x=y+3 (2)

Substituting (2) into (1) gives:

3(y+3)+2y= 14

3y+9+2y= 14

5y=5

y= 1

Substituting y= 1 into (2) gives:

x= 1+3=4

So the solution of the original equations is: x=4.

y= 1。

(2) 5x+2y=25 ( 1)

3x+4y= 15 (2)

( 1) x2-(2) De:

7x=35

x=5

Substituting x=5 into (1) gives:

25+2y=25

y=0

So the solution of the original equations is: x=5.

y=0。