Simultaneous Linear equation – Exercise 3.5.4 – Class IX

  1. Solve the following graphically:

(i) x + y = 7 and 2x – 3 = 9

Solution:

x + y = 7

If x = 0 then  y = 7 – 0 = 7

If x = 1 then y = 7 – 1 = 6

If x = 2 then y = 7 – 2 = 5

IF x = 3 then y = 7 – 3 = 4

If x = 4 then y = 7 – 4 = 3

x 0 1 2 3 4
y 7 6 5 4 3

consider, 2x – 3y = 9 , y = 9 – 2x/-3

If x = 0 then y = 9/-3 = -3

If x = 3 then  y = 9-2(3)/-3 = 9-6/-3 = 3/-3 = -1

If x = 6 then y = 9-2(6)/-3 = 9-12/-3 = -3/-3 = 1

if x = 9 then y = 9-2(9)/-3 = 9-18/-3 = -9/-3 = 3

x 0 3 6 9
y -3 -1 1 3

 

Simultaneous Linear equation – Exercise 3.5.3 – Class IX

Two straight lines intersect at E. The coordinates of E are (6, 1). Hence x = 6 and y = 1 is the solution of the system.


(ii)  2x +  y = 6

x – 2y = -2

Solution:

Consider, 2x +  y = 6 ; y =  6 – 2x

If  x = 0 then y = 6

If x  = 1 then y = 4

If  x  = 2 then y = 2

If x = 3 then y = 0

If x = 4 then y = -2

x 0 1 2 3 4
y 6 4 2 0 -2

Now, consider x – 2y = -2 ; y = -2-x/-2

If x = 0 then y = 1

If  x  =  2 then y = 0

If x = 4 then  y = 3

If x = 6 then y = 4

If x = 8 then y = 5

x 0 2 4 6 8
y 1 0 3 4 5

Simultaneous Linear equation – Exercise 3.5.3 – Class IX

Two straight lines intersect at E. The coordinates of E are (2, 2). Hence x = 2 and y = 2 is the solution of the system.


(iii) x – y = -2

x – y = 1

[hint : parallel line]

Solution:

Consider, x – y = -2  thus, -y = -2 –x ⇒ y = x + 2

If x  = 0 then y = 2

If x = 1 then y = 3

If x = 2 then y = 4

If x = -1 then y = 1

If x = -2 then y = 0

x 0 1 2 -1 -2
y 2 3 4 1 0

Now, x – y = 1 then, y = x – 1

If x = 0 then y = -1

If x = 1 then y  =  0

If x =   2 then y =  1

If x  = -1  then y = -2

If x =  -2 then  y = -3

x 0 1 2 -1 -2
y -1 0 1 -2 -3

Simultaneous Linear equation – Exercise 3.5.3 – Class IX

As x – y = -2 and x – y = 1 are parallel lines there is no solution for these two equations.


(iv) x – y = 1 and 2x – 2y = 2 [hint: represents the same line]

Solution:

x – y = 1 …………….(1)

2x – 2y = 2 ⇒(2x – 2y = 2)x1/2 ⇒ x – y = 1 …………(2)

Therefore equation (1) and equation (2) are same. Therefore, it represents the same line as follows:

Consider, x – y = 1, then y = x – 1

If x = 0 then y = -1

If x = 1 then y = 0

If x = 2 then y = 1

If x = 3 then y = 2

If x = -1 then y = -2

If x = -2 then y = -3

x 0 1 2 3 -1 -2
y -1 0 1 2 -2 -3

Simultaneous Linear equation – Exercise 3.5.3 – Class IX

As x – y = 1 and 2x – 2y = 2 represents the same line there are infinitely many solutions.


(v) x + y = 3 and 2x +  5y = 12

Solution:

Consider x + y = 3 where y = 3 – x

If x = 0 then y = 3

IF x = 1 then y = 2

If x = 2 then y = 1

If x = 3 then y = 0

x 0 1 2 3
y 3 2 1 0

Consider 2x + 5y = 12 where y = 12-2x/5

If x = 1 then y = 2

If x = 6 then y = 0

If x = -4 then y = 12-2(-4)/5= 12+8/5 = 20/5 = 4

If x = -9 then y = 12-2(-9)/5 = 12+18/5 = 30/5 = 6

x 1 6 -4 -9
y 2 0 4 6

Simultaneous Linear equation – Exercise 3.5.3 – Class IX

Two straight lines intersect at E. The coordinates of E are (1, 2). Hence x = 1 and y = 2 is the solution of the system.


(vi) 3x – y – 2 = 0 and 2x + y – 8 = 0

Solution:

3x – y – 2 = 0 ……………..(1)

2x + y – 8 = 0………………(2)

Consider (1), 3x – y – 2 = 0 where y = 3x – 2

If x = 0 and y = -2

If x = 1 then y = 1

If x = 2 then y = 4

IF x = 3 then y = 7

x 0 1 2 3
y -2 1 4 7

Consider (2) 2x + y – 8 = 0 where, y = 8 – 2x

If x = 0 then y = 8

If x = 1 then y = 6

If x = 2 then y = 4

If x = 3 then y = 2

x 0 1 2 3
y 8 6 4 2

Simultaneous Linear equation – Exercise 3.5.3 – Class IX

Two straight lines intersect at E. The coordinates of E are (2, 4). Hence x = 2 and y = 4 is the solution of the system.


  1. The cost of manufacturing x articles is (50 + 3x). The selling price of x articles is 4x. On a graph sheet by taking suitable scale draw two graphs, first for the cost of manufacturing against no. of articles and the second for the selling price against number of articles.

Use the graph to determine:

(i) Number 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.

Solution:

(i) The cost of manufacturing x articles is 50 + 3x i.e, y = 50 + 3x

If x = 0 then y = 50

if x = 10 then y = 80

if x = 20 then y = 110

if x = 50 then y = 200

if x = 60 then y = 230

x 0 10 20 50 60
y 50 80 110 200 230

Selling price of x articles is 4x i.e, y = 4x

if x = 0 then then y = 0

If x = 10 then y = 40

I x = 20 then y = 80

If x = 50 then y = 200

If x = 60 then y = 240

x 0 10 20 50 60
y 0 40 80 200 240

Simultaneous Linear equation – Exercise 3.5.3 – Class IX

The break-even point is 50.

 

(ii)

(a)The profit or loss made when 30 articles are manufactured,

The cost of manufacturing x articles is 50 + 3x, x = 30 ⇒ 50 + 3×30 = 50 + 90 = 140. Therefore, cost of manufacturing 30 articles is Rs. 140

The cost of selling x articles is 4x , x = 30 ⇒ 4x = 4(30) = 120. Therefore, the cost of selling 30 articles is Rs. 120

Thus, for selling 30 articles it is loss for Rs. 20.

(b)The profit or loss made when 60 articles are manufactured,

The cost of manufacturing x articles is 50 + 3x, x = 60 ⇒ 50 + 3×60 = 50 + 180 = 230. Therefore, cost of manufacturing 30 articles is Rs. 230

The cost of selling x articles is 4x , x = 60 ⇒ 4x = 4(60) = 240. Therefore, the cost of selling 60 articles is Rs. 240

Thus, for selling 60 articles it is profit for Rs. 10.

Simultaneous Linear equation – Exercise 3.5.3 – Class IX


 

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