Question

$$\begin{gathered} x+2y+1=0, \\ 2x-3y-12=0. \end{gathered}$$

Answer 1

The given equations are:
x + 2y + 1 = 0 ...(i)
2x − 3y − 12 = 0 ...(ii)
Here, a₁ = 1, b₁ = 2, c₁ = 1, a₂ = 2, b₂ = −3 and c₂ = −12
By cross multiplication, we have:

∴ $\frac{x}{\left[2\times \left(-12\right)-1\times \left(-3\right)\right]}=\frac{y}{\left[1\times 2-1\times \left(-12\right)\right]}=\frac{1}{\left[1\times \left(-3\right)-2\times 2\right]}$
⇒ $\frac{x}{\left(-24+3\right)}=\frac{y}{\left(2+12\right)}=\frac{1}{\left(-3-4\right)}$
⇒ $\frac{x}{\left(-21\right)}=\frac{y}{\left(14\right)}=\frac{1}{\left(-7\right)}$
⇒ $x=\frac{-21}{-7}=3,y=\frac{14}{-7}=-2$
Hence, x = 3 and y = −2 is the required solution.