SELINA Solution Class 9 Chapter 27 Graphical Solution (Solution of Simultaneous Linear Equations, Graphically) Exercise 27 B

Question 1.1

Solve, graphically, the following pairs of equation :
x - 5 = 0 
y + 4 = 0

Sol:

x - 5 = 0 ⇒ x = 5
y + 4 = 0 ⇒ y = - 4

Following is the graph of the two equations x = 5 and y = - 4:

Question 1.2

Solve, graphically, the following pairs of equation :
2x + y = 23
4x - y = 19

Sol:

2x + y = 23 ⇒ y = 23 - 2x
The table for y = 23 - 2x is

X	5	10	15
Y	13	3	- 7

Also, we have
4x - y = 19 ⇒ y = 4x - 19
The table for  y = 4x - 19 is

X	3	4	6
Y	- 7	- 3	5

Plotting the points we get the following required graph:

From the above graph, it is clear that the two lines y = 23 - 2x and y = 4x - 19 intersect at the point (7, 9)

Question 1.3 

Solve, graphically, the following pairs of equation :
3x + 7y = 27
8 - y = ${\displaystyle \frac{5}{2}x}$

Sol :

Solve, graphically, the following pairs of equation :
3x + 7y = 27
8 - y = ${\displaystyle \frac{5}{2}x}$

uestion 1.4

Solve, graphically, the following pairs of equations :
${\displaystyle \frac{x+1}{4}=\frac{2}{3}(1-2y)}$

Sol:

${\displaystyle \frac{x+1}{4}=\frac{2}{3}(1-2y)}$

⇒ ${\displaystyle \frac{x+1}{4}=\frac{2}{3}-\frac{4y}{3}}$

⇒ 12 x ${\displaystyle \frac{x+1}{4}=12\times \frac{2}{3}-12\times \frac{4y}{3}}$

⇒ 3(x + 1) = 8 - 16y
⇒ 3x + 3 = 8 - 16y
⇒ 3x + 3 - 8 = -16y
⇒ 3x - 5 = -16y

⇒ x = ${\displaystyle \frac{5-16y}{3}}$
The table for ${\displaystyle \frac{x+1}{4}=\frac{2}{3}(1-2y)}$ is

X	7	- 9	23
Y	- 1	2	- 4

Also, we have
${\displaystyle \frac{2+5y}{3}=\frac{x}{7}-2}$
⇒ 21 x ${\displaystyle \frac{2+5y}{3}=21\times \frac{x}{7}-21\times 2}$
⇒ 7(2 + 5y) = 3x - 42
⇒ 14 + 35y = 3x - 42
⇒ 3x = 14 + 35y + 42
⇒ 3x = 56 + 35y
⇒ x = ${\displaystyle \frac{56+35y}{3}}$
The table for ${\displaystyle \frac{2+5y}{3}=\frac{x}{7}-2}$ is

X	7	- 28	42
Y	- 1	- 4	2

Plotting the points we get the following required graph:

From the above graph, it is dear that the two lines ${\displaystyle \frac{x+1}{4}=\frac{2}{3}(1-2y)}$ and ${\displaystyle \frac{2+5y}{3}=\frac{x}{7}-2}$ intersect at the point (7, -1)

Question 2

Solve graphically the simultaneous equations given below. Take the scale as 2 cm = 1 unit on both the axes.
x - 2y - 4 = 0
2x + y = 3

Sol:

x - 2y - 4 = 0
⇒ x = 2y + 4
The table for x - 2y - 4 = 0 is

X	4	6	2
Y	0	1	- 1

Also we have
2x + y = 3
⇒ 2x = 3 - y
⇒ x = ${\displaystyle \frac{3-y}{2}}$
The table for 2x + y = 3 is

X	1	0	2
Y	1	3	- 1

Plotting the above points we get the following required graph:

From the above graph, it is dear that the two lines x - 2y - 4 = 0 and 2x + y = 3 intersect at the point (2, -1)

Question 3

Use graph paper for this question. Draw the graph of 2x - y - 1 = 0 and 2x + y = 9 on the same axes. Use 2 cm = 1 unit on both axes and plot only 3 points per line. Write down the coordinates of the point of intersection of the two lines.

Sol:

2x - y - 1 = 0
⇒ 2x = y + 1
⇒ x = ${\displaystyle \frac{y+1}{2}}$
This table for 2x - y - 1 = 0 is

X	2	1	0
Y	3	1	- 1

Also we have
2x + y = 9
⇒ 2x = 9 - y
⇒ x = ${\displaystyle \frac{9-y}{2}}$
The table for 2x + y = 9 is

X	4	3	5
Y	1	3	- 1

Plotting the above points we get the following required graph:

From the above graph, it is clear that the two lines 2x - y - 1 = 0 and 2x + y = 9 intersect at the point (2, 5, 4)

Question 4

Use graph paper for this question. Take 2 cm = 2 units on x-axis and 2 cm = 1 unit on y-axis.
Solve graphically the following equation:
3x + 5y = 12; 3x - 5y + 18 = 0 (Plot only three points per line)

Sol:

3x + 5y = 12
⇒ 3x = 12 - 5y
x = ${\displaystyle \frac{12-5y}{3}}$

The table for 3x + 5y = 12 is

X	4	- 1	- 6
Y	0	3	6

Also we have
3x - 5y + 18 = 0
⇒ 3x = 5y - 18
⇒ x = ${\displaystyle \frac{5y-18}{3}}$ 
The table for 3x - 5y + 18 = 0 is

X	- 6	4	- 1
Y	0	6	3

Plotting the above points we get the following required graph:

From the above graph, it is clear that the two lines 3x + 5y = 12 and 3x - 5y + 18 = 0 intersect at the point (-1, 3)

Question 5

Use graph paper for this question. Take 2 cm = 1 unit on both the axes.
(i) Draw the graphs of x + y + 3 = 0 and 3x - 2y + 4 = 0. Plot only three points per line.
(ii) Write down the coordinates of the point of intersection of the lines.
(iii) Measure and record the distance of the point of intersection of the lines from the origin in cm.

Sol:

(i)
x + y + 3 = 0
⇒ x = - 3 - y
The table for x + y = 3 = 0 is

X	1	0	- 2
Y	- 4	- 3	- 1

Also we have
3x - 2y + 4 = 0
⇒ 3x = 2y - 4
⇒ x = ${\displaystyle \frac{2y-4}{3}}$
The table for 3x - 2y + 4 = 0 is

X	0	- 2	${\displaystyle -\frac{2}{3}}$
Y	2	- 1	1

Plotting the above points we get the following required graph :

(ii) From the above graph, it is clear that the two lines x + y + 3 = 0 and 3x - 2y + 4 = 0 intersect at the point (-2, -1)

(iii) Applying Pythagoras Theorem,
the distance from the origin

= ${\displaystyle \sqrt{{(-2-0)}^2+{(-1-0)}^2}}$

= ${\displaystyle \sqrt{2^2+1^2}}$

= ${\displaystyle \sqrt{4+1}}$

= ${\displaystyle \sqrt{5}}$

= 2.2 cm (approx).

Question 6

The sides of a triangle are given by the equations y - 2 = 0; y + 1 = 3 (x - 2) and x + 2y = 0.
Find, graphically : 
(i) the area of a triangle;
(ii) the coordinates of the vertices of the triangle.

Sol:

y - 2 = 0
⇒ y = 2
y + 1 = 3(x - 2)
⇒ y + 1 = 3x - 6
⇒ y = 3x - 6 - 1
⇒ y = 3x - 7
The table for y + 1 = 3(x - 2) is 

X	1	2	3
Y	- 4	- 1	2

Also we have
x + 2y = 0
⇒ x = - 2y
The table for x + y = 0 is

X	- 4	4	- 6
Y	2	- 2	3

Plotting the above points we get the folllowing required graph:

(i) The area of the triangle ABC = ${\displaystyle \frac{1}{2}\times \text{AB}\times \text{CD}}$

= ${\displaystyle \frac{1}{2}\times 7\times 3}$

=  ${\displaystyle \frac{21}{2}}$

= 10.5 sq.units

(ii) The coordinates of the vertices of the triangle are ( - 4, 2), (3, 2) and (2, -1).

Question 7

By drawing a graph for each of the equations 3x + y + 5 = 0; 3y - x = 5 and 2x + 5y = 1 on the same graph paper; show that the lines given by these equations are concurrent (i.e. they pass through the same point). Take 2 cm = 1 unit on both the axes.

Sol:

3x + y + 5 = 0
⇒ y = - 3x  - 5
The table of 3x + y + 5 = 0 is

X	1	- 3	- 2
Y	- 8	4	1

3y - x = 5
⇒ x = 3y  - 5
The table of 3y - x = 5 is

X	- 2	1	7
Y	1	2	4

2x + 5y = 1
⇒ 2x = 1 - 5y
⇒ x = ${\displaystyle \frac{1-5y}{2}}$
The table of 2x + 5y = 1 is

X	3	- 7	- 2
Y	- 1	3	1

Plotting the above points, we get the following required graph:

The graph shows that the lines of these equations are concurrent.

Question 8

Using a scale of 1 cm to 1 unit for both the axes, draw the graphs of the following equations: 6y = 5x + 10, y = 5x - 15.
From the graph find :
(i) the coordinates of the point where the two lines intersect;
(ii) the area of the triangle between the lines and the x-axis.

SOl:

6y = 5x + 10
⇒ y = ${\displaystyle \frac{5x+10}{6}}$
The table of 6y = 5x + 10 is

X	4	- 2	- 8
Y	5	0	- 5

Also, we have
y = 5x - 15
The table of y = 5x - 15 is

X	3	4	5
Y	0	5	10

Plotting the points in a graph, we get the following graph.

(i)
The two lines intersect at (4, 5)
∴ AD ⊥ BC
AD = 5 units and BC = 5 units

(ii)

The area of the triangles = ${\displaystyle \frac{1}{2}\times \text{BC}\times \text{AD}}$

= ${\displaystyle \frac{1}{2}\times 5\times 5}$

= ${\displaystyle \frac{25}{2}\text{sq.units}}$

= 12.5 sq.units

Question 9.1

The cost of manufacturing x articles is Rs. (50 + 3x). The selling price of x articles is Rs. 4x.

On a graph sheet, with the same axes, and taking suitable scales draw two graphs, first for the cost of manufacturing against no. of articles and the second for the selling price against the number of articles.

Use your graph to determine:
No. of articles to be manufactured and sold to break even (no profit and no loss).

Sol:

Given that C.P. is 50 + 3x
Table of C.P.

X	0	10	20	30	40	50	60
C.P.	50	80	110	140	170	200	230

and S.P. = 4x
∴ Table of S.P.

X	0	10	20	30	40	50	60
S.P.	0	40	80	120	160	200	240

Now plot the points on a graph and we get the following required graph:

No. of articles to be manufactured and sold are 50 when there is no loss and no profit.
C.P. = S.P = Rs. 200.

Question 9.2

Given that C.P. is 50 + 3x
Table of C.P.

X	0	10	20	30	40	50	60
C.P.	50	80	110	140	170	200	230

and S.P. = 4x
∴ Table of S.P.

X	0	10	20	30	40	50	60
S.P.	0	40	80	120	160	200	240

Now plot the points on a graph and we get the following required graph:

No. of articles to be manufactured and sold are 50 when there is no loss and no profit.
C.P. = S.P = Rs. 200.

Sol:

Given that C.P. is 50 + 3x
Table of C.P.

X	0	10	20	30	40	50	60
C.P.	50	80	110	140	170	200	230

and S.P. = 4x
∴ Table of S.P.

X	0	10	20	30	40	50	60
S.P.	0	40	80	120	160	200	240

Now plot the points on a graph and we get the following required graph:

(a)
On article 30,
C.P. = Rs.140 and S.P. = 120
Therefore Loss = 140 - 120 = Rs. 20

(b)
On article 60,
C.P.= Rs. 230 and S.P.= Rs. 240
Therefore Profit = 240 - 230 = Rs.10

Question 10

Find graphically, the vertices of the triangle whose sides have the equations 2y - x = 8; 5y - x = 14 and y - 2x = 1 respectively. Take 1 cm = 1 unit on both the axes.

Sol:

2y - x = 8;
y = ${\displaystyle \frac{8+x}{2};}$
The table of 2y - x = 8 is

X	- 6	- 2	0
Y	1	3	4

5y - x = 14
⇒ x = 5y - 14
The table of x = 5y - 14 is

X	- 9	- 4	1
Y	1	2	3

y - 2x = 1
⇒ y = 1 + 2x
The table of y - 2x = 1 is

X	2	- 2	0
Y	5	- 3	1

Now plotting the points on a graph and we get the following required graph:

Thus, the verticles of the triangle ΔABC are: A(- 4, 2), B(1, 3) and C(2, 5).

Question 11

Using the same axes of co-ordinates and the same unit, solve graphically :
x + y = 0 and 3x - 2y = 10.
(Take at least 3 points for each line drawn).

Sol:

x + y = 0
y = - x;
The table of x + y = 0 is

X	5	2	- 5
Y	- 5	- 2	5

3x - 2y = 10
⇒ x = ${\displaystyle \frac{10+2y}{3}}$
The table of 3x - 2y = 10 is

X	4	6	2
Y	1	4	- 2

Now plotting the points on a graph and we get the following required graph:

The two lines intersect at (2, - 2)
∴ x = 2 and y = - 2

Question 12

Solve graphically, the following equations.
x + 2y = 4; 3x - 2y = 4.
Take 2 cm = 1 unit on each axis.
Also, find the area of the triangle formed by the lines and the x-axis.

Sol:

x + 2y = 4
⇒ x = 4 - 2y
The table of x + 2y = 4 is

X	2	- 4	12
Y	1	4	- 4

3x - 2y = 4
⇒ x = ${\displaystyle \frac{4+2y}{3}}$
The table of 3x - 2y = 4 is

X	2	4	6
Y	1	4	7

Now plotting the points on a graph and we get the following required graph:

Therefore the solution of the given system of equations is (2,1).
Thus the vertices of the triangle are:
A(2,1), B${\displaystyle (\frac{4}{3},0)}$and C(4,0)
AB ⊥ BC and D ≡ (2,0)
AD = 1 and BC = ${\displaystyle 2\frac{2}{3}\text{units}=\frac{8}{3}\text{units}}$
Area of the triangle ABC = ${\displaystyle \frac{1}{2}\times \text{AD}\times \text{BC}}$

= ${\displaystyle \frac{1}{2}\times 1\times \frac{8}{3}}$

= ${\displaystyle \frac{4}{3}\text{sq.units}}$

= ${\displaystyle 1\frac{1}{3}\text{sq.units}}$

Question 13

Use the graphical method to find the value of 'x' for which the expressions ${\displaystyle \frac{3x+2}{2}\text{and}\frac{3}{4}x-2}$

Sol:

y = ${\displaystyle \frac{3x+2}{2}}$
The table for y = ${\displaystyle \frac{3x+2}{2}}$ is

X	2	4	-2
Y	4	7	-2

y = ${\displaystyle \frac{3}{4}x-2}$
The table for y = ${\displaystyle \frac{3}{4}x-2}$ is

X	4	-4	8
Y	1	-5	4

Now plotting the points on a graph and we get the following required graph:

Thus, the value of 'x' is -4.

Question 14

The course of an enemy submarine, as plotted on rectangular co-ordinate axes, gives the equation 2x + 3y = 4. On the same axes, a destroyer's course is indicated by the graph x - y = 7. Use the graphical method to find the point at which the paths of the submarine and the destroyer intersect?

Sol:

2x + 3y = 4
⇒ x = ${\displaystyle \frac{4-3y}{2}}$
The table for 2x + 3y = 4 is

X	-1	-4	5
Y	2	4	-2

x - y = 7
⇒ x = y + 7
The table for x - y = 7 is

X	5	11	9
Y	-2	4	2

Now plot the points on a graph and we get the following required graph:

The point at which the paths of the submarine and the destroyer intersect are (5, -2)