Solution:
Given,
$a–2b=5$……. (i)
$2a+3b=10$……. (ii)
From equation (i) we get,
⇒ $b=(a–5)/2$
When we Put $b=0$, we get, $a=5$
When we Put $a=1$, we get, $b=-2$
Thus, we have the following table giving Points on the line $a–2b=5$
| a | $5$ | $1$ |
| b | $0$ | $-2$ |
From equation (ii),
Solve for b, we get:
⇒ $b=(10–2a)/3$
So, when we Put $a=5$
We get, $b=(10–2(5))/3=0$
And, when we Put $a=2$
⇒ $b=(10–2(2))/3=2$
Thus, we have the following table giving Points on the line $2a+3b=10$
| a | $5$ | $2$ |
| b | $0$ | $2$ |
Graph of the equations (i) and (ii) is given below:
From given graph, we can see that the two lines intersect at a single Point $P(5,0)$.
Hence, $a=5$ and $b=0$.