**NCERT Solutions Class 9 Mathematics**

**Exercise 4.3**

**Q.1.** Draw the graph of each of the following linear equations in two variables:

(i) x + y = 4 (ii) x - y = 2 (iii) y = 3x (iv) 3 = 2x + y

**Q.2.** Give the equations of two lines passing through (2, 14). How many more such lines are there, and why?

**Q.3.** If the point (3, 4) lies on the graph of the equation 3y = ax + 7, find the value of a.

**Q.4.** The taxi fare in a city is as follows: For the first kilometre, the fare is Rs 8 and for the subsequent distance it is Rs 5 per km. Taking the distance covered as x km and total fare as Rs y, write a linear equation for this information, and draw its graph.

**Q.5.** From the choices given below, choose the equation whose graphs are given in Fig. 4.6 and Fig. 4.7.

For Fig. 4. 6 |
For Fig. 4.7 | |||

(i) y = x |
(i) y = x + 2 | |||

(ii) x + y = 0 |
(ii) y = x - 2 | |||

(iii) y = 2x |
(iii) y = -x + 2 | |||

(iv) 2 + 3y = 7x |
(iv) x + 2y = 6 |

**Q.6.** If the work done by a body on application of a constant force is directly proportional to the distance travelled by the body, express this in the form of an equation in two variables and draw the graph of the same by taking the constant force as 5 units. Also read from the graph the work done when the distance travelled by the body is

(i) 2 units (ii) 0 unit

**Q.7**. Yamini and Fatima, two students of Class IX of a school, together contributed Rs 100 towards the Prime Minister’s Relief Fund to help the earthquake victims. Write a linear equation which satisfies this data. (You may take their contributions as Rs x and Rs y.) Draw the graph of the same.

F = $\left(\frac{9}{5}\right)$C + 32 - Draw the graph of the linear equation above using Celsius for x-axis and Fahrenheit for y-axis.
- If the temperature is 30°C, what is the temperature in Fahrenheit?
- If the temperature is 95°F, what is the temperature in Celsius?
- If the temperature is 0°C, what is the temperature in Fahrenheit and if the temperature is 0°F, what is the temperature in Celsius?
- Is there a temperature which is numerically the same in both Fahrenheit and Celsius? If yes, find it.
(i) x + y = 4 (ii) x - y = 2 (iii) y = 3x (iv) 3 = 2x + y
Given point is, x = 2 and y =14. Thus, x + y = 16 and also, y = 7x ⇒ 7x - y = 0 ⇒ The equations of two of the lines passing through (2, 14) are x + y = 16 and 7x - y = 0. - There will be infinitely many lines passing through the same point because infinite number of lines can pass through a given point.
The point (3, 4) lies on the graph of the equation 3y = ax + 7. Putting x = 3 and y = 4, we get, 3×4 = a×3 + 7 ⇒ 12 = 3a + 7 ⇒ 3a = 12 - 7 ⇒ a = $\frac{5}{3}$
Let total distance covered be x and the total fare be y.
In fig. 4.6, points on the line are (0, 0), (-1, 1) and (1, -1). If we put all these points in equation the four given equations, we see that LHS and RHS are equal for equation (ii) x + y = 0 for all the value of the points. Therefore this is the correct equation. In fig. 4.7, points on the line are (-1, 3), (0, 2) and (2, 0). If we put all these points in equation the four given equations, we see that LHS and RHS are equal for equation (iii) y = -x + 2 for all the value of the points. Therefore this is the correct equation.
(i) 2 units (ii) 0 unit
Let the distance traveled by the body be x and the work done by the force be y.
F = $\left(\frac{9}{5}\right)$C + 32 - Draw the graph of the linear equation above using Celsius for x-axis and Fahrenheit for y-axis.
- If the temperature is 30°C, what is the temperature in Fahrenheit?
- If the temperature is 95°F, what is the temperature in Celsius?
- If the temperature is 0°C, what is the temperature in Fahrenheit and if the temperature is 0°F, what is the temperature in Celsius?
- Is there a temperature which is numerically the same in both Fahrenheit and Celsius? If yes, find it.
(i) F = $\left(\frac{9}{5}\right)$C + 32 The solutions can be tabulated as follows,
Taking 1 unit = 10 on both x and y-axis (ii) Putting the value of C = 30 in F = $\left(\frac{9}{5}\right)\mathrm{}$C+ 32 , we get, F = $\left(\frac{9}{5}\right)$×30 + 32 = 54 + 32 = 86 (iii) Putting the value of F = 95 in F = $\left(\frac{9}{5}\right)\mathrm{}$C+ 32 , we get, 95 = $\left(\frac{9}{5}\right)\mathrm{}$C+ 32 ⇒C = (95 - 32)× $\left(\frac{5}{9}\right)$ = 63 × $\left(\frac{5}{9}\right)$ ⇒ C = 35 (iv) Putting the value of F = 0 in F = $\left(\frac{9}{5}\right)\mathrm{}$C+ 32 , we get 0 = $\left(\frac{9}{5}\right)\mathrm{}$C+ 32 ⇒ C = - 32× $\left(\frac{5}{9}\right)$ = 63 × $\left(\frac{5}{9}\right)$ ⇒ C = - $\frac{160}{9}$ Putting the value of C = 0 in F = $\left(\frac{9}{5}\right)\mathrm{}$C+ 32 , we get, F = $\left(\frac{9}{5}\right)\mathrm{}$× 0 + 32 = 32 (v) Putting F = C in F = $\left(\frac{9}{5}\right)\mathrm{}$C+ 32 , we get, F = $\left(\frac{9}{5}\right)\mathrm{}$F + 32 ⇒ F - $\left(\frac{9}{5}\right)\mathrm{}$F = 32 ⇒ - $\left(\frac{9}{5}\right)\mathrm{}$F = 32 ⇒ F = - 40 Therefore both Fahrenheit and Celsius numerically the same at -40 |