Question

解二元二次方程组 x²+y²=2 3x+4y-5=0

Answer 1

由第2式得：y=(5-3x)/4
代入1式得：x^2+(25-30x+9x^2)/16=2
即25x^2-30x-7=0
(5x-7)(5x+1)=0
x=7/5, -1/5
所以y1=(5-21/5)/4=1/5
y2=(5+3/5)/4=7/5
故解为(7/5, 1/5), (-1/5, 7/5)

Answer 2

3x+4y-5=0
4y=5-3x
x²+y²=2
16x²+16y²=32
16x²+（5-3x）²=32
16x²+9x²-30x+25=32
25x²-30x-7=0
(5x-7)(5x+1)=0
x1=1.4 x2=-0.2

4y=5-3x1.4=0.8
y1=0.2

4y=5+3x0.2=5.6
y2=1.4

Answer 3

3x+4y-5=0

y=5-3x/4
x²+y²=2
x²+(5-3x/4)²=2
16x²+(5-3x)²=32
16x²+25-30x+9x²=32
25x²-30x-7=0
(5x+1)(5x-7)=0
x=-1/5 x=7/5
当x=-1/5，y=9/10
当x=7/5，y=1/5