Question

$$\begin{gathered} 7x-2y=3, \\ 11x-\frac{3}{2}y=8. \end{gathered}$$

Answer 1

The given equations may be written as:
7x − 2y − 3 = 0 ...(i)
$11x-\frac{3}{2}y-8=0$ ...(ii)
Here, a₁ = 7, b₁ = −2 , c₁ = −3, a₂ = 11, b₂ = $-\frac{3}{2}$ and c₂ = −8
By cross multiplication, we have:

∴ $\frac{x}{\left[\left(-2\right)\times \left(-8\right)-\left(-{\displaystyle \frac{3}{2}}\right)\times \left(-3\right)\right]}=\frac{y}{\left[\left(-3\right)\times 11-\left(-8\right)\times 7\right]}=\frac{1}{\left[7\times \left({\displaystyle \frac{-3}{2}}\right)-11\times \left(-2\right)\right]}$
⇒ $\frac{x}{\left(16-{\displaystyle \frac{9}{2}}\right)}=\frac{y}{\left(-33+56\right)}=\frac{1}{\left(-{\displaystyle \frac{21}{2}}+22\right)}$
⇒ $\frac{x}{\left({\displaystyle \frac{23}{2}}\right)}=\frac{y}{23}=\frac{1}{\left({\displaystyle \frac{23}{2}}\right)}$
⇒ $x=\frac{{\displaystyle \frac{23}{2}}}{{\displaystyle \frac{23}{2}}}=1,y=\frac{23}{{\displaystyle \frac{23}{2}}}=2$
Hence, x = 1 and y = 2 is the required solution.