Given,
$2a+3b=4$……. (i)
$a–b+3=0$……. (ii)
From equation (i),
⇒ $b=(4–2a)/3$
When $a=-1$, we get $b=(4–2(-1))/3=2$
When $a=2$, we get $b=(4–2(2))/3=0$
Thus, we have the following table giving Points on the line $2a+3b=4$
| a | $-1$ | $2$ |
| b | $2$ | $0$ |
From equation (ii),
Solve for b:
⇒ $b=(a+3)$
So, when $a=0$
$b=(0+3)=3$
And, when $a=1$
⇒ $b=(1+3)=4$
Thus, we have the following table giving Points on the line $a–b+3=0$
| a | $0$ | $1$ |
| b | $3$ | $4$ |
Graph of the equations (i) and (ii) is given below:
From given graph, the two lines intersect at a single Point P $(-1,2)$
Thus, $a=-1$ and $b=2$.