing with $a x+b y+c=0$, we get

$$ \begin{align*} & a_1=1, b_1=1 \text{ and } c_1=-3 \\ & a_2=3, b_2=3 \text{ and } c_2=-9 \tag{ii}\\ & \dfrac{a_1}{a_2}=\dfrac{1}{3}, \dfrac{b_1}{b_2}=\dfrac{1}{3} \text{ and } \dfrac{c_1}{c_2}=\dfrac{-3}{-9}=\dfrac{1}{3} \\ & \dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=\dfrac{c_1}{c_2} \end{align*} $$

Here,

$\Rightarrow$

coincident. Therefore, these lines have infinitely many

Now, $\quad x+y=3 \Rightarrow y=3-x$

If $x=0$, then $y=3$, If $x=3$, then $y=0$

and

$ \begin{aligned} 3 x+3 y & =9 \Rightarrow \quad 3 y=9-3 x \\ y & =\dfrac{9-3 x}{3} \end{aligned} $

Plotting the points $A(0,3)$ and $B(3,0)$, we get the line $A B$. Again, plotting the points $C(0,3) D(1,2)$ and $E(3,0)$, we get the line $C D E$.

We observe that the lines represented by Eqs. (i) and (ii) are coincident.

Solution

The given pair of linear equations

Table for line $2 x+y=4$,

and table for line $2 x-y=4$,

Graphical representation of both lines. Here, both lines and $Y$-axis form a $\triangle A B C$.

Hence, the vertices of a $\triangle A B C$ are $A(0,4) B(2,0)$ and $C(0,-4)$.

$ \begin{aligned} \therefore \quad \text{ Required area of } \triangle A B C & =2 \times \text{ Area of } \triangle A O B \\ & =2 \times \dfrac{1}{2} \times 4 \times 2=8 \text{ sq units } \end{aligned} $

Hence, the required area of the triangle is 8 sq units.

Solution

Given pair of linear equations is and

$$ \begin{align*} & x+y-2=0 \tag{i}\\ & 2 x-y-1=0 \tag{ii} \end{align*} $$

On comparing with $a x+b y+c=0$, we get

$$ \begin{matrix} a_1=1, b_1=1 \text{ and } c_1=-2 & \text{ [from Eq. (i)] } \tag{i}\\ a_2=2, b_2=-1 \text{ and } c_2=-1 & \text{ [from Eq. (ii)] } \\ \dfrac{a_1}{a_2}=\dfrac{1}{2}, \dfrac{b_1}{b_2}=\dfrac{1}{-1} & \\ \dfrac{c_1}{c_2}=\dfrac{-2}{-1}=\dfrac{2}{1} \Rightarrow \dfrac{a_1}{a_2} \neq \dfrac{b_1}{b_2} & \end{matrix} $$

Here,

and

So, both lines intersect at a point. Therefore, the pair of equations has a unique Solution. Hence, these equations are consistent.

Now,

$x+y=2 \Rightarrow y=2-x$

If $x=0$, then $y=2$ and if $x=2$, then $y=0$

and

If $x=0$, then $y=-1$; if $x=\dfrac{1}{2}$, then $y=0$ and if $x=1$, then $y=1$

Plotting the points $A(2,0)$ and $B(0,2)$, we get the straight line $A B$. Plotting the points $C$ $(0,-1)$ and $D(1 / 2,0)$, we get the straight line $C D$. The lines $A B$ and $C D$ intersect at $E(1,1)$. Hence, infinite lines can pass through the intersection point of linear equations $x+y=2$ and $2 x-y=1$ i.e., $E(1,1)$ like as $y=x, 2 x+y=3, x+2 y=3$. so on.

Solution

Given that, $(x+1)$ is a factor of $f(x)=2 x^{s}+a x^{2}+2 b x+1$, then $f(-1)=0$.

$ \text{ [if }(x+\alpha) \text{ is a factor of } f(x)=a x^{2}+b x+c, \text{ then } f(-)=0 \text{ ] } $

$$ \begin{matrix} \Rightarrow & 2(-1)^{3}+a(-1)^{2}+2 b(-1)+1 & =0 \\ \Rightarrow & -2+a-2 b+1 & =0 \\ \Rightarrow & a-2 b-1 & =0 \tag{i}\\ \text{ Also, } & 2 a-3 b & =4 \\ \Rightarrow & 3 b & =2 a-4 \\ \Rightarrow & b & =\dfrac{2 a-4}{3} \end{matrix} $$

Now, put the value of $b$ in Eq. (i), we get

$ \begin{matrix} \Rightarrow & 3 a-2(2 a-4)-3=0 \\ \Rightarrow & 3 a-4 a+8-3=0 \\ \Rightarrow & -a+5=0 \\ \Rightarrow & a=5 \end{matrix} $

Now, put the value of a in Eq. (i), we get

$ \begin{aligned} \Rightarrow & 5-2 b-1 & =0 \\ \Rightarrow & 2 b & =4 \\ \Rightarrow & b & =2 \end{aligned} $

Hence, the required values of $a$ and $b$ are 5 and 2 , respectively.

Solution

Given that, $x, y$ and $40^{\circ}$ are the angles of a triangle.

$\therefore$

$\Rightarrow$

Also,

On adding Eqs. (i) and (ii), we get

$ x+y+40^{\circ}=180^{\circ} $

[since, the sum of all the angles of a triangle is $180^{\circ}$ ]

$$ \begin{equation*} x+y=140^{\circ} \tag{i} \end{equation*} $$

$$ \begin{equation*} x-y=30^{\circ} \tag{ii} \end{equation*} $$

$ \begin{matrix}& & 2 x & =170^{\circ} \\ \Rightarrow & x & =85^{\circ} \end{matrix} $

On putting $x=85^{\circ}$ in Eq. (i), we get

$ \begin{aligned} 85^{\circ}+y & =140^{\circ} \\ \hline & =55^{\circ} \end{aligned} $

$\Rightarrow \quad y=55^{\circ}$

Hence, the required values of $x$ and $y$ are $85^{\circ}$ and $55^{\circ}$, respectively.

Solution

Let Salim and his daughter’s age be $x$ and yyr respectively.

Now, by first condition

Two years ago, Salim was thrice as old as his daughter.

$$ \begin{align*} & \text{ i.e., } \quad x-2=3(y-2) \Rightarrow x-2=3 y-6 \\ & \Rightarrow \quad x-3 y=-4 \tag{i} \end{align*} $$

and by second condition, six years later. Salim will be four years older than twice her age.

$$ \begin{matrix} \Rightarrow & x+6=2(y+6)+4 \\ \Rightarrow & x+6=2 y+12+4 \\ \Rightarrow & x-2 y=16-6 \\ \Rightarrow & -2 y=10 \tag{ii} \end{matrix} $$

On subtracting Eq. (i) from Eq. (ii), we get

$$ \begin{array}{lr} & x-2y=10 \\ & \underset{-}{x}-\underset{+}{3y}=\underset{+}{-4} \\ \hline \\ & y=14 \end{array} $$

Put the value of $y$ in Eq. (ii), we get $$ \begin{aligned} & x-2 \times 14=10 \\ \Rightarrow & x=10+28 \Rightarrow x=38 \\ & \end{aligned} $$

Hence, Salim and his daughter’s age are $38 yr$ and $14 yr$, respectively.

Solution

Let the present age (in year) of father and his two children be $x$, $y$ and $z y r$, respectively. Now by given condition, and after $20 yr$,

$ \begin{aligned} x & =2(y+z) \\ (x+20) & =(y+20)+(z+20) \\ y+z+40 & =x+20 \end{aligned} $

$ \begin{aligned} & \Rightarrow \quad y+z+40=x+20 \\ & \Rightarrow \quad y+z=x-20 \\ & \text{ On putting the value of }(y+z) \text{ in Eq. (i) and get the present age of father } \\ & \therefore \quad x=2(x-20) \\ & \therefore \quad x=2 x-40=40 \end{aligned} $

Hence, the father’s age is $40 yr$.

Solution

Let the two numbers be $x$ and $y$.

Then, by first condition, ratio of these two numbers $=5: 6$

$$ \Rightarrow \quad \begin{align*} x: y & =5: 6 \\ \dfrac{x}{y} & =\dfrac{5}{6} \Rightarrow y=\dfrac{6 x}{5} \tag{i} \end{align*} $$

and by second condition, then, 8 is subtracted from each of the numbers, then ratio becomes $4: 5$.

$\begin{array}{rlrl} & & \dfrac{x-8}{y-8} & =\dfrac{4}{5} \\ \Rightarrow & & 5 x-40 & =4 y-32 \\ \Rightarrow & & 5 x-4 y & =8\end{array}$

Now, put the value of $y$ in Eq. (ii), we get

$ \begin{aligned} & 5 x-4 (\dfrac{6 x}{5})=8 \\ & \Rightarrow \quad 25 x-24 x=40 \\ & \Rightarrow \quad x=40 \\ & y=\dfrac{6}{5} \times 40 \\ & =6 \times 8=48 \end{aligned} $

Hence, the required numbers are 40 and 48.

Solution

Let the number of students in halls $A$ and $B$ are $x$ and $y$, respectively.

Now, by given condition,

$$ \begin{align*} x-10 & =y+10 \\ x-y & =20 \tag{i}\\ (x+20) & =2(y-20) \\ x-2 y & =-60 \tag{ii} \end{align*} $$

$ \begin{aligned} & \Rightarrow \\ & \text{ and } \\ & \Rightarrow \end{aligned} $

On subtracting Eq. (ii) from Eq. (i), we get

$ \begin{aligned} (x-y)-(x-2 y) & =20+60 \\ x-y-x+2 y & =80 \Rightarrow y=80 \end{aligned} $

On putting $y=80$ in Eq. (i), we get

$ x-80=20 \Rightarrow x=100 $

$ \text{ and } \quad y=80 $

Hence, 100 students are in hall $A$ and 80 students are in hall $B$.

Solution

Let Latika takes a fixed charge for the first two day is $\text {₹} x$ and additional charge for each day thereafter is ₹ $y$.

Now by first condition.

Latika paid ₹ 22 for a book kept for six days i.e.,

$$ \begin{equation*} x+4 y=22 \tag{i} \end{equation*} $$

and by second condition,

Anand paid ₹ 16 for a book kept for four days i.e.,

$$ \begin{equation*} x+2 y=16 \tag{ii} \end{equation*} $$

Now, subtracting Eq. (ii) from Eq. (i), we get

$ 2 y=6 \Rightarrow y=3 $

On putting the value of $y$ in Eq. (ii), we get

$ x+2 \times 3=16 $

$\therefore \quad x=16-6=10$

Hence, $\quad$ the fixed charge $= \text {₹} 10$

and $\quad$ the charge for each extra day $=\text {₹} 3$

Solution

Let $x$ be the number of correct answers of the questions in a competitive examination, then $(120-x)$ be the number of wrong answers of the questions.

Then, by given condition,

$ \begin{aligned} & & x \times 1-(120-x) \times \dfrac{1}{2} & =90 \\ \Rightarrow & & x-60+\dfrac{x}{2} & =90 \\ \Rightarrow & & \dfrac{3 x}{2} & =150 \\ \therefore & & x & =\dfrac{150 \times 2}{3}=50 \times 2=100 \end{aligned} $

Hence, Jayanti answered correctly 100 questions.

Solution

We know that, by property of cyclic quadrilateral,

Sum of opposite angles $=180^{\circ}$

$ \angle A+\angle C=(6 x+10)^{\circ}+(x+y)^{\circ}=180^{\circ} $

$\Rightarrow$

$7 x+y=170$

and

$$ \begin{equation*} \angle B+\angle D=(5 x)^{\circ}+(3 y-10)^{\circ}=180^{\circ} \tag{i} \end{equation*} $$

$[\because \angle B=(5 x)^{\circ}, \angle D=(3 y-10)^{\circ}.$, given $]$

$\Rightarrow \quad 5 x+3 y=190^{\circ}$

On multiplying Eq. (i) by 3 and then subtracting, we get

$$ \begin{align*} & & 3 \times(7 x+y)-(5 x+3 y) & =510^{\circ}-190^{\circ} \tag{ii}\\ \Rightarrow & & 21 x+3 y-5 x-3 y & =320^{\circ} \\ \Rightarrow & & 16 x & =320^{\circ} \\ \therefore & & x & =20^{\circ} \end{align*} $$

On putting $x=20^{\circ}$ in Eq. (i), we get

$ \begin{aligned} \Rightarrow \quad 7 \times 20+y & =170^{\circ} \\ \therefore \quad y & =170^{\circ}-140^{\circ} \Rightarrow y=30^{\circ} \\ \therefore \quad \angle A & =(6 x+10)^{\circ}=6 \times 20^{\circ}+10^{\circ} \\ & =120^{\circ}+10^{\circ}=130^{\circ} \\ \angle B & =(5 x)^{\circ}=5 \times 20^{\circ}=100^{\circ} \\ \angle C & =(x+y)^{\circ}=20^{\circ}+30^{\circ}=50^{\circ} \\ \angle D & =(3 y-10)^{\circ}=3 \times 30^{\circ}-10^{\circ} \\ & =90^{\circ}-10^{\circ}=80^{\circ} \end{aligned} $

Hence, the required values of $x$ and $y$ are $20^{\circ}$ and $30^{\circ}$ respectively and the values of the four angles i.e., $\angle A, \angle B, \angle C$ and $\angle D$ are $130^{\circ}, 100^{\circ}, 50^{\circ}$ and $80^{\circ}$, respectively.

Solution

Given equations are $2 x+y=6$ and $2 x-y+2=0$

Table for equation $2 x+y=6$,

Table for equation $2 x-y+2=0$,

Let $A_1$ and $A_2$ represent the areas of $\triangle A C E$ and $\triangle B D E$, respectively.

Now,

and

$ \begin{aligned} A_1=\text{ Area of } \triangle A C E & =\dfrac{1}{2} \times A C \times P E \\ & =\dfrac{1}{2} \times 4 \times 4=8 \end{aligned} $

$A_2=$ Area of $\triangle B D E=\dfrac{1}{2} \times B D \times Q E$

$ =\dfrac{1}{2} \times 4 \times 1=2 $

$ \therefore A_1: A_2=8: 2=4: 1 $

Hence, the pair of equations intersect graphically at point $E(1,4)$, i.e., $x=1$ and $y=4$.

Solution

Given linear equations are

For equation

$ x+y=8 \Rightarrow y=8-x $

If $x=0$, then $y=8$; if $x=8$, then $y=0$ and if $x=4$, then $y=4$

Table for line $x+y=8$.

Plotting the points $A(1,1)$ and $B(2,2)$, we get the straight line $A B$. Plotting the points $C(3,1)$ and $D(6,2)$, we get the straight line $C D$. Plotting the points $P(0,8), Q(4,4)$ and $R(8,0)$, we get the straight line $P Q R$. We see that lines $A B$ and $C D$ intersecting the line $P R$ on $Q$ and $D$, respectively.

So, $\triangle O Q D$ is formed by these lines. Hence, the vertices of the $\triangle O Q D$ formed by the given lines are $O(0,0), Q(4,4)$ and $D(6,2)$.

Solution

Given equation of lines $2 x-y-4=0, x=3$ and $x=5$

Table for line $2 x-y-4=0$,

Draw the points $P(0,-4)$ and $Q(2,0)$ and join these points and form a line $P Q$ also draw the lines $x=3$ and $x=5$.

$\therefore$ Area of quadrilateral $A B C D=\dfrac{1}{2} \times$ distance between parallel lines $(A B) \times(A D+B C)$

$ =\dfrac{1}{2} \times 2 \times(6+2) $

[since, quadrilateral $A B C D$ is a trapezium]

$B C=6]$

$ \because \quad A B=O B-O A=5-3=2, A D=2 $

$ =8 sq \text{ units } $

Hence, the required area of the quadrilateral formed by the lines and the $X$-axis is 8 sq units.

Solution

Let the cost of a pen be ₹ $x$ and the cost of a pencil box be ₹ $y$.

Then, by given condition,

and

$$ \begin{align*} 4 x+4 y & =100 \Rightarrow x+y=25 \tag{i}\\ 3 x & =y+15 \end{align*} $$

$\Rightarrow \quad 3 x-y=15$

On adding Eqs. (i) and (ii), we get

$ \begin{matrix}& & 4 x & =40 \\ \Rightarrow & x & =10 \end{matrix} $

By substituting $x=10$, in Eq. (i) we get

$ y=25-10=15 $

Hence, the cost of a pen and a pencil box are ₹ 10 and ₹ 15 , respectively.

Solution

ing lines (iii) and (i), we will get the intersecting point $A$.

On multiplying in Eq. (i) by 2 and then adding Eq. (iii), we get

$ \begin{gathered} 6 x-2 y=6 \\ x+2 y=8 \\ 7 x=14 \end{gathered} $

$ \Rightarrow \quad x=2 $

On putting the value of $x$ in Eq. (i), we get

$ \begin{aligned} & & 3 \times 2-y & =3 \\ \Rightarrow & & y & =6-3 \\ \Rightarrow & & y & =3 \end{aligned} $

So, the coordinate of point or vertex $A$ is $(2,3)$.

Hence, the vertices of the $\triangle A B C$ formed by the given lines are $A(2,3), B(1,0)$ and $C(4,2)$.

Solution

Let the speed of the rickshaw and the bus are $x$ and $y km / h$, respectively.

Now, she has taken time to travel $2 km$ by rickshaw , $t_1=\dfrac{2}{x} h . \quad \because$ speed $=\dfrac{\text{ distance }}{\text{ time }}$

and she has taken time to travel remaining distance i.e., $(14-2)=12 km$ by bus $=t_2=\dfrac{12}{y} h$.

By first condition, $\quad t_1+t_2=\dfrac{1}{2} \Rightarrow \dfrac{2}{x}+\dfrac{12}{y}=\dfrac{1}{2}$

Now, she has taken time to travel $4 km$ by rickshaw, $t_3=\dfrac{4}{x} h$

and she has taken time to travel remaining distance i.e., $(14-4)=10 km$ by bus $=t_4=\dfrac{10}{y} h$.

By second condition, $\quad t_3+t_4=\dfrac{1}{2}+\dfrac{9}{60}=\dfrac{1}{2}+\dfrac{3}{20}$

$\Rightarrow \quad \dfrac{4}{x}+\dfrac{10}{y}=\dfrac{13}{20}$

Let $\dfrac{1}{x}=u$ and $\dfrac{1}{y}=v$, then Eqs. (i) and (ii) becomes

and

$$ \begin{align*} & 2 u+12 v=\dfrac{1}{2} \tag{iii}\\ & 4 u+10 v=\dfrac{13}{20} \tag{iv} \end{align*} $$

On multiplying in Eq. (iii) by 2 and then subtracting, we get

$ \begin{aligned} & 4 u+24 v=1 \\ & 4 u+10 v=\dfrac{13}{20} \\ &-\quad-\quad \underline{-13} \\ & 14 v=1-\dfrac{7}{20}=\dfrac{7}{20} \\ & 2 v=\dfrac{1}{20} \Rightarrow v=\dfrac{1}{40} \end{aligned} $

Now, put the value of $v$ in Eq. (iii), we get

$ \begin{matrix} \Rightarrow & 2 u+12 \dfrac{1}{40}=\dfrac{1}{2} \\ \Rightarrow & 2 u=\dfrac{1}{2}-\dfrac{3}{10}=\dfrac{5-3}{10} \\ \therefore & 2 u=\dfrac{2}{10} \Rightarrow u=\dfrac{1}{10} \\ \Rightarrow & \dfrac{1}{x}=u \\ \text{ and } & \dfrac{1}{x}=\dfrac{1}{10} \Rightarrow x=10 km / h \\ \Rightarrow & \dfrac{1}{y}=v \Rightarrow \dfrac{1}{y}=\dfrac{1}{40} \\ & y=40 km / h \end{matrix} $

Hence, the speed of rickshaw and the bus are $10 km / h$ and $40 km / h$, respectively.

Solution

Let the speed of the stream be $v km / h$.

Given that, a person rowing in still water $=5 km / h$

The speed of a person rowing in downstream $=(5+v) km / h$

and the speed of a person has rowing in upstream $=(5-v) km / h$

Now, the person taken time to cover $40 km$ downstream,

$ t_1=\dfrac{40}{5+v} h \quad \because \text{ speed }=\dfrac{\text{ distance }}{\text{ time }} $

and the person has taken time to cover $40 km$ upstream,

$$ t_2=\dfrac{4 U}{5-v} \mathrm{~h} . $$

By condition,

$ \Rightarrow \hspace{29 mm} t_2 =t_1 \times 3 \\ \hspace{29 mm} \dfrac{40}{5-v} =\dfrac{40}{5+v} \times 3 $

$ \Rightarrow \hspace{29 mm } \dfrac{1}{5-v}=\dfrac{3}{5+v} $

$$ 5+v=15-3 v \Rightarrow 4 v=10 $$

$ \therefore \hspace{29 mm} v=\dfrac{10}{4}=2.5 \mathrm{~km} / \mathrm{h} $

Hence, the speed of the stream is $2.5 km / h$.

Solution

Let the speed of the motorboat in still water and the speed of the stream are $ukm / h$ and $v km / h$, respectively.

Then, a motorboat speed in downstream $=(u+v) km / h$

and a motorboat speed in upstream $=(u-v) km / h$.

Motorboat has taken time to travel $30 km$ upstream,

$ t_1=\dfrac{30}{u-v} h $

and motorboat has taken time to travel $28 km$ downstream,

$ t_2=\dfrac{28}{u+v} h $

By first condition, a motorboat can travel $30 km$ upstream and $28 km$ downstream in $7 h$

$ \begin{aligned} & \text{ i.e., } \quad t_1+t_2=7 h \\ & \Rightarrow \\ & \dfrac{30}{u-v}+\dfrac{28}{u+v}=7 \end{aligned} $

Now, motorboat has taken time to travel $21 km$ upstream and return i.e., $t_3=\dfrac{21}{u-v}$.

and

$ t_4=\dfrac{21}{u+v} $

[for upstream]

By second condition,

$ t_4+t_3=5 h $

$$ \begin{equation*} \Rightarrow \quad \dfrac{21}{u+v}+\dfrac{21}{u-v}=5 \tag{ii} \end{equation*} $$

[for downstream]

Let

Eqs. (i) and (ii) becomes

and

$\Rightarrow$

$$ \begin{gather*} x=\dfrac{1}{u+v} \text{ and } y=\dfrac{1}{u-v} \\ 30 x+28 y=7 \tag{iii}\\ 21 x+21 y=5 \\ x+y=\dfrac{5}{21} \tag{iv} \end{gather*} $$

Now, multiplying in Eq. (iv) by 28 and then subtracting from Eq. (iii), we get

$ \begin{aligned} & 30 x+28 y=7 \\ & 28 x+28 y=\dfrac{140}{21} \\ &-\quad-\quad \quad- \end{aligned} $

$ \Rightarrow \quad 2 x=\dfrac{1}{3} \Rightarrow x=\dfrac{1}{6} $

On putting the value of $x$ in Eq. (iv), we get

$$ \begin{matrix} \Rightarrow & \dfrac{1}{6}+y & =\dfrac{5}{21} \\ & & y=\dfrac{5}{21}-\dfrac{1}{6}=\dfrac{10-7}{42}=\dfrac{3}{42} \Rightarrow y=\dfrac{1}{14} \\ & \text{ and } & x & =\dfrac{1}{u+v}=\dfrac{1}{6} \Rightarrow u+v=6 \tag{v} \end{matrix} $$

$\Rightarrow \quad u-v=14$

Now, adding Eqs. (v) and (vi), we get

$ 2 u=20 \Rightarrow u=10 $

On putting the value of $u$ in Eq. (v), we get

$ \begin{aligned}& 10+v=6 \\ \Rightarrow & v=-4 \end{aligned} $

Hence, the speed of the motorboat in still water is $10 km / h$ and the speed of the stream $4 km / h$.

Solution

Let the two-digit number $=10 x+y$

Case I Multiplying the sum of the digits by 8 and then subtracting $5=$ two-digit number

$$ \begin{matrix} \Rightarrow & 8 \times(x+y)-5=10 x+y \\ \Rightarrow & 8 x+8 y-5=10 x+y \\ \Rightarrow & 2 x-7 y=-5 \tag{i} \end{matrix} $$

Case II Multiplying the difference of the digits by 16 and then adding $3=$ two-digit number

$$ \begin{matrix} \Rightarrow & 16 \times(x-y)+3=10 x+y \\ \Rightarrow & 16 x-16 y+3=10 x+y \\ \Rightarrow & 6 x-17 y=-3 \tag{ii} \end{matrix} $$

Now, multiplying in Eq. (i) by 3 and then subtracting from Eq. (ii), we get

$ \begin{array}{lr} & 6x-17y=-3 \\ & 6x-21y=-15 \\ \hline \\ & 4y=12 \Rightarrow y=3 \end{array} $

Now, put the value of $y$ in Eq. (i), we get

$ \hspace{34 mm} 2x-7x \times 3 = -5 \\ \Rightarrow \hspace{29 mm} 2x=21-5=16 \Rightarrow x=8 $

Hence, the required two-digit number

$ \begin{aligned} & =10 x+y \\ & =10 \times 8+3=80+3=83 \end{aligned} $

Solution

Let the cost of full and half first class fare be $\text {₹} x$ and $\text {₹} \dfrac{x}{2}$, respectively and reservation charges be ₹ y per ticket.

Case I The cost of one reserved first class ticket from the stations $A$ to $B$

$$ \begin{align*} & =\text {₹} 2530 \\ x+y & =2530 \tag{i} \end{align*} $$

$\Rightarrow$

Case II The cost of one reserved first class ticket and one reserved first class half ticket from stations $A$ to $B=\text {₹} 3810$

$$ \begin{matrix} \Rightarrow & x+y+\dfrac{x}{2}+y & =3810 \\ \Rightarrow & \dfrac{3 x}{2}+2 y & =3810 \\ \Rightarrow & 3 x+4 y & =7620 \tag{ii} \end{matrix} $$

Now, multiplying Eq. (i) by 4 and then subtracting from Eq. (ii), we get

$ \begin{aligned} & 3 x+4 y=7620 \\ & 4 x+4 y=10120 \\ &- \\ &-x=-2500 \end{aligned} $

$ \Rightarrow \quad x=2500 $

On putting the value of $x$ in Eq. (i), we get

$ \begin{aligned} & & 2500+y & =2530 \\ \Rightarrow & & y & =2530-2500 \\ \Rightarrow & & y & =30 \end{aligned} $

Hence, full first class fare from stations $A$ to $B$ is ₹ 2500 and the reservation for a ticket is ₹ 30 .

Solution

Let the cost price of the saree and the list price of the sweater be $\text {₹} x$ and $\text {₹} y$, respectively. Case I Sells a saree at $8 %$ profit + Sells a sweater at $10 %$ discount $=\text {₹} 1008$

$$ \begin{matrix}\Rightarrow & (100+8) \% \text{ of } x+(100-10) \% \text{ of } y & =1008 \\ \Rightarrow & 108 \% \text{ of } x+90 \% \text{ of } y & =1008 \\ \Rightarrow & 1.08 x+0.9 y & =1008 \tag{i} \end{matrix} $$

Case II Sold the saree at $10 \%$ profit + Sold the sweater at $8 \%$ discount $=\text {₹} 1028$

$$ \begin{matrix}\Rightarrow & (100+10) \% \text{ of } x+(100-8) \% \text{ of } y=1028 \\ \Rightarrow & 110 \% \text{ of } x+92 \% \text{ of } y=1028 \\ \Rightarrow & 1.1 x+0.92 y=1028 \tag{ii} \end{matrix} $$

On putting the value of $y$ from Eq. (i) into Eq. (ii), we get

$ \begin{matrix}& & 1.1 x+0.92 \dfrac{1008-1.08 x}{0.9} & =1028 \\ \Rightarrow & 1.1 \times 0.9 x+927.36-0.9936 x & =1028 \times 0.9 \\ \Rightarrow & 0.99 x-0.9936 x & =925.2-927.36 \\ \Rightarrow & -0.0036 x & =-2.16 \\ \therefore & x & =\dfrac{2.16}{0.0036}=600 \end{matrix} $

On putting the value of $x$ in Eq. (i), we get

$ 1.08 \times 600+0.9 y=1008 $

$ \begin{aligned} \Rightarrow & 108 \times 6+0.9 y & =1008 \\ \Rightarrow & 0.9 y & =1008-648 \\ \Rightarrow & 0.9 y & =360 \\ \Rightarrow & y & =\dfrac{360}{0.9}=400 \end{aligned} $

Hence, the cost price of the saree and the list price (price before discount) of the sweater are ₹ 600 and ₹ 400 , respectively.

Solution

Let the amount of investments in schemes $A$ and $B$ be $\text {₹} x$ and $\text {₹} y$, respectively.

Case I Interest at the rate of $8 %$ per annum on scheme $A+$ Interest at the rate of $9 %$ per annum on scheme $B=$ Total amount received

$$ \begin{matrix} \Rightarrow \quad & \dfrac{x \times 8 \times 1}{100}+\dfrac{y \times 9 \times 1}{100} & =\text {₹} 1860 \quad \because \text{ simple interest }=\dfrac{\text{ principal } \times \text{ rate } \times \text{ time }}{100} \\ \Rightarrow \quad 8 x+9 y & =186000 \tag{i} \end{matrix} $$

Case II Interest at the rate of $9 %$ per annum on scheme $A+$ Interest at the rate of $8 %$ per annum on scheme $B=\text {₹} 20$ more as annual interest

$ \begin{aligned} & \Rightarrow \quad \dfrac{x \times 9 \times 1}{100}+\dfrac{y \times 8 \times 1}{100}=\text {₹} 20+\text {₹} 1860 \\ & \Rightarrow \quad \dfrac{9 x}{100}+\dfrac{8 y}{100}=1880 \\ & \Rightarrow \quad 9 x+8 y=188000 \end{aligned} $

On multiplying Eq. (i) by 9 and Eq. (ii) by 8 and then subtracting them, we get

$ \begin{aligned} 72 x+81 y & =9 \times 186000 \\ 72 x+64 y & =8 \times 188000 \\ -\quad & \\ \hline 17 y & =1000[(9 \times 186)-(8 \times 188)] \\ & =1000(1674-1504)=1000 \times 170 \\ 17 y & =170000 \Rightarrow y=10000 \end{aligned} $

$ \Rightarrow $

On putting the value of $y$ in Eq. (i), we get

$ \begin{aligned} & & 8 x+9 \times 10000 & =186000 \\ \Rightarrow & & 8 x & =186000-90000 \\ \Rightarrow & & 8 x & =96000 \\ \Rightarrow & & x & =12000 \end{aligned} $

Hence, she invested ₹ 12000 and ₹ 10000 in two schemes A and B, respectively.

Solution

Let the number of bananas in lots $A$ and $B$ be $x$ and $y$, respectively

Case I Cost of the first lot at the rate of ₹ 2 for 3 bananas + Cost of the second lot at the rate of $\text {₹} 1$ per banana $=$ Amount received

$$ \begin{matrix} \Rightarrow & \dfrac{2}{3} x+y=400 \\ \Rightarrow & 2 x+3 y=1200 \tag{i} \end{matrix} $$

Case II Cost of the first lot at the rate of $\text {₹} 1$ per banana + Cost of the second lot at the rate of $\text {₹} 4$ for 5 bananas $=$ Amount received

$$ \begin{matrix} \Rightarrow & x+\dfrac{4}{5} y=460 \\ \Rightarrow & 5 x+4 y=2300 \tag{ii} \end{matrix} $$

On multiplying in Eq. (i) by 4 and Eq. (ii) by 3 and then subtracting them, we get

$ \begin{aligned} 8 x+12 y & =4800 \\ 15 x+12 y & =6900 \\ \hline-7 x & =-2100 \\ x & =300 \end{aligned} $

$\Rightarrow$

Now, put the value of $x$ in Eq. (i), we get

$ \begin{matrix} & 2 \times 300+3 y & =1200 \\ \Rightarrow & 600+3 y & =1200 \\ \Rightarrow & 3 y & =1200-600 \\ \Rightarrow & 3 y & =600 \\ \Rightarrow & y & =200 \end{matrix} $

$\therefore$ Total number of bananas $=$ Number of bana nas in lot $A+$ Number of bananas in lot $B$ $=x+y$

$ =300+200=500 $

Hence, he had 500 bananas.