(N/A) The pair of equations formed is:
$y = \frac{1}{2}x$
i.e.,$x - 2y = 0$ ......$(1)$
$3x + 4y = 20$ ......$(2)$
Let us represent these equations graphically. For this,we need at least two solutions for each equation. We give these solutions in the tables below:
| $x$ | $0$ | $2$ |
| $y = \frac{x}{2}$ | $0$ | $1$ |
| $x$ | $0$ | $\frac{20}{3}$ | $4$ |
| $y = \frac{20 - 3x}{4}$ | $5$ | $0$ | $2$ |
Recall from Class $IX$ that there are infinitely many solutions for each linear equation. So,each of you can choose any two values,which may not be the ones we have chosen. When one of the variables is zero,the equation reduces to a linear equation in one variable,which can be solved easily. For instance,putting $x = 0$ in Equation $(2)$,we get $4y = 20$,i.e.,$y = 5$. Similarly,putting $y = 0$ in Equation $(2)$,we get $3x = 20$,i.e.,$x = \frac{20}{3}$. But as $\frac{20}{3}$ is not an integer,it will not be easy to plot exactly on the graph paper. So,we choose $y = 2$,which gives $x = 4$,an integral value.
Plot the points $A(0, 0), B(2, 1)$ and $P(0, 5), Q(4, 2)$ corresponding to the solutions in the tables. Now draw the lines $AB$ and $PQ$,representing the equations $x - 2y = 0$ and $3x + 4y = 20$.