#1**+1 **

Solve the following system:

{2 x + 3 y = 13 | (equation 1)

12 x - 3 y = -2 | (equation 2)

Swap equation 1 with equation 2:

{12 x - 3 y = -2 | (equation 1)

2 x + 3 y = 13 | (equation 2)

Subtract 1/6 × (equation 1) from equation 2:

{12 x - 3 y = -2 | (equation 1)

0 x+(7 y)/2 = 40/3 | (equation 2)

Multiply equation 2 by 6:

{12 x - 3 y = -2 | (equation 1)

0 x+21 y = 80 | (equation 2)

Divide equation 2 by 21:

{12 x - 3 y = -2 | (equation 1)

0 x+y = 80/21 | (equation 2)

Add 3 × (equation 2) to equation 1:

{12 x+0 y = 66/7 | (equation 1)

0 x+y = 80/21 | (equation 2)

Divide equation 1 by 12:

{x+0 y = 11/14 | (equation 1)

0 x+y = 80/21 | (equation 2)

**x = 11/14 and y = 80/21**

Guest Sep 29, 2017

#2**+1 **

2x+3y=13 ⇒ 3y = 13 - 2x (1)

12x-3y=-2 ⇒ 3y = 2 + 12x (2)

Set (1) = (2)

13 - 2x = 2 + 12x add 2x to both sides, subtract 2 from both sides

11 = 14x divide both sides by 14

11/ 14 = x

Putting this into (1), we have

3y = 13 - 2 (11/ 14)

3y = 13 - 11/7 get a common denominator

3y = 91/7 - 11/7

3y = 80/7 divide both sides by 3

y = 80/ 21

CPhill Sep 29, 2017