Question

$$\begin{gathered} 6x-5y-16=0, \\ 7x-13y+10=0. \end{gathered}$$

Answer 1

The given equations are:
6x − 5y − 16 = 0 ...(i)
7x − 13y + 10 = 0 ...(ii)
Here, a₁ = 6, b₁ = −5 , c₁ = −16, a₂ = 7, b₂ = −13 and c₂ = 10
By cross multiplication, we have:

∴ $\frac{x}{\left[\left(-5\right)\times 10-\left(-16\right)\times \left(-13\right)\right]}=\frac{y}{\left[\left(-16\right)\times 7-10\times 6\right]}=\frac{1}{\left[6\times \left(-13\right)-\left(-5\right)\times 7\right]}$
⇒ $\frac{x}{\left(-50-208\right)}=\frac{y}{\left(-112-60\right)}=\frac{z}{\left(-78+35\right)}$
⇒ $\frac{x}{\left(-258\right)}=\frac{y}{\left(-172\right)}=\frac{1}{\left(-43\right)}$
⇒ $x=\frac{-258}{-43}=6,y=\frac{-172}{-43}=4$
Hence, x = 6 and y = 4 is the required solution.