Question

Show that each of the following systems of equations has a unique solution and solve it:

$\frac{x}{3}+\frac{y}{2}=3,x-2y=2.$

Answer 1

$\frac{x}{3}+\frac{y}{2}=3$

⇒$\frac{(2x+3y)}{6}=3$

$2x+3y–18=0$.....(1)

$x–2y–2=0$ ..........(2)

$a_1=2,b_1=3,c_1=-18$

$a_2=1,b_2=-2,c_2=-2$

Thus, $\frac{a_1}{a_2}\ne \frac{b_1}{b_2}$

⇒$\frac{2}{1}\ne \frac{3}{-2}$

Hence, the given system of equations has a unique solution.

The given equations are,

$2x+3y=18$ ...... (1)

$3x–6y=2$ ......... (2)

Multiplying (1) by 2, and (2) by 3, we get

$4x+6y=36$........ (3)

$3x–6y=6$.......... (4)

Adding (3) and (4), we get

$7x=42$

$x=6$

Putting $x=6$ in (1), we get

$\left(2\right)\left(6\right)+3y=18$

$3y=18–12$

$3y=6$

⇒ $y=\frac{6}{3}=2$

Hence, the solution is $x=6,y=2$