Question

$$\begin{gathered} \frac{5}{(x+y)}-\frac{2}{(x-y)}+1=0, \\ \frac{15}{(x+y)}+\frac{7}{(x-y)}-10=0 \\ (x\ne y,x\ne -y). \end{gathered}$$

Answer 1

Taking $\frac{1}{x+y}=u$ and $\frac{1}{x-y}=v$, the given equations become:
5u − 2v + 1 = 0 ...(i)
15u + 7v − 10 = 0 ...(ii)
Here, a₁ = 5, b₁ = −2, c₁ = 1, a₂ = 15, b₂ = −7 and c₂ = −10
By cross multiplication, we have:


∴ $\frac{u}{\left[-2\times \left(-10\right)-1\times 7\right]}=\frac{v}{\left[1\times 15-\left(-10\right)\times 5\right]}=\frac{1}{\left[35+30\right]}$
⇒ $\frac{u}{20-7}=\frac{v}{15+50}=\frac{1}{65}$
⇒ $\frac{u}{13}=\frac{v}{65}=\frac{1}{65}$
⇒ $u=\frac{13}{65}=\frac{1}{5},v=\frac{65}{65}=1$
⇒ $\frac{1}{x+y}=\frac{1}{5},\frac{1}{x-y}=1$
So, (x + y) = 5 ...(iii)
and (x − y) = 1 ...(iv)

Again, the above equations (iii) and (iv) may be written as:
x + y − 5 = 0 ...(v)
x − y − 1 = 0 ...(vi)
Here, a₁ = 1, b₁ = 1, c₁ = −5, a₂ = 1, b₂ = −1 and c₂ = −1
By cross multiplication, we have:

∴ $\frac{x}{\left[1\times \left(-1\right)-\left(-5\right)\times \left(-1\right)\right]}=\frac{y}{\left[\left(-5\right)\times 1-\left(-1\right)\times 1\right]}=\frac{1}{\left[1\times \left(-1\right)-1\times 1\right]}$

⇒ $\frac{x}{\left(-1-5\right)}=\frac{y}{\left(-5+1\right)}=\frac{1}{\left(-1-1\right)}$
⇒ $\frac{x}{-6}=\frac{y}{-4}=\frac{1}{-2}$
⇒ $x=\frac{-6}{-2}=3,y=\frac{-4}{-2}=2$
Hence, x = 3 and y = 2 is the required solution.