# Linear Equations

## Exercise 4.3

### Part 1

Question 1: Draw the graph of each of the following linear equation in two variables:

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

**Solution:** (i) The given equation is

x + y = 4

y = 4 – x …………….equation (1)

Now , putting the value x = 0 in equation (1)

y = 4 – 0 = 4. So the solution is (0, 4)

Putting the value x = 1 in equation (1)

y = 4 – 1 = 3. So the solution is (1, 3)

Putting the value x = 2 in equation (1)

y = 4 – 2 = 2. So the solution is (2, 2)

So, the table of the different solutions of the equation is

x | 0 | 1 | 2 |

y | 4 | 3 | 2 |

**Solution:**(ii) The given equation is

x - y = 2

x = 2 + y…………….equation (1)

Now , putting the value y = 0 in equation (1)

x = 2 + 0 = 2. So the solution is (2, 0)

Putting the value y = 1 in equation (1)

x = 2 + 1 = 3. So the solution is (3, 1)

Putting the value y = 2 in equation (1)

x = 2 + 2 = 4. So the solution is (4, 2)

So, the table of the different solutions of the equation is

**Solution:**(iii) The given equation is

y = 3x

y = 3x …………….equation (1)

Now , putting the value x = 0 in equation (1)

y = 3 x 0 = 0. So the solution is (0, 0)

Putting the value x = 1 in equation (1)

y = 3 x 1 = 3. So the solution is (1, 3)

Putting the value x = 2 in equation (1)

y = 3 x 2 = 6. So the solution is (2, 6)

So, the table of the different solutions of the equation is

**Solution:**(iv) The given equation is

3 = 2 x + y

2 x + y = 3

y = 3 – 2x …………….equation (1)

Now, putting the value x = 0 in equation (1)

y = 3 – 2 x 0

y = 3 – 0 = 3. So the solution is (0, 3)

Putting the value x = 1 in equation (1)

y = 3 – 2 x 1

y = 3 – 2 = 1. So the solution is (1, 1)

Putting the value x = 2 in equation (1)

y = 3 – 2 x 2

y = 3 – 4 = -1. So the solution is (2, -1)

So, the table of the different solutions of the equation is

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

**Solution:** Since the given solution is (2, 14)

Therefore, x = 2 and y = 14

Or, One equation is x + y = 2 + 14 = 16

x + y = 16

Second equation is x – y = 2 – 14 = -12

x - y = -12

Third equation is y = 7x

0 = 7x – y

7x - y = 0

In fact we can find infinite equations because infinite number of lines pass through one point.

Question 3: If the point (3, 4) lies on the graph of the equation 3y = ax + 7, find the value of a?

**Solution:** The given equation is

`3y = ax + 7` ……………….equation (1)

According to problem, point (3, 4) lie on it.

So, putting the value `x = 3` and `y = 4` in equation (1)

`3 xx 4 = a xx 3 + 7`

`12 = 3a + 7`

`12 – 7 = 3a`

`5 = 3a`

Or, `5/3=a`

Now, putting the value of ‘a’ in equation (1)

`3y=(5/3)x+7`

Now, for `x=3` and `y=4`

LHS `=3y=3xx4=12`

RHS, `(5/3)x+7=(5/3)xx3+7=5+7=12`

Hence, LHS=RHS

Or, `3y=(5/3)x+7`

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

**Solution:** Given, Taxi fare for first kilometer = Rs. 8

Taxi fare for subsequent distance = Rs. 5

Total distance covered = x

Total fare = y

Since the fare for first kilometer = Rs. 8

According to problem, Fare for (x – 1) kilometer = 5(x-1)

So, the total fare = 5(x-1) + 8

y = 5(x-1) + 8

y = 5x – 5 + 8

y = 5x + 3

Hence, y = 5x + 3 is the required linear equation.

Now the equation is; y = 5x + 3 …………….equation (1)

Now, putting the value x = 0 in equation (1)

y = 5 x 0 + 3

y = 0 + 3 = 3 So the solution is (0, 3)

Putting the value x = 1 in equation (1)

y = 5 x 1 + 3

y = 5 + 3 = 8. So the solution is (1, 8)

Putting the value x = 2 in equation (1)

y = 5 x 2 + 3

y = 10 + 3 = 13. So the solution is (2, 13)

So, the table of the different solutions of the equation is

Question 5: From the choices given below, choose the equation whose graphs are given in Fig. 1 and Fig. 2.

(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

**Solution:** From the given Fig. 1, the solutions of the equation are (-1, 1), (0, 0) and (1, -1)

Therefore the equation which satisfies these solutions is the correct equation.

Equation (ii) x + y = 0, satisfies these solutions.

**Proof:** Putting the value x = -1 and y = 1 in the equation x + y = 0

L.H.S = x + y = -1 + 1 = 0 = R.H.S

Putting the value x = 0 and y = 0

L.H.S = x + y = 0 + 0 = 0 = R.H.S

Putting the value x = 1 and y = -1

L.H.S = x + y = 1 + (-1) = 1 – 1 = 0 = R.H.S

Hence, option (ii) x + y = 0 is correct.

From the given Fig. 2, the solutions of the equation are (-1, 3), (0, 2) and (2, 0)

Therefore the equation which satisfies these solutions is the correct equation.

Equation (i) y = -x + 2 , satisfies these solutions.

**Proof:** Putting the value x = -1 and y = 3 in the equation y = -x + 2

x + y = 2

L.H.S = x + y = -1 + 3 = 2 = R.H.S

Putting the value x = 0 and y = 2

L.H.S = x + y = 0 + 2 = 2 = R.H.S

Putting the value x = 2 and y = 0

L.H.S = x + y = 2 + 0 = 2 = R.H.S

Hence, option (i) y = -x + 2 is correct.