Given:
2x^2-4x=y+12x2−4x=y+1
x+y=1x+y=1
y=1-xy=1−x
2x^2-4x=1-x+12x2−4x=1−x+1
2x^2-4x=2-x2x2−4x=2−x
2x^2-4x+x-2=02x2−4x+x−2=0
2x^2-3x-2=02x2−3x−2=0
2x^2-4x+x-2=02x2−4x+x−2=0
2x(x-2)+1(x-2)=02x(x−2)+1(x−2)=0
(2x+1)(x-2)=0(2x+1)(x−2)=0
2x+1=02x+1=0
2x=-12x=−1
x=-1/2x=−12
y=1-xy=1−x
y=1-(-1/2)y=1−(−12)
1+1/21+12
y=3/2y=32
Check:
2x^2-4x=y+12x2−4x=y+1----(1)
2(-1/2)^2-4xx(-1/2)=3/2+12(−12)2−4×(−12)=32+1
2xx1/4+4xx1/2=3/2+2/22×14+4×12=32+22
1/2+4/2=3/2+2/212+42=32+22
(1+4)/2=(3+2)/21+42=3+22
5/2=5/252=52
lhs=rhs
x+y=1x+y=1-----(2)
-1/2+3/2=1−12+32=1
(-1+3)/2=1−1+32=1
2/2=122=1
1=11=1
lhs=rhs
x-2=0x−2=0
x=2x=2
y=1-xy=1−x
y=1-2y=1−2
y=-1y=−1
Check:
2(2)^2-4(2)=-1+12(2)2−4(2)=−1+1-----(1)
2(4)-8=02(4)−8=0
8-8=08−8=0
0=00=0
lhs=rhs
x+y=1x+y=1-----(2)
2+(-1)=12+(−1)=1
2-1=12−1=1
1=11=1
lhs=rhs
Thus,
(x,y)=(-1/2,3/2)(x,y)=(−12,32)
and
(x,y)=(2,-1) form solutions(x,y)=(2,−1)formsolutions
