Chapter 4: Linear Equations in Two Variables

To draw a graph of linear equations in two variables, let us find out the points to plot.

To find out the points, we have to find the values which x and y can have, satisfying the equation.

Here,

x+y = 4

Substituting the values for x,

When x = 0,

x+y = 4

0+y = 4

y = 4

When x = 4,

x+y = 4

4+y = 4

y = 4–4

y = 0

The points to be plotted are (0, 4) and (4,0)

(ii) x–y = 2

To draw a graph of linear equations in two variables, let us find out the points to plot.

To find out the points, we have to find the values which x and y can have, satisfying the equation.

Here,

x–y = 2

Substituting the values for x,

When x = 0,

x–y = 2

0 – y = 2

y = – 2

When x = 2,

x–y = 2

2–y = 2

– y = 2–2

–y = 0

y = 0

The points to be plotted are (0, – 2) and (2, 0)

(iii) y=3x

To draw a graph of linear equations in two variables, let us find out the points to plot.

To find out the points, we have to find the values which x and y can have, satisfying the equation.

Here,

y = 3x

Substituting the values for x,

When x = 0,

y = 3x

y = 3×0

y = 0

When x = 1,

y = 3x

y = 3×1

y = 3

The points to be plotted are (0, 0) and (1, 3)

(iv) 3 = 2x+y

To draw a graph of linear equations in two variables, let us find out the points to plot.

To find out the points, we have to find the values which x and y can have, satisfying the equation.

Here,

3 = 2x+y

Substituting the values for x,

When x = 0,

3 = 2x+y

3 = 2×0+y

3 = 0+y

y = 3

When x = 1,

3= 2x+y

3 = 2×1+y

3 = 2+y

y = 3–2

y = 1

The points to be plotted are (0, 3) and (1, 1)

We know that an infinite number of lines pass through a point.

The equation of 2 lines passing through (2,14) should be in such a way that it satisfies the point.

Let the equation be 7x = y

7x–y = 0

When x = 2 and y = 14

(7×2)-14 = 0

14–14 = 0

0 = 0

L.H.S. = R.H.S.

Let another equation be 4x = y-6

4x-y+6 = 0

When x = 2 and y = 14

(4×2–14+6 = 0

8–14+6 = 0

0 = 0

L.H.S. = R.H.S.

Since both the equations satisfy the point (2,14), then we can say that the equations of two lines passing through (2, 14) are 7x = y and 4x = y-6

We know that an infinite number of line passes through one specific point. Since there is only one point (2,14) here, there can be infinite lines that pass through the point.

The given equation is

3y = ax+7

According to the question, x = 3 and y = 4

Now, substituting the values of x and y in the equation 3y = ax+7,

We get,

(3×4) = (a×3)+7

⟹ 12 = 3a+7

⟹ 3a = 12–7

⟹ 3a = 5

⟹ a = 5/3

The value of a, if the point (3,4) lies on the graph of the equation 3y = ax+7 is 5/3.

Given,

Total distance covered = x

Total fare = y

Fare for the first kilometre = 8 per km

Fare after the first 1km = 5 per km

If x is the total distance, then the distance after one km = (x-1)km

i.e., fare after the first km = 5(x-1)

According to the question,

The total fare = Fare of first km+ fare after the first km

y = 8+5(x-1)

y = 8+5(x-1)

y = 8+5x – 5

y = 5x+3

Solving the equation,

When x = 0,

y = 5x+3

y = 5×0+3

y = 3

When y = 0,

y = 5x+3

o = 5x+3

5x = -3

x = -3/5

The points to be plotted are (0, 3) and (-3/5, 0)

The points given in figure 4.6 are (0,0), (-1,1), (1,-1)

Substituting the values for x and y from these points in the equations, we get,

(i) y = x

(0,0) ⟹ 0 = 0

(-1, 1) ⟹ -1 ≠ 1 ————————— equation not satisfied

(1, -1) ⟹ 1≠ -1 ————————— equation not satisfied

(ii) x+y = 0

(0,0) ⟹ 0+0 = 0

(-1, 1) ⟹ -1+1 = 0

(1, -1) ⟹ 1+(-1) =0

(iii) y = 2x

(0,0) ⟹ 0 = 2×0

0 = 0

(-1, 1) ⟹ 1 = 2×(-1)

1≠ -2 ————————— equation not satisfied

(1, -1) ⟹ -1 = 2×1

-1 ≠ 2 ————————— equation not satisfied

(iv) 2+3y = 7x

(0,0) ⟹ 2+(30) = 7×0

2 ≠ 0 ————————— equation not satisfied

(-1, 1) ⟹ 2+(3×1) = 7×-1

5 ≠ -7 ————————— equation not satisfied

(1, -1) ⟹ 2+(3×-1) = 7×1

-1 ≠ 7 ————————— equation not satisfied

Since only equation x+y = 0 satisfies all the points, the equation whose graphs are given in Fig. 4.6 is

x+y = 0

For Fig. 4. 7

(i) y = x+2

(ii) y = x–2

(iii) y = –x+2

(iv) x+2y = 6

The points given in figure 4.7 are (0,2), (2,0), (-1,3)

Substituting the values for x and y from these points in the equations, we get,

(i) y = x+2

(0,2) ⟹2 = 0+2

2 = 2

(2, 0) ⟹ 0= 2+2

0 ≠ 4 ————————— equation not satisfied

(-1, 3) ⟹ 3 = -1+2

3 ≠ 1 ————————— equation not satisfied

(ii) y = x–2

(0,2) ⟹ 2 = 0–2

2 ≠ -2 ————————— equation not satisfied

(2, 0) ⟹ 0 = 2–2

0= 0

(-1, 3) ⟹ 3= –1–2

3 ≠ –3 ————————— equation not satisfied

(iii) y = –x+2

(0,2) ⟹ 2 = -0+2

2 = 2

(2, 0) ⟹ 0 = -2+2

0 = 0

(-1, 3) ⟹ 3= -(-1)+2

3 = 3

(iv) x+2y = 6

(0,2) ⟹ 0+(2×2) = 6

4 ≠ 6 ————————— equation not satisfied

(2, 0) ⟹ 2+(2×0) = 6

2 ≠ 6 ————————— equation not satisfied

(-1, 3) ⟹ -1+(2×3) = 6

5 ≠ 6 ————————— equation not satisfied

Since only equation y = –x+2 satisfies all the points, the equation whose graphs are given in Fig. 4.7 is

y = –x+2

Let the distance travelled by the body be x and the force applied on the body be y.

It is given that,

The work done by a body is directly proportional to the distance travelled by the body.

According to the question,

y ∝ x

y = 5x (5 is a constant of proportionality)

Solving the equation,

(i) when x = 2 units,

then y = 5×2 = 10 units

(2, 10)

(ii) when x = 0 units,

then y = 5×0 = 0 units.

(0, 0)

The points to be plotted are (2, 10) and (0, 0)

Let Yamini’s donation be ₹x and Fatima’s donation be ₹y

According to the question,

x+y = 100

We know that,

when x  = 0 , y = 100

when x = 50, y = 50

when x = 100, y = 0

The points to be plotted are (0,100), (50,50), (100,0)

(i) According to the question,

F = (9/5)C + 32

Solving the equation,

We get,

When C = 0, F = 32

When C = -10 , F = 14

The points to be plotted are (0, 32), (-10, 14)

(ii) When C = 30,

F = (9/5)C +32

F = (9×30)/5+32

= (9×6)+32

= 54+32

= 86^oF

(iii) When F = 95,

95 = (9/5)C +32

(9/5)C = 95-32

(9/5)C =63

C = (63×5)/9

=35^oC

(iv) When C = 0,

F = (9/5)C +32

F = (9×0)/5 +32

=0+32

=32^oF

When F = 0,

0 = (9/5)C+32

(9/5)C = 0-32

(9/5)C = -32

C = (-32×5)/9

=-17.7777

=-17.8^oC

(v) When F = C,

C = (9/5)C+32

C – (9/5)C = 32

(5-9)C/5 =32

(-4/5)C = 32

(-4/5)C = (-32×5)/4

= – 40^oC

Hence, -40^o is the temperature which is numerically the same in both Fahrenheit and Celsius.