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