Question
Solve :
$\frac{4}{x-y}+\frac{1}{x+y}=3 ; \frac{2}{x-y}-\frac{3}{x+y}=5$
$\frac{4}{x-y}+\frac{1}{x+y}=3 ; \frac{2}{x-y}-\frac{3}{x+y}=5$
✓
Answer
$\begin{aligned}
\frac{4}{x-y}+\frac{1}{x+y} & =3 ; \frac{2}{x-y}-\frac{3}{x+y}=5 \\
4\left(\frac{1}{x-y}\right)+1\left(\frac{1}{x+y}\right) & =3 \ldots \text { (I) } \\
2\left(\frac{1}{x-y}\right)-3\left(\frac{1}{x+y}\right) & =5 \ldots \text { (II) }
\end{aligned}$
Replacing $\left(\frac{1}{x-y}\right)$ by $a$ and $\left(\frac{1}{x+y}\right)$ by $b$ we get
$\begin{aligned}
& 4 a+b=3 \ldots \text { (III) } \\
& 2 a-3 b=5 \ldots \text { (IV) }
\end{aligned}$
On solving these equations we get, $a=1 b=-1$
But $a=\left(\frac{1}{x-y}\right), b=\left(\frac{1}{x+y}\right)$
$\begin{aligned}
\therefore\left(\frac{1}{x-y}\right) & =1,\left(\frac{1}{x+y}\right)=-1 \\
\therefore \quad x-y & =1 \ldots( V ) \\
x+y & =-1 \ldots( VI )
\end{aligned}$
Solving equation (V) and (VI) we get $x=0, y=-1$
$\therefore$ Solution of the given equations is $(x, y)=(0,-1)$
\frac{4}{x-y}+\frac{1}{x+y} & =3 ; \frac{2}{x-y}-\frac{3}{x+y}=5 \\
4\left(\frac{1}{x-y}\right)+1\left(\frac{1}{x+y}\right) & =3 \ldots \text { (I) } \\
2\left(\frac{1}{x-y}\right)-3\left(\frac{1}{x+y}\right) & =5 \ldots \text { (II) }
\end{aligned}$
Replacing $\left(\frac{1}{x-y}\right)$ by $a$ and $\left(\frac{1}{x+y}\right)$ by $b$ we get
$\begin{aligned}
& 4 a+b=3 \ldots \text { (III) } \\
& 2 a-3 b=5 \ldots \text { (IV) }
\end{aligned}$
On solving these equations we get, $a=1 b=-1$
But $a=\left(\frac{1}{x-y}\right), b=\left(\frac{1}{x+y}\right)$
$\begin{aligned}
\therefore\left(\frac{1}{x-y}\right) & =1,\left(\frac{1}{x+y}\right)=-1 \\
\therefore \quad x-y & =1 \ldots( V ) \\
x+y & =-1 \ldots( VI )
\end{aligned}$
Solving equation (V) and (VI) we get $x=0, y=-1$
$\therefore$ Solution of the given equations is $(x, y)=(0,-1)$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
A bag contains 5 black, 7 red and 3 white balls. A ball is drawn from the bag at random. Find the probability that the ball drawn is:
- Red.
- Black or white.
- Not black.
In the figure 7.45, if A is the centre of the circle. ∠ PAR = 30°, AP = 7.5, find the area of the segment PQR. (π = 3.14)
If $\alpha$ and $\beta$ are the zeroes of the quadratic polynomial $p(s) = 3s^2 - 6s + 4$, find the value of $\frac{\alpha}{\beta}+\frac{\beta}{\alpha}+2\Big(\frac{1}{\alpha}+\frac{1}{\beta}\Big)+3\alpha\beta.$
→If $\alpha$ and $\beta$ are the roots of $x^2+5 x-1=0$ then find :
1. $\alpha^3+\beta^3$
2. $\alpha^2+\beta^2$.
→1. $\alpha^3+\beta^3$
2. $\alpha^2+\beta^2$.
Find the mean of each of the following frequency distributions:
→|
Class interval
|
$25-35$
|
$35-45$
|
$45-55$
|
$55-65$
|
$65-75$
|
|
Frequency
|
$6$
|
$10$
|
$8$
|
$12$
|
$4$
|
A fraction becomes $\frac{1}{3}$ if 1 is subtracted from both its numerator and denominator. It 1 is added to both the numerator and denominator, it becomes $\frac{1}{2}.$ Find the fraction.
→Prove the following:$\frac{\cos(90^\circ-\text{A})\sin(90^\circ-\text{A})}{\tan(90^\circ-\text{A})}-\sin^2\text{A}=0$
→Evaluate the following:
$\sin 45^{\circ} \sin30^{\circ} +\cos 45^{\circ} \cos30^{\circ}$
→$\sin 45^{\circ} \sin30^{\circ} +\cos 45^{\circ} \cos30^{\circ}$
In the adjoining figure, chord AD $\cong$ chord BC. m(arc ADC) = 100°, m(arc CD) = 60°. Find m(arc AB) and m(arc BC).
→The sum of a number and its reciprocal is $\frac{17}{4}.$ Find the number.
→