2元一次方程

2x - 5y + 1 = 3y + 5x - 13 = 0

可寫為
2x - 5y + 1 = 0 ……. eqn 1
3y + 5x - 13 = 0……. eqn 2

從 eqn 1,
2x - 5y + 1 = 0
x = (5y -1)/2

把 x=(5y -1)/2 代入 eqn 2
3y + 5[(5y -1)/2] – 13 = 0
6y + 25y – 5 – 26 = 0
31y = 31
y = 1

把 y = 1代入x = (5y -1)/2
x = [5(1) – 1)/2
x = 2

得出 x= 2, y=1

We can both use subsitution and elimation to solve. Let me show you both.
Subsitution:

THe simultaneous equation can be written as the following,

2x-5y+1=0 equation1
3y+5x-13=0 equation2

From1,
2x-5y=-1
2x =-1+5y
x = (-1+5y)/2

Sub. x = (-1+5y)/2 into equation 2.
3y+5[(-1+5y)/2]-13=0
3y+ (-5+25y)/2 = 13
(-5+25y)/2 = 13-3y
-5+25y = 26-6y
25y = 31-6y
31y =31
y =1

Sub. y=1 into 1
2x-5+1=0
2x-4 =0
x =2

The solution is x=2,y=1.

Elimalation:

THe simultaneous equation can be written as the following,
2x-5y+1=0 equation1
3y+5x-13=0 equation2

5^equation1, 10x-25y+5=0 equation3
2^equation2, 6y+10x-26=0 equation4

equation4-equation3,
(6y+10x-26)-(10x-25y+5)=0
31y-31 =0
y =1

Sub. y=1 into equation1
2x-5+1=0
2x-4 =0
x =2
The solution is x=2,y=1.

Both methods can solve the question, the key point is to break up the equation into two. Then, you can solve the problem.(Hope you don't mind that I use English to answer the question, since I study in an EMI school., sorry. The answer is the same.)
P.S. 代入is subsitution.