Chapter 13 Linear Equations in Two Variables RD Sharma Solutions Exercise 13.3 Class 9 Maths
Chapter Name  RD Sharma Chapter 13 Linear Equations in Two Variables Exercise 13.3 
Book Name  RD Sharma Mathematics for Class 10 
Other Exercises 

Related Study  NCERT Solutions for Class 10 Maths 
Exercise 13.3 Solutions
1. Draw the graph of each of the following linear equation in two variables.
(i) x + y = 4
(ii) x  y = 2
(iii) x + y = 6
(iv) y = 2x
(v) 3x + 5y = 15
(vi) x/2  y/3 = 3
(vii) (x  2)/3 = y  3
(viii) 2y = x + 1
Solution
(i) We have x + y = 4
x = 4  y
Putting y = 0, we get x = 4  0 = 4
Putting y = 3, we get x = 4  3 = 1
Thus, we get the following table giving the two points on the line represented by the equation x + y = 4
Graph for the equation x + y = 4
(ii) We have,
x  y = 2
x = 2 +y ...(i)
Putting y = 0, we get x = 2 + 0 = 2
Putting y = 2, we get x = 2  2 = 0
Thus, we get the following table giving the two points on the line represented by the equation x  y = 2
Graph for the equation x  y = 2
(iii) We have
x + y = 6
⇒ x = 6 +x
Putting y = 4, we get y = 6  4 = 2
Putting x = 3 we get y = 6  3 = 3
Thus, we get the following table giving the two points on the line represented by the equation x + y = 6
Graph for the equation x + y = 6.
(iv) We have
y = 2x ...(i)
Putting x = 0, we get y = 2 × 0 = 0
Putting x = 1 we get y = 2×1 = 2
Thus, we get the following table giving the two points on the line represented by the equation y =2x
Graph for the equation y = 2x
(v) We have
3x + 5y = 15
3x = 15  5y
x = (15 5y)/3
Putting y = 0, we get x = (15  5×0)/3 = 5
Putting y = 3 we get x = 15  5× 3)/3 = 0
Thus, we get the following table giving the two points on the line represented by the equation 3x + 5y  15
Graph for the equation 3x + 5y  15
(vi) We have
Putting y = 3, we get x = [12 + 2(3)]/3 = 2
Putting y = 0 we get x = (12 + 0)/3 = 4
Thus, we get the following table giving the two points on the line represented by the equation x/2  y/3 = 2
Graph for the equation x/2  y/3 = 3
(vii) We have,
Thus, we get the following table giving the two points on the line represented by the equation (x  2)/y = y  3
Graph for the equation (x  2)/y = y  3
(viii) We have
2y = x + 1
⇒ x  1 = 2y ...(1)
Putting y = 0, we get x = 1  2×0 = 1
Putting y = 1, we get x = 1  2(1) = 3
Thus, we have the following table giving the two points on the line represented by the equation
2y = x + 3
2y = x + 1
Graph for the equation 2y = x + 1
2. Given the equations of two lines passing through (3, 12). How many more such lines are there, and why ?
Solution
The equation of two lines passing through (3, 12) are
4x  y = 0
3x  y + 3 = 0 ...(i)
There are infinitely many lines passing through (3, 12)
3. A three  wheeler scooter charges Rs 15 for first kilometer and Rs 8 each for every subsequent kilometer. For a distance of x km, an amount of Rs y is paid. Write the linear equation representing the above information.
Solution
Total fare of Rs. y for covering distance of x kilometers is given by
y = 15 + 8(x 1)
⇒ y = 15 + 8x  8
⇒ y = 8x + 7
This is the required linear equation for the given information.
4. A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Aarushi paid Rs 27 for a book kept for seven days. If fixed charges are Rs x and per day charges are Rs y. Write the linear equation representing the above information.
Solution
Total charges paid by Aarushi is given by
27 = x + 4y
⇒ x + 4y = 27
This is the required linear equation for the given information.
5. A number is 27 more than the number obtained by reversing its digits. If its unit's and ten's digit are x and y respectively, write the linear equation representing the above statement.
Solution
Total original number is 10y + x
The new number is obtained after reversing the order of digits is 10x + y
According to question
⇒ 10y + x = 10x + y + 27
⇒ 9y  9x = 27
⇒ y x = 3
⇒ x  y + 3 = 0
This is the required linear equation for the given information.
6. The sum of a two digit number and the number obtained by reversing the order of its digits is 121. If units and ten's digit of the number are x and y respectively then write the linear equation representing the above statement.
Solution
Total original number is 10y + x
The new number is obtained after reversing the order of digits is (10x y)
According to problem
(10y +x) + (10x +y) = 121
⇒ 11x + 11y = 121
⇒ 11(x + y) = 121
⇒ x + y = 11
Thus is the required linear equation for the given information.
7. Plot the points (3, 5) and (1, 3) on a graph paper and verify that the straight line passing through these points also passes through the point (1, 4).
Solution
The points given in the graph :
It is clear from the graph the straight lines passes through these points also pass a through (1, 4).
8. From the choices given below, choose the equation whose graph is given in Fig. below.
(i) y = x
(ii) x + y = 0
(iii) y = 2x
(iv) 2 + 3y = 7x
[Hint : Clearly, (1, 1) and (1,  1) satisfy the equation x + y = 0 ]
Solution
Clearly (1, 1) and (1, 1) satisfy the equation x + y = 0
∴ The equation whose graph is given by x + y = 0
9. From the choices given below, choose the equation whose graph is given in fig. below.
(i) y = x + 2
(ii) y = x  2
(iii) y = x + 2
(iv) x + 2y = 6
[Hint : Clearly, (2, 0) and (1, 3) satisfy the equation y = x +2 ]
Solution
Clearly (2, 0) and (1, 3) satisfy the equation y = x + 2
∴ The equation whose graph is given by y = x + 2
10. If the point (2, 2) lies on the graph of the linear equation 5x + ky = 4 , find the value of k.
Solution
It is given that (2, 2) is a solution of the equation 5x + ky = 4
∴ 5 × 2 + k ×(2) = 4
⇒ 10  2k = 4
⇒ 2k = 4  10
⇒ 2k 6
⇒ k = 3
11. Draw the graph of the equation 2x + 3y = 12. From the graph, find the coordinates of the point:
(i) whose ycoordinates is 3.
(ii) whose x  coordinate is 3.
Solution
Graph of the equation 2x + 3y = 12.
We have,
2x + 3y = 12
⇒ 2x = 12  3y
⇒ x = (12  3y)/2
Putting y = 2, we get x = (12 3×2)/2 = 3
Putting y = 4, we get x = (12  3×4)/2 = 0
Thus, (3, 0) and (0, 4) are two points on the line 2x + 3y = 12
The graph of line represents by the equation 2x + 3y = 12
X 
0 
3 
Y 
4 
2 
(ii) x + 4y = 8
(iii) 2x + y = 6
(iv) 3x + 2y + 6 = 0
⇒ 3(2x  y) = 12
⇒ 2x  y = 4
⇒ 2x  4 = y
⇒ y = 2x  4 ...(i)
Putting x = 0 in (i), we get y = 4
Putting x = 2 in (i), we get y = 0
Thus, we obtain the following table giving coordinates of two points on the line represented by the equation 6x  3y = 12.
The graph of the line 6x  3y = 12
⇒ 4y  8 = x
⇒x = 4y = 8
Putting y = 1 in (i), we get x = 4 × 1  8 = 4
Putting y = 2 in (i), we get x = 4 ×2  8 = 0
Thus, we obtain the following table giving coordinates of two points on the line represented by the equation x + 4y = 8
Graph of the equation x + 4y = 8
2x + y = 6
⇒ y = 6 2x ...(i)
Putting x = 3 in (i), we get y = 6 = 2 × 3 = 0
Putting x = 4 in (i), we get y = 6  2× 4 = 2
Thus, we obtain the following table giving coordinate of two points on the line represented by the equation 2x + y = 6
Graph of the equation 2x + y = 6
⇒ 2y = 6  3x
⇒ y = (6  3x)/2
Putting x = 2 in (i), we get x = [6  3(2)]/2 = 0
Putting x = 4 in (i), we get y = [6  3(4)]/2 = 3
Thus, we obtain the following table giving coordinates of two points on the line represented by the equation 3x + 2y + 6 = 0
Graph of the equation 3x  2y + 6 = 0
y = 6  2x ...(i)
Putting x = 3 in (i), we get y = 6  2 × 3 = 0
Putting x = 0 in (i), we get y = 6  2 × 0 = 6
Thus, we obtained the following table giving coordinates of two points on the line represented by the equation 2x + y = 6
X 
3 
0 
Y 
0 
6 
⇒ 4x + 3y = 12
⇒ 4x = 12 3y
⇒ x = (12  3y)/4
Putting y = 0 in (i), we get x = (12  3×0)/4 = 3
Thus, we obtained the following table giving coordinate of two points on the line represents by the equation x/3 + y/4 = 1.
X  0  3 
Y  4  0 
Putting x = 0, we get y = 0
Putting x = 2, we get y = 2
Putting x = 2, we get y = 12
X 
0 
2 
2 
Y 
0 
2 
2 
Putting x = 0, we get y = 2 ...(ii)
Putting x = 1, we get y = 3
Putting x = 1, we get y = 3
X  0  1  1 
Y  2  3  3 
Find the coordinates of the vertices of the triangle formed by the two straight lines and the y  axis. Also, find the area of the triangle.
We have
⇒ x = (12  3y)/2
Putting y = 4, we get x = (12  3× 4)/2 = 0
Putting y = 2, we get x = (12  3×2)/2 = 3
Thus, we have the following table for the P table for the points on the line 2x + 3y = 12
X  0  3 
Y  4  2 
we obtain graph of the equation.
Thus, we have the following table for the points the line x  y = 1
X  1  0 
Y  0  1 
The graph of time 2x + 3y = 12 intersect with y  axis at B(0, 4) and the graph of the line x  y = 1 intersect with y  axis at C(0, 1).
So, the vertices of the triangle formed by thee two straight lines and y  axis are A(3, 2) and B(0, 4) and C(0, 1)
Now,
⇒ 4x  3y = 4
Putting y = 0, we get x = (3 × 0  4)/4 =  1
Putting y = 4, we get x = (3 × 4  4)/ 4 = 2
Thus, we have the following table for the p table for the points on the line 4x  3y + 4 = 0
X  1  2 
Y  0  4 
⇒ 4x = 20  3y
Putting y = 4, we get x = (20  3 × 4)/4 = 2.
Thus, we have the following table for the p table for the points on the line 4x  3y  20 = 0
X  0  2 
Y  0  4 
Clearly, two lines intersect at A(2, 4)
The graph of the lines 4x  3y + 4 = 0 and 4x + 3y  20 = 0 intersect with y  axis at a + b(1, 0) and c(5, 0) respectively
⇒ 3x = 12  4y
⇒ 3x = (12  4y)/3
⇒ 7x  49 = y  7
⇒ 7x  42 = y ...(i)
It is also given that after three years from now Ravish shall be three times a sold as her daughter
Now, y = 7x  42 [using (i)]
Putting x = 5, we get y = 7× 5  42 = 7
Thus, we have following table for the points on the
(i) 2(1/2) Hours
We know that speed= distance/speed
⇒ y = 60x
Putting x = 1, we get y = 60
Putting x = 2, we get y = 120
Thus, we have the following table for the points on the line y = 60x