Question

Solve for x and y: $\frac{35}{x+y}+\frac{14}{x-y}=19,\frac{14}{x+y}+\frac{35}{x-y}=37$

Answer 1

We have:
$\frac{35}{x+y}+\frac{14}{x-y}=19\mathrm{and}\frac{14}{x+y}+\frac{35}{x-y}=37$
Taking $\frac{1}{x+y}=u$ and $\frac{1}{x-y}=v$:
35u + 14v − 19 = 0 ....(i)
14u + 35v − 37 = 0 ....(ii)
Here, a₁ = 35, b₁ = 14, c₁ = −19, a₂ = 14, b₂ = 35, c₂ = −37
By cross multiplication, we have:


∴ $\frac{u}{\left[14\times \left(-37\right)-35\times \left(-19\right)\right]}=\frac{v}{\left[\left(-19\right)\times 14-\left(-37\right)\times \left(35\right)\right]}=\frac{1}{\left[35\times 35-14\times 14\right]}$
⇒ $\frac{u}{-518+665}=\frac{v}{-266+1295}=\frac{1}{1225-196}$
⇒ $\frac{u}{147}=\frac{v}{1029}=\frac{1}{1029}$
⇒ $u=\frac{147}{1029}=\frac{1}{7},v=\frac{1029}{1029}=1$
⇒ $\frac{1}{x+y}=\frac{1}{7},\frac{1}{x-y}=1$
∴ (x + y) = 7 ....(iii)
And, (x − y) = 1 ....(iv)

Again, the equations (iii) and (iv) can be written as follows:
x + y − 7 = 0 ....(v)
x − y − 1 = 0 ....(vi)
Here, a₁ = 1, b₁ = 1, c₁ = −7 , a₂ = 1 , b₂ = −1 , c₂ = −1
By cross multiplication, we have:

⇒ $\frac{x}{\left[1\times \left(-1\right)-\left(-1\right)\times \left(-7\right)\right]}=\frac{y}{\left[\left(-7\right)\times 1-\left(-1\right)\times 1\right]}=\frac{1}{\left[1\times \left(-1\right)-1\times 1\right]}$

⇒ $\frac{x}{-1-7}=\frac{y}{-7+1}=\frac{1}{-1-1}$
⇒ $\frac{x}{-8}=\frac{y}{-6}=\frac{1}{-2}$
⇒ $x=\frac{-8}{-2}=4,y=\frac{-6}{-2}=3$
Hence, x = 4 and y = 3 is the required solution.