(i) Let the number of girls be \(x\) and the number of boys be \(y.\)
According to the question, the algebraic representation is \(x + y = 10 x − y = 4\) For \(x + y = 10, x = 10 − y;\)
| \(x\) | \(5\) | \(4\) | \(6\) |
| \(y\) | \(5\) | \(6\) | \(4\) |
For \(x − y = 4, x = 4 + y\)
| \(x\) | \(5\) | \(4\) | \(3\) |
| \(y\) | \(1\) | \(0\) | \(-1\) |
Hence, the graphic representation is as follows.

From the figure, it can be observed that these lines intersect each other at points \((7, 3)\).
Therefore, the number of girls and boys in the class is \(7\) and \(3\) respectively.
(ii) Let the cost of \(1\) pencil be Rs \(x\) and the cost of \(1\) pen be Rs \(y\). According to the question, the algebraic representation is
\(5x + 7y = 50\)
\(7x + 5y = 46\)
For \(5x + 7y = 50,\)
\(x= 50-\dfrac{7y}{5}\)
| \(x\) | \(3\) | \(10\) | \(-4\) |
| \(y\) | \(5\) | \(0\) | \(10\) |
\(7x + 5y = 46\)
\(x= 46-\dfrac{5y}{7}\)
| \(x\) | \(8\) | \(3\) | \(-2\) |
| \(y\) | \(-2\) | \(5\) | \(12\) |
Hence, the graphic representation is as follows.

From the figure, it can be observed that these lines intersect each other at point \((3, 5).\)
Therefore, the cost of a pencil and a pen are Rs \(3\) and Rs \(5\) respectively.
Graphical representation needs the plotting of the x and y in graph paper. Plotting the x and y values of the equation on the coordinate plane. For plotting the graph you will be needed at least 3 sets of points. It is very much important for those points to fall in a straight line. If the points are haphazardly placed, it will indicate some fault in your work.

On the graph paper, draw the x and y-axis. Simply, the x and y-axis are two number lines that are placed perpendicular to each other at the (0,0) point. This point is known to be the origin. Use the aforesaid values for plotting the points. Join all of these points with a straight line.
