Question

The cost of 7 erasers and 5 pencils is Rs.58 and the cost of 5 erasers and 6 pencils is Rs.56. Formulate this problem algebraically and solve it graphically.

Answer 1

Hint: Suppose the cost of one eraser is x and the cost of one pencil is y.
Given that, the cost of 7 erasers and 5 pencils is Rs.58 and the cost of 5 erasers and 6 pencils is Rs.56. Formulating these two statements algebraically, we get two equations. Next, we plot them graphically and look for the intersection point of the two straight lines. The x coordinate of point will be the cost of one eraser and y coordinate of point will be the cost of one pencil.

Complete step-by-step answer:
Let the cost of one eraser be $x$ and the cost of one pencil be$y$
Given that, the cost of 7 erasers and 5 pencils is Rs.58
  \[ \Rightarrow 7x + 5y = 58{\text{ }}...(i)\]
Again, cost of 5 erasers and 6 pencils is Rs.56
 \[ \Rightarrow 5x + 6y = 56{\text{ }}...(ii)\]

To solve (i) and (ii) graphically, we will take one equation at a time and plot the solutions for x and y to find its corresponding straight line.

For equation (i),
 \[7x + 5y = 58\]
We know that to plot a equation of line we need at least two points
So let us find for y=6 and y=-1
On solving for x we get,
   $
   \Rightarrow 7x = 58 - 5y \\
   \Rightarrow x = \dfrac{{58 - 5y}}{7} \\
 $
On Taking, y=6 we get,
 $x = \dfrac{{58 - 5(6)}}{7}$
On simplification we get,
$ \Rightarrow x = \dfrac{{58 - 30}}{7}$
$ \Rightarrow x = \dfrac{{28}}{7}$
On further simplification we get,
$ \Rightarrow x = 4$

Now, on taking \[y = ( - 1)\] we get,
$x = \dfrac{{58 - 5( - 1)}}{7}$
On simplification we get,
$ \Rightarrow x = \dfrac{{58 + 5}}{7}$
$ \Rightarrow x = \dfrac{{63}}{7}$
On further simplification we get,
$ \Rightarrow x = 9$

Therefore the solutions of equation (i) are,

x	4	9
y	6	−1

Again, for equation (ii),
\[5x + 6y = 56\]
We know that to plot a equation of line we need at least two points
So let us find for y=1 and y=6

On solving for x we get,
 $
   \Rightarrow 5x = 56 - 6y \\
   \Rightarrow x = \dfrac{{56 - 6y}}{5} \\
 $
On Taking y=1 we get,
 $x = \dfrac{{56 - 6(1)}}{5}$
On simplification we get,
 $ \Rightarrow x = \dfrac{{56 - 6}}{5}$
 $ \Rightarrow x = \dfrac{{50}}{5}$
On further simplification we get,
 $ \Rightarrow x = 10$

Now, on taking y=6 we get,
 $x = \dfrac{{56 - 6(6)}}{5}$
On simplification we get,
 $ \Rightarrow x = \dfrac{{56 - 36}}{5}$
 $ \Rightarrow x = \dfrac{{20}}{5}$
On further simplification we get,
 $ \Rightarrow x = 4$

Therefore, the solutions of equation (ii) are,

x	10	4
y	1	6

Next we plot the graphs of the equations to find the intersection point of the straight lines.

We can see that the straight lines intersect at the point \[\;\left( {4,{\text{ }}6} \right)\]

Therefore the solution of the two equations is
\[x = 4,y = 6\]

So cost of one eraser is Rs.4 and that of a pencil is Rs.6

Note: To solve a problem graphically, one needs to formulate the problem into equations first. Next, we plot the graphs and look for the intersection point. The solution of these equations is given by the coordinates of their point of intersection.
Here, we were asked to solve using a graphical method, so while making the graph most of the students could not find the nature of the equation, but we can see that both the equations are linear in x and y hence they will form straight lines.