Ex-3.2 Pair Of Linear Equations In Two Variables (Part - 3), Class 10, Math RD Sharma Solutions | Extra Documents, Videos & Tests for Class 10 PDF Download

Q26. Solve graphically the system of linear equation:

4x-3y + 4 = 0 and 4x + 3y-20 = 0

Find the area bounded by these lines and x-axis.

Sol.

The given system of equation is 4x-3y + 4 = 0 and 4x + 3y-20 = 0

Now, 4x-3y + 4 = 0

When y = 0, then x = -1

When y = 4, then x = 2

Thus, we have the following table:

We have,

4x + 3y-2 = 0

When y = 0, then x = 5

When y = 4, then x = 2

Thus, we have the following table:

Graph of the given system is:

Clearly, the two lines intersect at A(2,4)

We also observe that the lines meet x-axis B(-1,0) and C(5,0)

Thus x = 2 and y = 4 is the solution of the given system of equations.

AD is drawn perpendicular A on x-axis. Clearly we have,

AD = y-coordinate point A(2,4)

AD = 3 and BC = 5-(-1) = 4 + 1 = 6

Area of the shaded region =  1/2×base×altitude

=  1/2×6×4

= 12 sq. units

Q27. Solve the following system of linear equations graphically:

3x + y-11 = 0 and x-y-1 = 0

Shade the region bounded by these lines and y- axis. Also find the area of the region bounded by these lines and y-axis.

Sol. The given system of equations is 3x + y-11 = 0 and x-y-1 = 0

Now, 3x + y-11 = 0

y = 11-3x

When x = 0, then y = 11

When x = 3, then y = 2

Thus, we have the following table:

We have

x-y-1 = 0

y = x-1

When x = 0, then y = -1

When x = 3, then y = 2

Thus, we have the following table:

Graph of the given system is:

Clearly, the two lines intersect at A (3,2)

We also observe that the lines meet y-axis B(0,11) and C(0,-1)

Thus x = 3 and y = 2 is the solution of the given system of equations.

AD is drawn perpendicular A on x-axis. Clearly we have,

AD = y-coordinate point A(2,4)

AD = 3 and BC = 11-(-1) = 11 + 1 = 12

Area of the shaded region =  1/2×base×altitude

=  1/2×12×3

= 18 sq. units

Q29. Draw the graph of the following equation:

2x-3y + 6 = 0

2x + 3y-18 = 0

y-2 = 0

Sol. Now,2x-3y + 6 = 0

When y = 0, then x = -3

When y = 2, then x = 0

Thus, we have the following table:

We have

2x + 3y-18 = 0

When y = 2, then x = 6

When y = 6, then x = 0

Thus, we have the following table:

We have

y-2 = 0

y = -2

Graph of the given system of equations:

From the graph of three equations, we find that the three lines taken in pairs intersect each other at points A(3,4) , B(0,2) and C(6,2)

Hence, the vertices of the required triangle are (3,4) , (0,2) and (6,2)

From the graph, we have

AD = 4-2 = 2

BC = 6-0 = +

Area of the shaded region =  1/2×base×altitude

=  1/2×6×2

= 6 sq. units

Q30. Solve the following system of equations graphically:

2x-3y + 6 = 0 and 2x + 3y-18 = 0

Also, find the area of the region bounded by these lines and y-axis.

Sol. The given system of equations:

2x-3y + 6 = 0 and 2x + 3y-18 = 0

Now,2x-3y + 6 = 0

When x = 0, then y = 2

When x = -3, then y = 0

Thus, we have the following table:

We have

2x + 3y-18 = 0

When y = 2, then x = 6

When y = 6, then x = 0

Thus, we have the following table:

Graph of the given system of equations:

Clearly, the two lines intersect at A(3,4) .Hence x = 3 and y = 4 is the solution of the given system of equations.

From the graph, we have

AD = x-coordinate point A(3,4) = 3

BC = 6-2 = 4

Area of the shaded region =  12×base×altitude

=  12×4×3

= 6 sq. units

Q31. Solve the following system of linear equation graphically;

4x-5y-20 = 0 and 3x + 5y-15 = 0

Sol. Now,4x-5y-20 = 0

When y = 0, then x = 5

When y = -4, then x = 0

Thus, we have the following table:

We have

3x + 5y-15 = 0

When y = 0, then x = 5

When y = -4, then x = 0

Thus, we have the following table:

Graph of the given system of equations:

Clearly, the two lines intersect at A(5,0). Hence x-5 , y-0 is the solution of the given system of equations.

The lines meet y-axis at B(0,-4) and C(0,3) respectively.

The vertices of the triangle are (5,0) , (0,-4) and (0,3)

Q32: Draw the graphs of the equations 5x-y = 5 and 3x-y = 3. Determine the coordinates of the vertices of the triangle forms by these lines and y-axis. Calculate the area of the triangle forms.

Sol. 5x-y = 5

= > y = 5x-5

Three solutions of this equation can be written as follows:

3x-y = 3

Y = 3x-3

The graphical representation of the two lines will be as follows:

It can observed that the required triangle is ABC

The coordinates of its vertices A (1,0) B(0,-3) and C(0,-5)

Q33. Form the pair of linear equation in the following problems, and find their solution graphically:

(i) 10 students of class X took part in 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 costs Rs.50, whereas 7 pencils and 5 pens together cost Rs.46.

Find the cost of one pencil and one pen.

(iii) Champa went to a sale to purchase some pants and skirts. When her friends asked her how many of each kind she had bought, she answered, “the number of skirts is two less than twice the number of pants purchase”. Also,”the number of skirts is four less than four times the number of pants purchased.”Help her friends to find how many pants and skirts champa bought.

Sol. (i) Let the number of girls and boys in the class be x and y respectively.

According to the Q.,

x + y = 10 and x-y = 4 are the given equations

Now, x + y = 10

x = 10-y

Three solutions of this equation can be written as follows:

x-y = 4

x = 4 + y

Three solutions of this equation can be written as follows:

The graphical representation is as follows:

From the graph , it can be observed that the two lines intersect each other at point (7,3).

So x = 7 and y = 3

The number of girls and boys in the class are 7 and 3 respectively.

(ii) Let the cost of one pencil and one pen Rs. x and Rs. y respectively.

According to the Q., we have,

5x + 7y = 50

7x + 5y = 50

Now, 5x + 7y = 50

Three solutions of this equation can be written as follows:

7x + 5y = 46

Three solutions of this equation can be written as follows:

The graphical representation is as follows:

From the graph it can be observed that the two lines intersect each other at the point (3,5)

So x = 3 and y = 5

Therefore the cost of the pencil and the pen are 3 and 5 respectively.

(iii) Let us denote the number of pants by x and the number of skirts be y . Then the equations formed are :

y = 2x?2 …………… (i)

y = 4x?2 ……………..(ii)

let us draw the graphs of the equations (i) and (ii) by finding the two solutions for each of the equations.

The lines intersect at point (1,0)

The value of x = 1 and y = 0

Q34. Solve the following system of equations graphically:

Shade the region between the lines and y-axis

(i) 3x-4y = 7 and 5x + 2y = 3

(ii) 4x-y = 4 and 3x + 2y = 14

Sol.

(i) 3x-4y = 7 and 5x + 2y = 3

The given system of linear equation is 3x-4y = 7 and 5x + 2y = 3

Now, 3x-4y = 7

When x = 1 then, y = -1

When x = -3 then y = -4

Thus, we have the following table

Now, 5x + 2y = 3

When x = 1 then, y = -1

When x = 3 then y = -6

Thus, we have the following table

Graph of the given system of equations are :

Clearly the two lines intersect at A(1,-1)

Hence x = 1 and y = -1 is the solution of the given system of equations.

(ii) 4x-y = 4 and 3x + 2y = 14

The given system of linear equation is 4x-y = 4 and 3x + 2y = 14

Now, 4x-y = 4

y = 4x-4

When x = 0 then, y = -4

When x = -1 then y = -8

Thus, we have the following table

Now, 3x + 2y = 14

When x = 0then, y = 7

When x = 4 then y = 1

Thus, we have the following table

Graph of the given system of equations are:

Clearly the two lines intersect at A (2, 4)

Hence x = 2 and y = 4 is the solution of the given system of equations.

Q35. Represent the following pair of equations graphically and write the coordinates of points where the lines intersects y-axis

x + 3y = 6 and 2x-3y = 12

Sol. The given systems of equations are:

x + 3y = 6 and 2x-3y = 12

Now, x + 3y = 6

When x = 0 then, y = 2

When x = 3 then y = 1

Thus, we have the following table

Now, 2x-3y = 12

When x = 0 then, y = -4

When x = 6 then y = 0

Thus, we have the following table

Graph of the given system of equations are :

Clearly the two lines meet y-axis at B(0,2) and C(0,-4) respectively.

Hence the required coordinates are (0,2) and (0,-4)

Q36. Given the linear equation 2x + 3y-8 = 0 , write another in two variables in two variables such that the geometrical representation of the pair so formed is (i)intersecting lines (ii) parallel lines (iii) coincident lines

Sol.

(i) For the two lines a₁x + b₁x + c₁ = 0  and a₂x + b₂x + c₂ = 0 to be intersecting. We must have

a₁a₂ ≠ b₁b₂

So the other linear equation can be 5x + 6y-16 = 0

a₁a₂ =  2/5

b₁/b₂ =  3/6  = 1/2

c₁/c₂ =   − 8/ − 1/6  = 1/2

(ii) For the two lines a₁x + b₁x + c₁ = 0  and a₂x + b₂x + c₂ = 0 to be parallel we must have

a₁a₂ = b₁b₂ ≠ c₁c₂

So, the other linear equation can be 6x + 9y + 24 = 0

a₁/a₂ =  2/6  = 1/3

b₁/b₂ =  3/9  = 1/3

c₁/c₂ =  \(\frac{-8}{-24}/\( = 13\( = \(\frac{1}{3}\]”>

(iii) For the two lines a₁x + b₁x + c₁ = 0  and a₂x + b₂x + c₂ = 0 to be coincident we must have

a₁/a₂ = b₁/b₂ = c₁/c₂

So, the other linear equation can be 6x + 9y + 24 = 0

a₁/a₂ =  \(\frac{2}{8}/\( = 1/4\( = \(\frac{1}{4}\]”>

b₁/b₂ =  \(\frac{3}{12}/\( = 1/4\( = \(\frac{1}{4}\]”>

c₁/c₂ =  \(\frac{-8}{-32}/\( = 1/4\( = \(\frac{1}{4}\]”>