# ICSE Solutions for Selina Concise Chapter 27 Graphical Solution of Simultaneous Linear Equations Class 9 Maths

### Exercise 27(A)

1. Draw the graph for the equation, given below :

(i) x = 5
(ii) x +5 =0
(iii) y = 7
(iv) y+7=0
(v) 2x + 3y =0
(vi) 3x + 2y = 6
(vii) x-5y+4=0
(viii) 5x + y +5=0

(i) The graph x= 5 in the following figure is a straight line AB which is parallel to y axis at a distance of 5 units from it.

(ii)
x+5=0
x = -5
The graph x= -5 in the following figure is a straight line AB which is parallel to y axis at a distance of 5 units from it in the negative x direction.

(iii) The graph y = 7 in the following figure is a straight line AB which is parallel to x axis at a distance of 7 units from it.

(iv) y + 7 = 0
y = -7
The graph y = -7 in the following figure is a straight line AB which is parallel to x axis at a distance of 7 units from it in the negative y direction.

(v) 2x + 3y = 0
⇒ 3y = -2x

Plotting these points we get the required graph as shown below :

(vi) 3x + 2y = 6
⇒ 2y = 6- 3x

Plotting these points we get the required graph as shown below :

(vii) x - 5y + 4 = 0
⇒ 5y = 4 + x

Plotting these points we get the required graph as shown below :
(viii) 5x +y + 5= 0
⇒ y = -5x - 5
When x = 0; y = -5× x - 5 = -0 - 5 = - 5
When x = -1 ; y = -5×(-1) - 5 = 5 - 5 = 0
When x = -2 ; y = -5×(-2) - 5 = 10 - 5 = 5

Plotting these points we get the required graph as shown below :

2. Draw the graph for the equation given below; hence find the co-ordinates of the points where the graph is drawn meets the co-ordinates axes:

(i) x/3 + y/5 =1
(ii) (2x + 15) /3 = y – 1

Plotting these points we get the required graph as shown below :

From the figure it is clear that, the graph meets the coordinate axes at (3, 0) and (0, 5).
(ii) (2x + 15)/3 = y - 1
⇒ 2x + 15 = 3(y-1)
⇒ 2x + 15 = 3y - 3
⇒ 2x -3y = -15 - 3
⇒ 2x - 3y = -18
⇒ -3y = -18 - 2x

Plotting these points we get the required graph as shown below :

From the figure it is clear that, the graph meets the coordinate axes at (-9, 0) and (0, 6)

3. Draw the graph of the straight line given by the equation 4x – 3y + 36 = 0

Calculate the area of the triangle formed by the line drawn and the co-ordinate axes.

4x - 3y + 36 = 0
⇒ 4x -3y = -36
⇒ -3y = -36 - 4x
⇒ 3y = 36 + 4x

Plotting these points we get the required graph as shown below :
The straight line cuts the co - ordinate axis at A(0, 12) and B(-9, 0).
∴ The triangle △AOB is formed.
Area of the triangle  AOB = (1/2)× AO × OB
= (1/2)× 12 × 9
= 54 sq. units
∴ Area of the triangle is  54 sq. units .

4.Draw the graph of the equation 2x – 3y – 5 = 0
From the graph, find:
(i) x1, the value of x, when y = 7
(ii) x2, the value of x, when y = – 5.

2x - 3y - 5 = 0
⇒ 2x = 3y + 5
Plotting these points we get the required graph as shown below :
The value of x, when y = 7:
We have the equation of the line as

5.Draw the graph of the equation
4x + 3y + 6 = 0
From the graph, find :
(i) y1, the value of y, when x = 12.
(ii) y2, the value of y, when x = – 6.

4x + 3y + 6= 0
⇒ 3y = -4x - 6

Plotting these points we get the required graph as shown below :
The value of y, when x = 12 :
We have the equation of the line as

6. Use the table given below to draw the graph.
 X –5 –1 3 b 13 Y –2 a 2 5 7

From your graph, find the values of ‘a’ and ‘b’.
State a linear relationship between the variables x and y.

The table is  :

Plotting the points as shown in the above table,
we get the following required graph :

When x = -1 , then  y = 0
⇒ a = 0
When y = 5, then x = 9
⇒ b = 9
Let y = px + q  ...(1)
be a linear relation between x and y
Substitute x= 9 and y = 5 in the equation (1), we have ,
5 = 9p + q  ...(2)
Substitute  x = -1 and y =0 in the equation (1), we have,
0 = - p + q  ...(3)
Subtracting (3) from  (2), we have,
5 = 10 p

7. Draw the graph obtained from the table below:
 X a 3 – 5 5 c – 1 Y – 1 2 b 3 4 0

Use the graph to find the values of a, b and c. State a linear relation between the variables x and y.

The table is :

Plotting the points as shown in the above table,
we get the following required graph :

When y = -1, then x = -3
⇒ a = -3
When x = -5, then y = -2
⇒ b = - 2
When  y = -4, then x = 7
⇒ c = 7
Let y = px + q   ... (1)
be a linear relation between x  and y
Substitute x = -3 and y = -1 in the equation (1), we have,
-1 = -3p + q ...(2)
Substitute x = -5 and  y = -2 in the equation (1), we have,
-2 = -5p + q ...(3)
Subtracting (3) from (2), we have,
1 = 2p
⇒ p = 1/2
From (3), we have,
-2 = 5p + q

8. A straight line passes through the points (2, 4) and (5, –2). Taking 1 cm = 1 unit; mark these points on a graph paper and draw the straight line through these points. If points (m, –4) and (3, n) lie on the line drawn; find the values of m and n.

The table is:

Plotting the points as shown in the above table,
we get the following required graph :
Plotting the points in the graph we get the above required graph.
Now draw a line x = 3, parallel to y - axis to meet the line
It meets the line at y = 2 ad therefore, n = 2
Now draw a line y = -4, parallel to x - axis to meet the line
It meets the line at x = 6 and therefore, m = 6
Thus the values of m and n are 6 and 2 respectively.

9. Draw the graph (straight line) given by equation x – 3y = 18. If the straight line is drawn passes through the points (m, –5) and (6, n); find the values of m and n

Consider the equation
x -3y = 18
⇒ -3y = 18 -x
⇒ 3y = x -18
⇒ y = (x - 18)/3
The table for x - 3y = 18 is

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

From the above figure, we have
m = 3 and  n = -4

10. Use the graphical method to find the value of k, if:
(i) (k, -3) lies on the straight line 2x + 3y = 1
(ii) (5, k – 2) lies on the straight line x – 2y + 1 = 0

(i) 2x + 3y = 1
⇒ 3y = 1 -2x
⇒ y = (1-2x)/3
The table for 2x + 3y = 1 is

Plotting the above points in a graph, we get the following graph :

From the above graph, it is clear that k = 5

(ii) x - 2y + 1 = 0
⇒ 2y =  x+ 1
⇒ y = (x +1)/2
The table for x - 2y + 1 = 0 is

Plotting the above points in a graph, we get the following graph :
From the above graph, it is clear that
k - 2 = 3
⇒ k = 5

### Exercise 27(B)

1. Solve, graphically, the following pairs of equation :

(i) x – 5 = 0, y + 4 = 0
(ii) 2x + y = 23, 4x – y = 19
(iii) 3x + 7y = 27, 8 – y = 5x/2
(iv) (x+1)/4 = (2/3)(1 - 2y), (2+ 5y)/3 = (x/7)  - 2

(i) x -5 = 0 ⇒ x = 5
y + 4 = 0 ⇒ y = -4
Following is the graph of the two equations
x = 5 and y = -4 :

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

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

Plotting the points we get the following required graph :

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

(iii) 3x +7 y = 27⇒ 3x = 27 - 7y
⇒ x = (27 - 7y)/3
The table for 3x +7y = 27 is

Plotting the points we get the following required graph :
From the above graph, it is dear
that the two lines 3x + 7y = 27 and 8- y = (5/2) x
intersect at the point (2, 3)

⇒ 3(x + 1) = 8 - 16y
⇒ 3x + 3 = 8 - 16y
⇒ 3x + 3 - 8 = -16y
⇒ 3x - 5 = -16y
⇒ 7(2 +5y) = 3x - 42
⇒ 14 + 35y = 3x - 42
⇒ 3x = 14 + 35y + 42
⇒ 3x = 56 + 35y

Plotting the points we get the following required graph :
From the above graph, it is dear that the two lines (x + 1)/4 = (2/3)(1 - 2y) and (2 +5y)/3 = (x/7) - 2
intersect at the point (7, -1)

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

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

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)

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.

2x - y-1 = 0
⇒ 2x = y + 1

Plotting the above points we get the following required graph :

From the above graph, it is dear that the two lines 2x -y - 1= 0 and 2x +y = 9
intersect at the point (2.5 , 4)

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)

3x + 5y = 12
⇒ 3x = 12 - 5y

Also we have
3x - 5y + 18 = 0
⇒ 3x = 5y - 18

Plotting the above points we get the following required graph :

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

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.

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

Also we have
3x -2y + 4 = 0
⇒ 3x = 2y - 4

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,

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.

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

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

Plotting the above points we get the following required graph :
The area of the triangle ABC = (1/2)×AB×CD
= (1/2)× 7×3
= 21/2
= 10.5 sq. units
(ii) The coordinates of the vertices of the triangle are (-4, 2) , (3, 2) and (2, -1)

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.

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

Plotting the above points, we get the following required graph :
The graph shows that the lines of these equations are concurrent.

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.

Plotting the points in a graph, we get the following graph.
(i) The two lines intersect at (4, 5)
AD = 5 units and BC = 5 units
(ii) The area of the triangle = (1/2)×BC ×AD
= (1/2) ×5×5
= 25/2 sq. units
= 12.5 sq. units

9. 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.

(i) No. of articles to be manufactured and sold to break even (no profit and no loss).

(ii) The profit or loss made when (a) 30 (b) 60 articles are manufactured and sold.

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

Now plotting the points on a graph and we get the following required graph :
(i) No. of articles to be manufactured and sold are 50 when there is no loss and no profit.
C.P = S.P = Rs. 200
(ii) (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

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.

Now plotting the points on a graph and we get the following required graph :
Thus, the vertices of the triangle △ABC are :
A(-4, 2) , B(1, 3) and C(2, 5)

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).

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

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

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.

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

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(4/3 , 0) and C(4, 0 )
AD ⊥ BC and D ≡ (2, 0)

13. Use the graphical method to find the value of ‘x’ for which the expressions (3x+2)/2 and 3x/4 – 2