2x + y = 12 4x - 3y = 2

Given the equations:

2x+ y = 12

4x - 3y = 2

Solve for the systems of linear equation.

Solution:

Start first by finding the value of y for each equation. For the first equation, transpose 2x on the other side of the equation, hence we must take note that in transposing numbers, the sign changes.

2x + y = 12

y = 12 - 2x

For the second equation, transpose -3y and 2 to the opposite sides of the equation. Then divide both sides by 3 to cancel out 3y, leaving behind y.

4x - 3y =2

3y = 4x - 2

3y/3 = (4x - 2)/3

y = 4/3x - ⅔

Equate the two solutions into one single equation.

4/3x - ⅔ =12 - 2x

Transpose 2x and ⅔ to the opposite sides of the equation.

4/3x  + 2x = 12 + ⅔

4/3x + 6/3x = 36/3 + ⅔

10/3x = 38/3

Multiply both sides by the reciprocal of 10/3 which 3/10 to cancel out 10/3x, leaving behind x.

10/3x (3/10) = (3/10)(38/3)

x = 114/30

x = 3.8

Final answer:

x = 3.8