# Conditions For Solvability (Or Consistency) Of System Of Equation

## Pair Of Linear Equations In Two Variables of Class 10

### UNIQUE SOLUTION:

Two lines a_{1 }+ b_{1}y + c_{1} = 0 and a_{2}x + b_{0}y + c_{2} = 0, if the denominator a_{1}b_{2} – a_{2}b_{1} ≠ 0 then the given system of equations has unique solution (i.e. only one solution) and the solutions are said to be consistent.

∴ a_{1}b_{2} – a_{2}b_{1} ≠ 0

⇒

**Question** **1.** Find the value of ‘P’ for which the given system of equations has only one solution (i.e. unique solution).

Px – y = 2 ....(i)

6x – 2y = 3 ....(ii)

**Solution: ** a_{1} = P, b_{1 }= –1, c_{1 }= –2

a_{1} = 6 b_{2} = –2, c_{2} = –3.

Conditions for unique solution is

⇒

∴ P can have all real values except 3.

### NO SOLUTION:

Two lines a_{1}x + b_{1}y + c_{1} = 0 and a_{2}x + b_{2}y + c_{2} = 0, if the denominator a_{1}b_{2 }– a_{2}b_{1} = 0 then the given system of equations has no solution and solutions are said to be consistent.

∴

**Question 1.** Determine the value of k so that the following linear equations has no solution.

(3x + 1) x + 3y - 2 = 0

(k2 + 1) x + (k - 2) y – 5 = 0.

Sol. Here a_{1} = 3k + 1, b_{1 }= 3 and c_{1} = –2

a_{2} = k_{2} + 1, b_{2} = k – 2 and c_{2} = –5

For no solution, condition is

.

⇒ (3k + 1) (k – 2) = 3(k_{2} + 1)

⇒ 3k_{2} – 5k – 2 = 3k_{2} + 3

⇒ –5k – 2 = 3

⇒ –5k = 5

⇒ k = –1

Clearly, 3/k - 2 ≠ 2/5 for k = –1.

Hence, the given system of equations will has no solution for k = - 1.

### MANY SOLUTION (INFINITE SOLUTIONS):

Two lines a_{1}x + b_{1}y + c_{1} = 0 and a_{2}x + b_{2}y + c_{2} = 0, if then system of equations has many solution and solutions are said to be consistent.

**Question 1.** Find the value of k for which the system of linear equation

kx + 4y = k – 4

16x + ky = k has infinite solution.

**Solution: **a_{1 }= k, b_{1 }= 4, c_{1} = –(k – 4)

a_{2} = 16, b_{2 }= k, c2 = – k.

Here condition is

⇒ k/16 = 4/k = (k - 4)/(k)

⇒ k/16 = 4/k also 4/k = k - 4/k

⇒ k_{2} = 64 ⇒ 4k = k_{2} – 4k

⇒ k = ± 8 ⇒ k(k – 8) = 0.

k = 0 or k = 8 but k = 0 is not possible otherwise equation will be one variable.

∴ k = 8 is correct value for infinite solution.