Find out the value of x and y 3+x+xy=03+x+xy=0 2x+5y=82x+5y=8 ???
Thanks in advance !
Thanks in advance !
1 Answer
Explanation:
3+x+xy=0to(1)3+x+xy=0→(1)
2x+5y=8to(2)2x+5y=8→(2)
"rearrange equation "(2)" in terms of x or y"rearrange equation (2) in terms of x or y
rArrx=1/2(8-5y)=4-5/2yto(3)⇒x=12(8−5y)=4−52y→(3)
"substitute "x=4-5/2y" in "(1)substitute x=4−52y in (1)
rArr3+4-5/2y+4y-5/2y^2=0⇒3+4−52y+4y−52y2=0
rArr-5/2y^2+3/2y+7=0⇒−52y2+32y+7=0
"multiply through by "-2multiply through by −2
rArr5y^2-3y-14=0⇒5y2−3y−14=0
rArr5y^2-10y+7y-14=0larr" split middle term"⇒5y2−10y+7y−14=0← split middle term
"factorise by grouping"factorise by grouping
color(red)(5y)(y-2)color(red)(+7)(y-2)=05y(y−2)+7(y−2)=0
"factor out "(y-2)factor out (y−2)
rArr(y-2)(5y+7)=0⇒(y−2)(5y+7)=0
"equate each factor to zero and solve for y"equate each factor to zero and solve for y
y-2=0rArry=2y−2=0⇒y=2
5y+7=0rArry=-7/55y+7=0⇒y=−75
"substitute these values into equation "(3)substitute these values into equation (3)
"for corresponding values of x"for corresponding values of x
y=2tox=4-5=-1y=2→x=4−5=−1
y=-7/5tox=4+7/2=15/2y=−75→x=4+72=152
rArr"points of intersection "(-1,2)" and "(15/2,-7/5)⇒points of intersection (−1,2) and (152,−75)
graph{(3+x+xy)(y+2/5x-8/5)((x+1)^2+(y-2)^2-0.04)((x-15/2)^2+(y+7/5)^2-0.04)=0 [-10, 10, -5, 5]}
