Solve for x and y: 3 x+y +2 x-y =2, 9 x+y -4 x-y =1 - Vidyadip

Question

Solve for x and y:
$\frac{3}{\text{x}+\text{y}}+\frac{2}{\text{x}-\text{y}}=2,$
$\frac{9}{\text{x}+\text{y}}-\frac{4}{\text{x}-\text{y}}=1$

✓

Answer

Putting $\frac{1}{\text{x}+\text{y}}=\text{u}$ and $\frac{1}{\text{x}-\text{y}}=\text{v}$
3u + 2v = 2 ...(1)
9u - 4v = 1 ...(2)
Multiply (1) by 2 and (2) by 1, we get
6u + 4v = 4 ...(3)
9u - 4v = 1 ...(4)
Adding (3) and (4), we get
$\text{15u}=5,$
$\text{u}=\frac{5}{15}=\frac{1}{3}$
Putting $\text{u}=\frac{1}{3}$ in (i), we get
$3\times\frac{1}{3}+\text{2v}=2$
$\Rightarrow1+\text{2v}=2$
$\Rightarrow\text{2v}=1$
$\text{v}=\frac{1}{2}$
Now, $\text{u}=\frac{1}{3}$
$\Rightarrow\frac{1}{\text{x}+\text{y}}=\frac{1}{3}$
$\Rightarrow\text{x}+\text{y}=3\ \dots(5)$
and $\text{v}=\frac{1}{2}$
$\Rightarrow\frac{1}{\text{x}-\text{y}}=\frac{1}{2}$
$\Rightarrow\text{x}-\text{y}=2\ \dots(6)$
Adding (5) and (6), we get
$\text{2x}=5$
$\Rightarrow\text{x}= \frac{5}{2}$
Putting $\text{x}=\frac{5}{2}$ in (5), we get
$\frac{5}{2}+\text{y}=3$
$\Rightarrow\text{y}=3-\frac{5}{2}=\frac{6-5}{2}=\frac{1}{2}$
$\therefore$ the solution is $\text{x}=\frac{5}{2}$ and $\text{y}=\frac{1}2{}$

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