Express the following linear equations in the form ax + by + c = 0 and indicate the values of a, b and c in each case:

Express the following linear equations in the form ax + by + c = 0 and indicate the values of a, b and c in each case:

(i) 2x + 3y = 9.35 (ii) x - y/5 - 10 = 0 (iii) -2x + 3y = 6 (iv) x = 3y

(v) 2x = -5y (vi) 3x + 2 = 0 (vii) y - 2 = 0 (viii) 5 = 2x

**Answer**

(i) 2x + 3y = 9.35

⇒ 2x + 3y - 9.35 = 0

On comparing this equation with ax + by + c = 0, we get

a = 2x, b = 3 and c = -9.35

(ii) x - y/5 - 10 = 0

On comparing this equation with ax + by + c = 0, we get

a = 1, b = -1/5 and c = -10

(iii) -2x + 3y = 6

⇒ -2x + 3y - 6 = 0

On comparing this equation with ax + by + c = 0, we get

a = -2, b = 3 and c = -6

(iv) x = 3y

⇒ x - 3y = 0

On comparing this equation with ax + by + c = 0, we get

a = 1, b = -3 and c = 0

(v) 2x = -5y

⇒ 2x + 5y = 0

On comparing this equation with ax + by + c = 0, we get

a = 2, b = 5 and c = 0

(vi) 3x + 2 = 0

⇒ 3x + 0y + 2 = 0

On comparing this equation with ax + by + c = 0, we get

a = 3, b = 0 and c = 2

(vii) y - 2 = 0

⇒ 0x + y - 2 = 0

On comparing this equation with ax + by + c = 0, we get

a = 0, b = 1 and c = -2

(viii) 5 = 2x

⇒ -2x + 0y + 5 = 0

On comparing this equation with ax + by + c = 0, we get

a = -2, b = 0 and c = 5

y = 3x + 5 has

(i) a unique solution, (ii) only two solutions, (iii) infinitely many solutions

**Answer**

Since the equation, y = 3x + 5 is a linear equation in two variables. It will have (iii) infinitely many solutions.

(i) 2x + y = 7 (ii) πx + y = 9 (iii) x = 4y

**Answer**

(i) 2x + y = 7

⇒ y = 7 - 2x

→ Put x = 0,

y = 7 - 2 × 0 ⇒ y = 7

(0, 7) is the solution.

→ Now, put x = 1

y = 7 - 2 × 1 ⇒ y = 5

(1, 5) is the solution.

→ Now, put x = 2

y = 7 - 2 × 2 ⇒ y = 3

(2, 3) is the solution.

→ Now, put x = -1

y = 7 - 2 × -1 ⇒ y = 9

(-1, 9) is the solution.

The four solutions of the equation 2x + y = 7 are (0, 7), (1, 5), (2, 3) and (-1, 9).

(ii) πx + y = 9

⇒ y = 9 - πx

→ Put x = 0,

y = 9 - π×0 ⇒ y = 9

(0, 9) is the solution.

→ Now, put x = 1

y = 9 - π×1 ⇒ y = 9-π

(1, 9-π) is the solution.

→ Now, put x = 2

y = 9 - π×2 ⇒ y = 9-2π

(2, 9-2π) is the solution.

→ Now, put x = -1

y = 9 - π× -1 ⇒ y = 9+π

(-1, 9+π) is the solution.

The four solutions of the equation πx + y = 9 are (0, 9), (1, 9-π), (2, 9-2π) and (-1, 9+π).

(iii) x = 4y

→ Put x = 0,

0 = 4y ⇒ y = 0

(0, 0) is the solution.

→ Now, put x = 1

1 = 4y ⇒ y = 1/4

(1, 1/4) is the solution.

→ Now, put x = 4

4 = 4y ⇒ y = 1

(4, 1) is the solution.

→ Now, put x = 8

8 = 4y ⇒ y = 2

(8, 2) is the solution.

The four solutions of the equation πx + y = 9 are (0, 0), (1, 1/4), (4, 1) and (8, 2).

Check which of the following are solutions of the equation x - 2y = 4 and which are not:

Check which of the following are solutions of the equation x - 2y = 4 and which are not:

(i) (0, 2) (ii) (2, 0) (iii) (4, 0) (iv) (√2, 4√2) (v) (1, 1)

**Answer**

(i) Put x = 0 and y = 2 in the equation x - 2y = 4.

0 - 2×2 = 4

⇒ -4 ≠ 4

∴ (0, 2) is not a solution of the given equation.

(ii) Put x = 2 and y = 0 in the equation x - 2y = 4.

2 - 2×0 = 4

⇒ 2 ≠ 4

∴ (2, 0) is not a solution of the given equation.

(iii) Put x = 4 and y = 0 in the equation x - 2y = 4.

4 - 2×0 = 4

⇒ 4 = 4

∴ (4, 0) is a solution of the given equation.

(iv) Put x = √2 and y = 4√2 in the equation x - 2y = 4.

√2 - 2×4√2 = 4 ⇒ √2 - 8√2 = 4 ⇒ √2(1 - 8) = 4

⇒ -7√2 ≠ 4

∴ (√2, 4√2) is not a solution of the given equation.

(v) Put x = 1 and y = 1 in the equation x - 2y = 4.

1 - 2×1 = 4

⇒ -1 ≠ 4

∴ (1, 1) is not a solution of the given equation.