解方程组｛2分之2x+1+2分之4y-3=2 3(2x+1)-2(4y-3)=5用代入法

令 2x+1=u, 4y-3=v. 则原方程组变为: u/2 + v/2 =2, 3u - 2v = 5, 即
u + v =4, 3u - 2v = 5. 由第一个方程得 v = 4 - u, 带入第二个方程：3u - 2(4 - u) = 5,
3u - 8 + 2u = 5, 5u = 13, u = 13/5. 带入 v = 4-u 得：v = 4 - 13/5 = 7/5. 所以
2x + 1 = 13/5, 4y - 3 = 7/5, x = 4/5, y = 11/10.

x=4/5,y=11/10(可将上式化简)

(2x+1)/2 + (4y-3)/2 = 2 ...(1)
3(2x+1) - 2(4y-3) = 5 ...(2)
由（1）得：
(2x+1) + (4y-3)= 4
(4y-3)= 4 - (2x+1) ...(3)
将（3）代入（2）：
3(2x+1) - 2*{4 - (2x+1)} = 5
3(2x+1) - 8 + 2 (2x+1) = 5
5(2x+1) = 13
2x+1 = 13/5 .(4)
x = 4/5
将（4）代入（3）：
4y-3 = 4 - (2x+1) = 4 - 13/5 = 7/5
4y=3+7/5=22/5
y=11/10