Question

用代入法解方程组4(x+2)+5y=1，2x+3(y+2)=3？

Answer 1

∵ 4(x+2)+5y=1
∴ 5y = 1 - 4(x+2)
= 1 - 4x - 8
= -4x-7
那么
2x+3(y+2)=3
10x + 5*3(y+2) = 15
10x + 3*5y + 30 = 15
10x + 3*(-4x-7) + 30 = 15
10x - 12x - 21 + 30 = 15
-2x + 9 = 15
-2x = 6
x = -3
则
5y = -4*(-3) - 7 = 5
y = 1

Answer 2

4(x+2)+5y=1 4x+8+5y=1 x=(-7-5y)/4
代入2x+3(y+2)=3
2(-7-5y)/4+3(y+2)=3
2(-7-5y)+12(y+2)=12
-14-10y+12y+24=12
2y=2
y=1
x=(-7-5)/4=-3

Answer 3

用代入法解方程组4(x+2)+5y=1，2x+3(y+2)=3？
4(x+2)+5y=1
2x+3(y+2)=3
4x+8+5y=1
2x+3y+6=3
4x+5y=-7---------1
2x+3y=-3---------2
2式乘-2得：
-4x-6y=6----------3
1式+3式得：
y=1代入2式得
x=-3
x=-3
y=1