Question

I. If $x-y=xy=1-x-y,$ then $x+y$ is $\frac{5}{6}$
II. The system of equations $3x+2y=a$ and $5x+by=4$ has infinitely many solutions for $x$ and $y$, then $a=4$, $b=3$
III. If $\frac{x}{a}+\frac{y}{b}=2$ and $ax-by=a^2-b^2$, then $x=a,y=b$
Which is true?

• A. I only
• B. II only
• C. III only
• D. None of these

Answer 1

Answer: (A), (C)

The correct options are
A I only
C III only

Solution:-

(I) Given:-

$x-y=1-x-y$ $⟶{eq}^{n}1$

$xy=1-x-y$ $⟶{eq}^{n}2$

From ${eq}^{n}1$, we have

$x-y=1-x-y$

$\Rightarrow$ $x=1-x$

$\Rightarrow$ $2x=1$

$\Rightarrow$ $x=$ $\frac{1}{2}$

On putting the value of $x$ in ${eq}^{n}2$ we get

$\frac{1}{2}y=1-\frac{1}{2}-y$

$\Rightarrow \frac{1}{2}y+y=\frac{1}{2}$

$\Rightarrow \frac{3}{2}y=\frac{1}{2}$

$\Rightarrow y=\frac{1}{3}$

$\therefore x+y=\frac{1}{2}+\frac{1}{3}=\frac{5}{6}$

Hence, statement I is correct.

(II) Given:-

$3x+2y=a$ $⟶{eq}^{n}1$

$5x+by=4$ $⟶{eq}^{n}2$

For infinite many solutions,

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

Here,

$a_1=3$

$a_2=5$

$b_1=2$

$b_2=b$

$c_1=-a$

$c_2=-4$

$\frac{3}{5}=\frac{2}{b}=\frac{a}{4}⟶{eq}^{n}3$

On solving, ${eq}^{n}3$, we get

$a=\frac{12}{5}\&b=\frac{10}{3}$

Hence, statement II is incorrect.

(III) Given:-

$\frac{x}{a}+\frac{y}{b}=2⟶{eq}^{n}1$

$ax-by=a^2-b^2⟶{eq}^{n}2$

From ${eq}^{n}1$, we have

$\frac{x}{a}+\frac{y}{b}=2$

$\Rightarrow bx+ay=2ab$

Multiplying both sides by $a$, we get

$\Rightarrow abx+a^{2}y=2a^{2}b⟶{eq}^{n}3$

Multiplying both sides in ${eq}^{n}2$ by $b$, we get

$\Rightarrow abx-b^{2}y=a^{2}b-b^3⟶{eq}^{n}4$

On subtracting ${eq}^{n}4$ from ${eq}^{n}3$, we get

$a^{2}y+b^{2}y=2a^{2}b-a^{2}b+b^3$

$\Rightarrow y(a^2+b^2)=b(a^2+b^2)$

$\Rightarrow y=b$

On putting the value of y in ${eq}^{n}2$, we get

$ax-b^2=a^2-b^2$

$\Rightarrow ax=a^2$

$\Rightarrow x=a$

Hence, statement III is correct.