Solving a pair of equations indicate that those pairs are simultaneous linear equations. By solving the equations, you can say that those equations are the system of linear equations and consistent or not. Get the solved example questions along with solutions in the below-mentioned sections of this page.

## Pair of Equations Problems

**Example 1.**

If solvable, solve the following pairs of equations:

(i) 3x + 2y = 5, x + 4y = 7

(ii) 3x – 2y = 4, 18x – 12y – 24 = 0

**Solution:**

(i) 3x + 2y = 5, x + 4y = 7

Here, comparing the co-efficient of x, y we get;

3/1 ≠ 2/4

Therefore, subtracting the two equations, we get the general solution.

6x – 10 = x – 7

5x = -7 + 10

5x = 3

x = 3/5

Substitute x = 3/5 in x + 4y = 7

3/5 + 4y = 7

4y = 7 – 3/5 = 32/5

y = 32/20

Therefore, the required solution is x = 3/5, y = 32/20.

(ii) 3x – 2y = 4, 18x – 12y – 24 = 0

Here, comparing the co-efficient of x, y we get;

3/18 = -2/-12 = -4/-24

Therefore, two equations are, in fact, same.

Assuming x = c in 3x – 2y = 4 we get;

y = (3c – 4)/2

Therefore, required solution: x = c

y = (3c – 4)/2 for any real value of c.

**Example 2.**

From the following pairs of equations find the pair or pairs representing simultaneous equations:

(i) 5x + 2y – 6 = 0, 2x + y – 7 = 0

(ii) 6x + 4y + 9 = 0, 3x + 2y +7 = 0

**Solution:**

(i) 5x + 2y – 6 = 0, 2x + y – 7 = 0

5/2 ≠ 2/1 ≠ -6/-7

So the two equations represent simultaneous equations; in this case they have only one solution.

(ii) 6x + 4y + 9 = 0, 3x + 2y +7 = 0

6/3 = 4/2 ≠9/7

So, the given equations are not simultaneous equations.

**Example 3.**

For which value of k, kx + y = 5 and x + ky = 1 are inconsistent?

**Solution:**

The two equations will be inconsistent if k/1 = 1/k ≠ 5/1 that means, k² = 1 or k = ±1

Therefore, the two given equations will be inconsistent if k = ±1.

**Example 4.**

From the following pairs of equations find the pair or pairs representing simultaneous equations:

(i) x + y = 5, x – 2y = 11

(ii) 3x – 4y + 2 = 0, 12x – 16y + 8 = 0

**Solution:**

(i) x + y = 5, x – 2y = 11

1/1 ≠ 1/-2 ≠ -5/-11

So the two equations represent simultaneous equations; in this case they have only one solution.

(ii) 3x – 4y + 2 = 0, 12x – 16y + 8 = 0

3/12 = -2/-16 = 2/8

So, these are simultaneous equations and have infinite solutions.

**Example 5.**

If solvable, solve the following pairs of equations:

(i) 5x + 7y = 9, x – 2y = 3

(ii) 3x – 2y = 2, 9x – 6y = 6

**Solution:**

(i) 5x + 7y = 9, x – 2y = 3

Here, comparing co-efficient of x and y, we get;

5/1 ≠ 7/-2

Therefeore, solve the value of x,

x = 39/17

Put x = 39/17 in x – 2y = 3

39/17 – 3 = 2y

(39 – 51)/17 = 2y

2y = -12/17

y = -12/34

Therefore, the required solution is x = 39/17, y = -12/34

(ii) 3x – 2y = 2, 9x – 6y = 6

Here, comparing co-efficient of x and y, we get;

3/9 = -2/-6 = 2/6

Therefore, two equations are, in fact, same.

Assuming x = c in 3x – 2y = 2

3c – 2y = 2

(3c – 2) = 2y

y = (3c – 2)/2

y = (3c – 2)/2 for any real value of c.