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10th class > Pair of Linear Functions in Two Variables > Exercise 3.1

Exercise 3.1

Form the pair of linear equations in the following problems, and find their solutions.

(i)

(i) 10 students of Class X took part in a Mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls who took part in the quiz.

Solution

Given that,

Total number of students who took part in the the quiz =

Let, the number of boys who took part in the quiz = x

and, number of girls who took part in the quiz = y

According to question,

x + y = ...........(1)

y = x + ...........(2)

By putting value of x = 10 − y in equation(2), we get

y = 10 - + 4

or, 2y =

So, y = 142 =

and x = 10 − 7 =

Therefore, Number of boys who took part in the quiz = 3

Number of girls who took part in the class = 7

(ii)

(ii)5 pencils and 7 pens together cost ₹50, whereas 7 pencils and 5 pens together cost ₹46. Find the cost of one pencil and that of one pen.

Solution

Let the cost of one pencil be Rs. x and the cost of one pen be Rs. y

According to the first condition

5x + 7y = ...........(1)

⇒ x = 50-7y5

According to the second condition

7x + 5y = ...........(2)

Put the value of x in eq(2)

7(50-7y)5 + 5y =

⇒ 350 − 49y + 25y = 230 ⇒ y =

Resubstitute the value of y to find the value of x

⇒ x= 50-7(5)5 = 50-355

⇒ x =

cost of one pencil = Rs. 3 and cost of one pen = Rs. 5

On comparing the ratios a1a2, b1b2 and c1c2 find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincident:

(i)

(i) 5x - 4y + 8 = 0 and 7x + 6 - 9 = 0

Solution

a₁ = , b₁ = , c₁ =

a₂ = , b₂ = , c₂ =

a1a2 = ...(1)

b1b2 = 46 = ...(2)

From (1) and (2)

a1a2 b1b2

Therefore, they are at a point.

(ii)

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

Solution

a1 = , b₁ = , c₁ =

a2 = , b2 = , c2 =

a1a2 = 918 = ...(1)

b1b2 = 36 = ...(2)

c1c2 = 1224 = ...(3)

From (1), (2) and (3)

a1a2 = b1b2 = c1c2 =

Therefore, they are lines.

(iii)

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

Solution

a₁ = , b₁ = , c₁ =

a₂ = , b₂ = , c₂ =

a1a2 = 62 = ...(1)

b1b2 = 31 = ...(2)

c1c2 = ...(3)

From (1), (2) and (3)

a1a2 b1b2 c1c2

Therefore, they are lines.

On comparing the ratios a1a2, b1b2 and c1c2 find out whether the following pair of linear equations are consistent, or inconsistent.

(i)

(i) 3x + 2y = 5; 2x - 3y = 7

Solution

3x + 2y = 5; 2x - 3y = 7

a1a2 =

b1b2 =

c1c2 = -5(-7) =

From the above,

a1a2 b1b2

Therefore, lines are and have a solution,

Hence, the pair of equations is .

(ii)

(ii) 2x - 3y = 8; 4x - 6y = 9

Solution

2x - 3y = ; 4x - 6y =

a1a2 = 24 =

b1b2 = 36 =

c1c2 = 89 =

From the above,

a1a2 = b1b2 c1c2

Therefore, these lines are and have solution,

Hence, the pair of equations is .

(iii)

(iii) 32x + 53y = 7; 9x -10y = 14

Solution

a1a2 = (3/2)9 = 32 × 19 =

b1b2 = (5/3)(-10) = 53 × 110 = 16 =

From the above,

a1a2 b1b2

Therefore, lines are and have a solution.

Hence, they are .

(iv)

(iv) 5x - 3y = 11; -10x + 6y = -22

Solution

5x - 3y = ; -10x + 6y =

a1a2 = 5(-10) =

b1b2 = 36 =

c1c2 = 1122 =

From the above,

a1a2 b1b2 c1c2

Therefore, lines are and have solutions.

Hence, they are .

(v)

(v)4/3x + 2y = 8; 2x + 3y = 12

Solution

43x + 2y = ; 2x + 3y =

a1a2 = (4/3)2 = 43 × 12 =

b1b2 =

c1c2 = -8(-12) =

From the above,

a1a2 b1b2 c1c2

Therefore, lines are and have solutions.

Hence, they are .

Half the perimeter of a rectangular garden, whose length is 4 m more than its width, is 36 m. Find the dimensions of the garden.

Solution

Perimeter of a rectangle = 2( + )

Let the width(w) of the garden = x meter

Then length(l) = meter

Half perimeter = m

So perimeter of garden =(2 × 36) = meters

According to the question

⇒ 2(l+b) =

⇒ 2(x + x + ) = 72

⇒ 2x + + 4 =

⇒ 4x =

⇒ x = meters

Hence,the width(w) of the garden = 16 meters

The length(l) of the garden = (16 + 4) = meters

Given the linear equation 2x + 3y – 8 = 0, write another linear equation in two variables such that the geometrical representation of the pair so formed is:

(i) intersecting lines

(ii) parallel lines

(iii) coincident lines

(i)

(i) Intersecting lines

Solution: For intersecting line, the linear equations should meet following condition:

a1a2b1b2

For getting another equation to meet this criterion, multiply the coefficient of x with any number and multiply the coefficient of y with any other number. A possible equation can be as follows:

4x + 9y − = 0

(ii)

(ii)Parallel lines

Solution: For parallel lines, the linear equations should meet following condition:

a1a2 = b1b2c1c2

For getting another equation to meet this criterion, multiply the coefficients of x and y with the same number and multiply the constant term with any other number. A possible equation can be as follows:

4x + 6y − = 0

(iii)

(iii) Coincident lines

Solution: For getting coincident lines, the equations should meet following condition;

a1a2 = b1b2 = c1c2

For getting another equation to meet this criterion, multiply the whole equation with any number. A possible equation can be as follows:

4x + 6y − = 0