Ans. (i) Let number of boys who took part in the quiz = x 

Let number of girls who took part in the quiz = y

According to given conditions, we have

x + y = 10… (1)

And, y = x + 4

⇒ x – y = −4 … (2)

For equation x + y = 10, we have following points which lie on the line

For equation x – y = –4, we have following points which lie on the line

We plot the points for both of the equations to find the solution.

We can clearly see that the intersection point of two lines is (3, 7).

Therefore, number of boys who took park in the quiz = 3 and, number of girls who took part in the quiz = 7.

(ii) Let cost of one pencil = Rs x and Let cost of one pen = Rs y

According to given conditions, we have

5x + 7y = 50… (1)

7x + 5y = 46… (2)

For equation 5x + 7y = 50, we have following points which lie on the line

For equation 7x + 5y = 46, we have following points which lie on the line

We can clearly see that the intersection point of two lines is (3, 5).

Therefore, cost of pencil = Rs 3 and, cost of pen = Rs 5

Ans. (i) 5x − 4y + 8 = 0, 7x + 6y – 9 = 0 

Comparing equation 5x − 4y + 8 = 0 with  and 7x + 6y – 9 = 0 with 

We get, , , , , , 

We have because 

Hence, lines have unique solution which means they intersect at one point.

(ii) 9x + 3y + 12 = 0, 18x + 6y + 24 = 0

Comparing equation 9x + 3y + 12 = 0 with  and 18x + 6y + 24 = 0 with ,

We get, , , , , , 

We have  because  ⇒12=12=12⇒12=12=12

Hence, lines are coincide.

(iii) 6x − 3y + 10 = 0, 2x – y + 9 = 0

Comparing equation 6x − 3y + 10 = 0 with  and 2x – y + 9 = 0 with ,

We get, , , , , , 

We have  because  ⇒31=31≠109⇒31=31≠109

Hence, lines are parallel to each other.

Ans. (i) 3x + 2y = 5, 2x − 3y = 7 

Comparing equation 3x + 2y = 5 with  and 2x − 3y = 7 with ,

We get, a1=3,b1=2,c1=−5,a1=3,b1=2,c1=−5, 

 and b1b2=2−3b1b2=2−3

Here which means equations have unique solution.

Hence they are consistent.

(ii) 2x − 3y = 8, 4x − 6y = 9

Comparing equation 2x − 3y = 8 with  and 4x − 6y = 9 with ,

We get, 

Here because 24=−3−6≠−8−9⇒24=−3−6≠−8−9⇒

Therefore, equations have no solution because they are parallel.

Hence, they are inconsistent.

(iii)  9x − 10y = 14

Comparing equation with  and 9x − 10y = 14 with ,

We get, , , c1=−7,a2=9,c1=−7,a2=9, b2=−10,c2=−14b2=−10,c2=−14

 and 

Here 

Therefore, equations have unique solution.

Hence, they are consistent.

(iv) 5x − 3y = 11, −10x + 6y = −22

Comparing equation 5x − 3y = 11 with  and −10x + 6y = −22 with ,

We get, 

  and c1c2=−1122=−12c1c2=−1122=−12

Here 

Therefore, the lines have infinite many solutions.

Hence, they are consistent.

(v) 43x+2y=8;  2x+3y=1243x+2y=8;  2x+3y=12

Comparing equation 43x+2y=843x+2y=8 with  and 2x+3y=122x+3y=12 with ,

We get, a1=43,b1=2,c1=−8,a1=43,b1=2,c1=−8, a2=2,b2=3,c2=−12a2=2,b2=3,c2=−12

a1a2=432=23,a1a2=432=23,b1b2=23b1b2=23 and c1c2=812=23c1c2=812=23

Here 

Therefore, the lines have infinite many solutions.

Hence, they are consistent.

Ans. (i) x + y = 5, 2x + 2y = 10 

For equation x + y = 5  we have following points which lie on the line

For equation 2x + 2y – 10 = 0, we have following points which lie on the line

We can see that both of the lines coincide. Hence, there are infinite many solutions. Any point which lies on one line also lies on the other. Hence, by using equation (x + y – 5 = 0), we can say that x = 5 − y

We can assume any random values for y and can find the corresponding value of x using the above equation. All such points will lie on both lines and there will be infinite number of such points.

(ii) x – y = 8, 3x − 3y = 16

For x – y = 8, the coordinates are:

And for 3x – 3y = 16, the coordinates

Plotting these points on the graph, it is clear that both lines are parallel. So the two lines have no common point. Hence the given equations have no solution and lines are inconsistent.

(iii) 2x + y – 6 = 0, 4x − 2y – 4 = 0

For equation 2x + y – 6 = 0, we have following points which lie on the line

For equation 4x – 2y – 4 = 0, we have following points which lie on the line

We can clearly see that lines are intersecting at (2, 2) which is the solution.

Hence x = 2 and y = 2 and lines are consistent.

(iv) 2x − 2y – 2 = 0, 4x − 4y – 5 = 0

For 2x – 2y – 2 = 0, the coordinates are:

And for 4x – 4y – 5 = 0, the coordinates

Plotting these points on the graph, it is clear that both lines are parallel. So the two lines have no common point. Hence the given equations have no solution and lines are inconsistent.

Ans. Let length of rectangular garden = x metres 

Let width of rectangular garden = y metres

According to given conditions, perimeter = 36 m

 12[2(x+y)]=3612[2(x+y)]=36

x + y = 36 ……(i)

And x = y + 4

⇒ x – y = 4……..(ii)

Adding eq. (i) and (ii),

2x = 40

 x = 20 m

Subtracting eq. (ii) from eq. (i),

2y = 32

 y = 16 m

Hence, length = 20 m and width = 16 m

Ans. (i) Let the second line be equal to  

Comparing given line 2x + 3y – 8 = 0 with ,

We get 

Two lines intersect with each other if 

So, second equation can be x + 2y = 3 because 

(ii) Let the second line be equal to 

Comparing given line 2x + 3y – 8 = 0 with ,

We get 

Two lines are parallel to each other if 

So, second equation can be 2x + 3y – 2 = 0 because 

(iii) Let the second line be equal to 

Comparing given line 2x + 3y – 8 = 0 with ,

We get 

Two lines are coincident if 

So, second equation can be 4x + 6y – 16 = 0 because 

Ans. For equation x – y + 1 = 0, we have following points which lie on the line 

For equation 3x + 2y – 12 = 0, we have following points which lie on the line

We can see from the graphs that points of intersection of the lines with the x–axis are (–1, 0), (2, 3) and (4, 0).