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。
Summary of school-based activities 1
In the first semester of the 20xx school year, under the unified leadership of the school, with the guidance and h