Question

Show that the following system of equations has a unique solution:
$\frac{x}{3}+\frac{y}{2}=3,x-2y=2.$
Also, find the solution of the given system of equations.

Answer 1

The given system of equations are:
$\frac{x}{3}+\frac{y}{2}=3$
⇒ $\frac{2x+3y}{6}=3$
2x + 3y = 18
⇒ 2x + 3y − 18 = 0 ...(i)
and
x − 2y = 2
x − 2y − 2 = 0 ...(ii)
These equations are of the forms:
a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0
where, a₁ = 2, b₁= 3, c₁ = −18 and a₂ = 1, b₂ = −2, c₂ = −2
For a unique solution, we must have:
$\frac{a_1}{a_2}\ne \frac{b_1}{b_2}$, i.e., $\frac{2}{1}\ne \frac{3}{-2}$
Hence, the given system of equations has a unique solution.

Again, the given equations are:
2x + 3y − 18 = 0 ...(iii)
x − 2y − 2 = 0 ...(iv)
On multiplying (i) by 2 and (ii) by 3, we get:
4x + 6y − 36 = 0 ...(v)
3x − 6y − 6 = 0 ...(vi)
On adding (v) and (vi), we get:
7x = 42
⇒ x = 6
On substituting x = 6 in (iii), we get:
2(6) + 3y = 18
⇒ 3y = (18 − 12) = 6
⇒ y = 2
Hence, x = 6 and y = 2 is the required solution.