Pair of Linear Equations in Two Variables Class 10 Exercise 3.1 Solution.

Pair of Linear Equations in Two Variables Class 10 Exercise 3.1 Solution.

Introduction:

        दो चर वाले रैखिक समीकरणों का युग्म (Pair of Linear Equations in Two Variables) कक्षा 10 गणित का एक महत्वपूर्ण अध्याय है। इसमें हम ऐसे समीकरणों का अध्ययन करते हैं जिनमें दो चर (जैसे $x$ और $y$) होते हैं और जिनका सामान्य रूप $ax + by + c = 0$होता है।

        इस अध्याय में हम सीखते हैं कि दो रैखिक समीकरणों को हल करके उनके सामान्य हल (solution) कैसे प्राप्त करें। Exercise 3.1 में मुख्य रूप से हम यह समझते हैं कि दिए गए समीकरणों का युग्म संगत (consistent) है या असंगत (inconsistent) तथा उनके ग्राफ के माध्यम से हल निकालते हैं।

1) From 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.

Solution:

Step 1: Let Variables

Let:

• Number of boys = x
• Number of girls = y

Step 2: Form Equations

1st Condition:

Total students = 10

$x + y = 10$

2nd Condition:

Girls are 4 more than boys

$\mathrm{<span\; style="font-family:\; arial;">y\; =\; x\; +\; 4</span>}$

Step 3: Graphical Method

Lets find points to draw a graph:

Equation 1: x + y = 10

x	0	10
y	10	0

Equation 2: y = x + 4

x	0	2
y	4	6

Hence the graphical Representation is as follows,

From the graph, it is can be seen that the given lines intersect each other at point (3, 7).

So the Number of Boys are 3 and Girls are 7

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

Solution:

Step 1: Let Variables

Let:

• Cost of one pencil = x
• Cost of one pen = y

Step 2: Form Equations

1st Condition:

$5x + 7y = 50$

2nd Condition:

$$7x + 5y = 46

Step 3: Find Points for Graph

$\mathrm{<span\; style="font-family:\; arial;">\; </span>}$

Equation 1: 5x + 7y = 50

x	1	5
y	45/7	25/7

Equation 2: 7x + 5y = 46

x	3	1
y	5	39/5

Hence the graphical Representation is as follows,

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

So the Cost of one pencil is 3 and that of one pen is 5

2. On comparing the ratios a₁ / a₂ , b₁ / b₂ and c₁ / c₂ , find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincident:

i) 5x − 4y + 8 = 0
   7x + 6y − 9 = 0

Solution:

Comparing the given equation with,

        a₁x + b₁y + c₁ = 0

        a₂x + b₂y + c₂ = 0 

We get, 

For 5x − 4y + 8 = 0 Here, a₁ = 5 , b₁ = -4 and c₁ = 8

For 7x + 6y − 9 = 0 Here, a₂ = 7 , b₂ = 6 and c₂ = -9

Ratios:

As a₁ / a₂ ≠ b₁ / b₂

$\frac{5}{7} \neq \frac{-4}{6}$

Therefore, Lines are intersecting (one solution)

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

Ratios:

$\frac{9}{18} = \frac{3}{6} = \frac{12}{24}$

Therefore, Lines are coincident (infinite solutions)

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

Ratios:

$\frac{6}{2} = \frac{-3}{-1} \neq \frac{10}{9}$

Therefore, Lines are parallel (no solution)

3.On comparing the ratios a₁ / a₂ , b₁ / b₂ and c₁ / c₂ , find out whether the following pair of linear equations are consistent , or inconsistent.

Criterion for Consistent & Inconsistent Equations

Pair of linear equations ka general form hota hai:

$$                             a1x+b1y+c1=0                                a2x+b2y+c2=0​

Conditions

1) Consistent Equations

• When Solution is exists.

(a) Unique Solution (Intersecting lines)

        a₁ / a₂ ≠ b₁ / b₂

• Lines intersect at a point

(b) Infinite Solutions (Coincident lines)

    a₁ / a₂ = b₁ / b₂ = c₁ / c₂

• Both equation represent the same line

2) Inconsistent Equations

• When Solution is not exists.

Condition:

    a₁ / a₂ = b₁ / b₂ ≠ c₁ / c₂

• Lines are parallel (Never intersect each other)

i)  3x + 2y = 5
    2x − 3y = 7

Solution:

Comparing the given equation with,

        a₁x + b₁y + c₁ = 0

        a₂x + b₂y + c₂ = 0 

We get, 

For 3x + 2y - 5 = 0 Here, a₁ = 3 , b₁ = 2 and c₁ = -5

For 2x - 3y − 7 = 0 Here, a₂ = 2 , b₂ = -3 and c₂ = -7

     a₁ / a₂ = 3 / 2 

     b₁ / b₂ = 2 / -3

     c₁ / c₂ = -5 / -7

As, 

     a₁ / a₂ ≠ b₁ / b₂

_{Therefore, the given lines are intersect at a point (unique solution)}

_{So the given pair of linear equation is consistent.}

ii) 2x − 3y = 8
     4x − 6y = 9

Solution:

By comparing the given equation,

we get,

      a₁ / a₂ = 2 / 4 

     b₁ / b₂ = -3 / -6

     c₁ / c₂ = -8 / -9

As,

     a₁ / a₂ = b₁ / b₂ ≠ c₁ / c₂

_{Therefore, the given lines are parallel.}

_{So the given pair of linear equation is inconsistent.}

iii) 3/2 x + 5/3 y = 7 

      9x − 10y = 14

Solution:

$$32x+53y=7 and 9x − 10y = 14

By comparing the given equation,

we get,

      a₁ / a₂ = 3/2 / 9 = 1 / 6

     b₁ / b₂ = 5/3 / -10 = - 1 / 6

     c₁ / c₂ = -7 / -14 = 1 / 2

As,

         a₁ / a₂ ≠ b₁ / b₂

_{Therefore, the given lines are intersect at a point (unique solution)}

So the given pair of linear equation is consistent.

iv) 5x − 3y = 11
  −10x + 6y = −22

Solution:

By comparing the given equation,

we get,

      a₁ / a₂ = 5 / -10 = - 1 / 2

      b₁ / b₂ = -3 / 6 = - 1 / 2

      c₁ / c₂ = -11 / 22 = - 1 / 2

As, 

        a₁ / a₂ = b₁ / b₂ = c₁ / c₂

Therefore, both the equation represent the same line.

So the given pair of linear equation is consistent.

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

Solution:

$\frac{4}{3}x + 2y = 8$2x + 3y = 12

By comparing the given equation,

we get,

      a₁ / a₂ = 4/3 / 2 = 2 / 3

     b₁ / b₂ = 2 / 3 = 2 / 3

     c₁ / c₂ = -8 / -12 = 2 / 3

As, 

        a₁ / a₂ = b₁ / b₂ = c₁ / c₂

Therefore, both the equation represent the same line.

So the given pair of linear equation is consistent.

4. Which of the following pairs of linear equations are consistent/inconsistent? If consistent, obtain the solution graphically.

i) x + y = 5

  2x + 2y = 10

Solution:

By comparing the given equation,

we get the ratios as,

 a₁ / a₂ = 1 / 2

 b₁ / b₂ = 1 / 2  

 c₁ / c₂  = 1 / 2   

As, 

        a₁ / a₂ = b₁ / b₂ = c₁ / c₂

Therefore, both the equation represent the same line (conincident).

So the given pair of linear equation is consistent.

Algebraic representation,

i) Solution table for equation, x + y = 5

x	4	3	2
y	1	2	3

ii) Solution table for equation, 2x + 2y = 10

x	4	3	2
y	1	2	3

The graphical solution is as given below,

ii) x − y = 8

   3x − 3y = 16

Solution:

By comparing the given equation,

we get,

      a₁ / a₂ = 1 / 3 

     b₁ / b₂ = -1 / -3 = 1 / 3

     c₁ / c₂ = -8 / -16 = 1 / 2

As,

     a₁ / a₂ = b₁ / b₂ ≠ c₁ / c₂

_{Therefore, the given lines are parallel (No solution exist).}

_{So the given pair of linear equation is inconsistent.}

iii) 2x + y − 6 = 0

     4x − 2y − 4 = 0

Solution:

By comparing the given equation,

we get,

      a₁ / a₂ = 2 / 4 

     b₁ / b₂ = 1 / -2

     c₁ / c₂ = -6 / -4 = -3 / 2

As,

         a₁ / a₂ ≠ b₁ / b₂

_{Therefore, the given lines are intersect at a point (unique solution)}

So the given pair of linear equation is consistent.

Algebraic representation,

i) Solution table for equation, 2x + y - 6 = 0

x	0	1	2
y	6	4	2

ii) Solution table for equation, 4x - 2y - 4 = 0

x	1	2	3
y	0	2	4

The graphical solution is as given below,

iv) 2x − 2y − 2 = 0
     4x − 4y − 5 = 0

Solution:

By comparing the given equation,

we get,

      a₁ / a₂ = 2 / 4 = 1 / 2

     b₁ / b₂ = -2 / -4 = 1 / 2

     c₁ / c₂ = -2 / -5

As,

     a₁ / a₂ = b₁ / b₂ ≠ c₁ / c₂

_{Therefore, the given lines are parallel (No solution exist).}

_{So the given pair of linear equation is inconsistent.}

_{5. 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:

Step 1: Assume variables

Let

• Length = $x$ m
• Width = $y$ m

Step 2: Form equations

Given: Length is 4 m more than width

$$x = y + 4$$
$$x - y = 4 \quad \text{(Equation 1)}$$

Given: Half perimeter = 36 m

$$$$x+y=36(Equation 2)

x + y = 36 \quad \text{(Equation 2)}

Step 3: Write equations for graph

Equation 1:

$$x - y = 4$$

Equation 2:

$$x + y = 36$$ Step 4: Find points to draw graph

For $x - y = 4$

x	4	8
y	0	4

For $x + y = 36$

x	0	36
y	36	0

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