**Question 1: **Solve the system of linear equations:

3x + 4y = 12

4x + 2y = 10.

** Solution: **

Given equations are

3x + 4y = 12 ......(1)

4x + 2y = 10 ......(2)

Multiply equation (1) by 4 and equation (2) by 3 to get

12x + 16y = 48

12x + 6y = 30

$\overline{0 + 10 y = 18}$

=> y = $\frac{18}{10}$ or y = $\frac{9}{5}$.

Substitute the value of y in equation (1) to get

3x + 4 $\frac{9}{5}$ = 12

3x + $\frac{36}{5}$ = 12

3x = 12 - $\frac{36}{5}$

= $\frac{12 \times 5 - 36}{5}$

= $\frac{60 - 36}{5}$

3x = $\frac{24}{5}$

or

x = $\frac{24}{15}$

The value of variable x = $\frac{24}{5}$ and y = $\frac{9}{5}$ or (x,y) = (1.6, 1.8).

**Question 2: **Solve the system of linear equations:

5x + 2y = 10

3x + 4y = 6

** Solution: **

Given equations are

5x + 2y = 10 .........(1)

3x + 4y = 6 ...........(2)

Multiply equation (1) by 3 and equation (2) by 5 to get

15x + 6y = 30

15x + 20y = 30

$\overline{\ 0 - 14y = 0}$
=> y = 0.

Substitute the value of y in equation (1) to get

5x + 0 = 10
or
x = $\frac{10}{5}$
= 2

The value of variable x = 2 and y = 0. The solution is (x,y) = (2,0).