Solve for x and y x^3-y^3=7 x^2+xy+y^2=7
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calendarEducator since 2008
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x^3 - y^3 = 7..........(1)
x^2 + xy + y^2 = 7 ..............(2)
We need to solve the system.
First we will rewrite equation (1) as a difference between cubes.
==> x^3 - y^2 = (x-y)(x^2 + xy+y^2)
But from equation (2) , we know that x^2 + xy+ y^2 = 7
==> 7* (x-y) = 7
Now we will divide by 7.
==> x-y = 1
==> x= y+1 ............(3)
Now we will substitute with either equations.
==> x^3 - y^3 = 7
==> (y+1)^3 - y^3 = 7
==> y^2 + 2y + 1)(y+1)
==> y^3 + 3y^2 +3y + 1 - y^3 = 7
We will reduce similar.
==> 3y^2 + + 3y -6 = 0
We will divide by 3.
==> y^2 + y -2 = 0
Now we will factor.
==> (y+2)(y-1) = 0
==> y1= -2 ==> x1= -2+1 = -1
==> y2= 1 ==> x2= 1+1 = 2
Then we have two solutions: ( -1, -2) and (2, 1)
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calendarEducator since 2010
write12,544 answers
starTop subjects are Math, Science, and Business
We have the system of equations:
x^3 - y^3 = 7
x^2 + xy + y^2 = 7
x^3 - y^3 = 7
=> (x - y)(x^2 + xy + y^2) = 7
=> (x - y) *7 = 7
=> x - y = 1
= x = 1 + y
x^2 + xy + y^2 = 7
=> (1 + y)^2 + y(1 + y) + y^2 = 7
=> 1 + y^2 + 2y + y + y^2 + y^2 = 7
=> 3y^2 + 3y - 6 = 0
=> y^2 + y - 2 = 0
=> y^2 + 2y - y - 2 = 0
=> y(y +2) - 1(y +2) = 0
we get y = 1 and -2
x = 1 + y = 2 and -1
The solution is (2, 1) and (-1, -2)
We notice that the first equation is a difference of cubes and it could be written as:
x^3 - y^3 = (x-y)(x^2+xy+y^2)
We'll re-write the first equation:
(x-y)(x^2+xy+y^2) = 7
We notice that the 2nd factor of the product is the 2nd equation.
But x^2+xy+y^2 = 7
We'll re-write the first equation:
7(x-y) = 7
We'll divide by 7
x - y = 1
x = 1 + y (3)
We'll substitute (3) in (2):
(1+y)^2 + (1+y)*y + y^2 = 7
We'll expand the square and we'll remove the brackets:
1 + 2y + y^2 + y + y^2 + y^2 = 7
We'll combine like terms:
3y^2 + 3y + 1 - 7 = 0
3y^2 + 3y - 6 = 0
We'll divide by 3:
y^2 + y - 2 = 0
y1 = [-1+sqrt(1+8)]/2
y1 = (-1+3)/2
y1 = 1
y2 = (-1-3)/2
y2 = -2
x1 = 1 + 1
x1 = 2
x2 = -1
The solutions of the system are: {(2 ; 1) ; (-1 ; -2)}.
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