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**Q.1. Form the pair of linear equations in the following problems, and find their solutions graphically.(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.(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:**

x + y = 10

x – y = 4

Now, for x + y = 10 or x = 10 − y, the solutions are;

For x – y = 4 or x = 4 + y, the solutions are;

According to the question, the algebraic expression cab be represented as;

5x + 7y = 50

7x + 5y = 46

For, 5x + 7y = 50 or x = (50 - 7y) / 5, the solutions are;

For 7x + 5y = 46 or x = (46 - 5y) / 7, the solutions are;

From the graph, it is can be seen that the given lines cross each other at point (3, 5).

So, the cost of a pencil is 3/- and cost of a pen is 5/-.

(i) 5x – 4y + 8 = 0

7x + 6y – 9 = 0

(ii) 9x + 3y + 12 = 0

18x + 6y + 24 = 0

(iii) 6x – 3y + 10 = 0

2x – y + 9 = 0

Solution:

5x − 4y + 8 = 0

7x + 6y − 9 = 0

Comparing these equations with a1x + b1y + c1 = 0

And a2x + b2y + c2 = 0

We get,

a1 = 5, b1 = -4, c1 = 8

a2 = 7, b2 = 6, c2 = -9

(a1 / a2) = 5 / 7

(b1 / b2) = -4 / 6 = -2 / 3

(c1 / c2) = 8 / -9

Since, (a1 / a2) ≠ (b1 / b2)

So, the pairs of equations given in the question have a unique solution and the lines cross each other at exactly one point.

9x + 3y + 12 = 0

18x + 6y + 24 = 0

Comparing these equations with a1x + b1y + c1 = 0

And a2x + b2y + c2 = 0

We get,

a1 = 9, b1 = 3, c1 = 12

a2 = 18, b2 = 6, c2 = 24

(a1 / a2) = 9 / 18 = 1 / 2

(b1 / b2) = 3 / 6 = 1 / 2

(c1 / c2) = 12 / 24 = 1 / 2

Since (a1 / a2) = (b1 / b2) = (c1 / c2)

So, the pairs of equations given in the question have infinite possible solutions and the lines are coincident.

6x – 3y + 10 = 0

2x – y + 9 = 0

Comparing these equations with

a1x + b1y + c1 = 0

And a2x + b2y + c2 = 0

We get,

a1 = 6, b1 = -3, c1 = 10

a2 = 2, b2 = -1, c2 = 9

(a1 / a2) = 6 / 2 = 3 / 1

(b1 / b2) = -3 / -1 = 3 / 1

(c1 / c2) = 10 / 9

Since (a1 / a2) = (b1 / b2) ≠ (c1 / c2)

So, the pairs of equations given in the question are parallel to each other and the lines never intersect each other at any point and there is no possible solution for the given pair of equations.

(i) 3x + 2y = 5 ; 2x – 3y = 7

(ii) 2x – 3y = 8 ; 4x – 6y = 9

(iii) (3 / 2)x + (5 / 3)y = 7; 9x – 10y = 14

(iv) 5x – 3y = 11 ; – 10x + 6y = –22

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

Solution:

and 2x – 3y = 7 or 2x – 3y -7 = 0

Comparing these equations with a1x + b1y + c1 = 0

And a2x + b2y + c2 = 0

We get,

a1 = 3, b1 = 2, c1 = -5

a2 = 2, b2 = -3, c2 = -7

(a1 / a2) = 3 / 2

(b1 / b2) = 2 / -3

(c1 / c2) = -5 / -7 = 5 / 7

Since, (a1 / a2) ≠ (b1 / b2)

So, the given equations intersect each other at one point and they have only one possible solution. The equations are consistent.

Therefore,

a1 = 2, b1 = -3, c1 = -8

a2 = 4, b2 = -6, c2 = -9

(a1 / a2) = 2 / 4 = 1 / 2

(b1 / b2) = -3 / -6 = 1 / 2

(c1 / c2) = -8 / -9 = 8 / 9

Since , (a1 / a2) = (b1 / b2) ≠ (c1 / c2)

So, the equations are parallel to each other and they have no possible solution. Hence, the equations are inconsistent.

Therefore,

a1 = 3 / 2, b1 = 5 / 3, c1 = -7

a2 = 9, b2 = -10, c2 = -14

(a1 / a2) = 3 / (2 × 9) = 1 / 6

(b1 / b2) = 5 / (3× - 10)= -1 / 6

(c1 / c2) = -7 / -14 = 1 / 2

Since, (a1 / a2) ≠ (b1 / b2)

So, the equations are intersecting each other at one point and they have only one possible solution. Hence, the equations are consistent.

Therefore,

a1 = 5, b1 = -3, c1 = -11

a2 = -10, b2 = 6, c2 = 22

(a1 / a2) = 5 / (-10) = -5 / 10 = -1 / 2

(b1 / b2) = -3 / 6 = -1 / 2

(c1 / c2) = -11 / 22 = -1 / 2

Since (a1 / a2) = (b1 / b2) = (c1 / c2)

These linear equations are coincident lines and have infinite number of possible solutions. Hence, the equations are consistent.

a1 = 4 / 3 , b1 = 2 , c1 = -8

a2 = 2, b2 = 3 , c2 = -12

(a1 / a2) = 4 / (3 × 2)= 4 / 6 = 2 / 3

(b1 / b2) = 2 / 3

(c1 / c2) = -8 / -12 = 2 / 3

Since (a1 / a2) = (b1 / b2) = (c1 / c2)

These linear equations are coincident lines and have infinite number of possible solutions. Hence, the equations are consistent.

(i) x + y = 5, 2x + 2y = 10

(ii) x – y = 8, 3x – 3y = 16

(iii) 2x + y – 6 = 0, 4x – 2y – 4 = 0

(iv) 2x – 2y – 2 = 0, 4x – 4y – 5 = 0

Solution: (i)

(a1 / a2) = 1 / 2

(b1 / b2) = 1 / 2

(c1 / c2) = 1 / 2

Since (a1 / a2) = (b1 / b2) = (c1 / c2)

∴The equations are coincident and they have infinite number of possible solutions.

So, the equations are consistent.

For, x + y = 5 or x = 5 – y

For 2x + 2y = 10 or x = (10 - 2y) / 2

From the figure, we can see, that the lines are overlapping each other.

Therefore, the equations have infinite possible solutions.

(a1 / a2) = 1 / 3

(b1 / b2) = -1 / -3 = 1 / 3

(c1 / c2) = 8 / 16 = 1 / 2

Since, (a1 / a2) = (b1 / b2) ≠ (c1 / c2)

The equations are parallel to each other and have no solutions. Hence, the pair of linear equations is inconsistent.

(a1 / a2) = 2 / 4 = 1 / 2

(b1 / b2) = 1 / -2

(c1 / c2) = -6 / -4 = 3 / 2

Since, (a1 / a2) ≠ (b1 / b2)

The given linear equations are intersecting each other at one point and have only one solution. Hence, the pair of linear equations is consistent.

Now, for 2x + y – 6 = 0 or y = 6 – 2x

And for 4x – 2y – 4 = 0 or y = (4x - 4) / 2

From the graph, it can be seen that these lines are intersecting each other at only one point,(2,2).

(a1 / a2) = 2 / 4 = 1 / 2

(b1 / b2) = -2 / -4 = 1 / 2

(c1 / c2) = 2 / 5

Since, a1 / a2 = b1 / b2 ≠ c1 / c2

Thus, these linear equations have parallel and have no possible solutions. Hence, the pair of linear equations are inconsistent.

Solution:

The width of the garden is x and length is y.

Now, according to the question, we can express the given condition as;

y – x = 4

and

y + x = 36

Now, taking y – x = 4 or y = x + 4

For y + x = 36, y = 36 – x

From the graph you can see, the lines intersects each other at a point(16, 20). Hence, the width of the garden is 16 and length is 20.

(i) Intersecting lines

(ii) Parallel lines

(iii) Coincident lines

Solution: (i)

To find another linear equation in two variables such that the geometrical representation of the pair so formed is intersecting lines, it should satisfy below condition;

(a1 / a2) ≠ (b1 / b2)

Thus, another equation could be 2x – 7y + 9 = 0, such that;

(a1 / a2) = 2 / 2 = 1 and (b1 / b2) = 3 / -7

Clearly, you can see another equation satisfies the condition.

To find another linear equation in two variables such that the geometrical representation of the pair so formed is parallel lines, it should satisfy below condition;

(a1 / a2) = (b1 / b2) ≠ (c1 / c2)

Thus, another equation could be 6x + 9y + 9 = 0, such that;

(a1 / a2) = 2 / 6 = 1 / 3

(b1 / b2) = 3 / 9= 1 / 3

(c1 / c2) = -8 / 9

Clearly, you can see another equation satisfies the condition.

(a1 / a2) = (b1 / b2) = (c1 / c2)

Thus, another equation could be 4x + 6y – 16 = 0, such that;

(a1 / a2) = 2 / 4 = 1 / 2 ,(b1 / b2) = 3 / 6 = 1 / 2, (c1 / c2) = -8 / -16 = 1 / 2

Clearly, you can see another equation satisfies the condition.

Solution:

For, x – y + 1 = 0 or x = 1 + y

For, 3x + 2y – 12 = 0 or x = (12 - 2y) / 3

From the figure, it can be seen that these lines are intersecting each other at point (2, 3) and x-axis at (−1, 0) and (4, 0). Therefore, the vertices of the triangle are (2, 3), (−1, 0), and (4, 0).

**Check out the NCERT Solutions of all the exercises of Linear Equations in Two Variables:**

Ex 3.1 NCERT Solutions: Pair of Linear Equations in Two Variables

Ex 3.3 NCERT Solutions: Pair of Linear Equations in Two Variables

Ex 3.4 NCERT Solutions: Pair of Linear Equations in Two Variables

Ex 3.5 NCERT Solutions: Pair of Linear Equations in Two Variables

Ex 3.6 NCERT Solutions: Pair of Linear Equations in Two Variables

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- Ex 3.5 NCERT Solutions: Pair of Linear Equations in Two Variables