You have to isolate one variable in the first equation first because its easier, then substitute it into the second one:
x=2-y
Now plug this in for x in the second equation:
3(2-y)^2+4(2-y)y+y^2=6
Now solve for y, then plug your answer back into:
x=2-y
3(2-y)(2-y)+8y-4y^2+y^2=6
3(4-4y+y^2)+8y-3y^2=6
12-12y+3y^2+8y-3y^2=6
-4y+12=6
-4y=6-12
y=-1.5
Now plug it in:
x=2-(-1.5)
x=3.5
Hope that helped! :)
how do i solve for x and y x+y=2 and 3x^2+4xy+y^2=6
3 answers
my bad, the 1.5 should be positive! so y=1.5 and x=0.5
x + y = 2 Subtract x to both sides
x + y - x = 2 - x
y = 2 - x
3 x ^ 2 + 4 x y + y ^ 2 = 6
3 x ^ 2 + 4 x ( 2 - x ) + ( 2 - x ) ^ 2 = 6
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Remark :
( a - b ) ^ 2 = a ^ 2 - 2 a b + b ^ 2
So :
( 2 - x ) ^ 2 = 2 ^ 2 - 2 * 2 *x + x ^ 2
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3 x ^ 2 + 4 x * 2 + 4 x * ( - x ) + 2 ^ 2 - 2 * 2 * x + x ^ 2 = 6
3 x ^ 2 + 8 x - 4 x ^ 2 + 4 - 4 x + x ^ 2 = 6
3 x ^ 2 - 4 x ^ 2 + x ^ 2 + 8 x + 4 - 4 x = 6
- x ^ 2 + x ^ 2 + 8 x - 4 x + 4 = 6
4 x + 4 = 6 Subtract 4 to both sides
4 x + 4 - 4 = 6 - 4
4 x = 2 Divide both sides by 4
4 x / 4 = 2 / 4
x = 2 / ( 2 * 2 )
x = 1 / 2
y = 2 - x
y = 2 - 1 / 2
y = 4 / 2 - 1 / 2
y = 3 / 2
The solutions are :
x = 1 / 2 , y = 3 / 2
x + y - x = 2 - x
y = 2 - x
3 x ^ 2 + 4 x y + y ^ 2 = 6
3 x ^ 2 + 4 x ( 2 - x ) + ( 2 - x ) ^ 2 = 6
________________________________
Remark :
( a - b ) ^ 2 = a ^ 2 - 2 a b + b ^ 2
So :
( 2 - x ) ^ 2 = 2 ^ 2 - 2 * 2 *x + x ^ 2
________________________________
3 x ^ 2 + 4 x * 2 + 4 x * ( - x ) + 2 ^ 2 - 2 * 2 * x + x ^ 2 = 6
3 x ^ 2 + 8 x - 4 x ^ 2 + 4 - 4 x + x ^ 2 = 6
3 x ^ 2 - 4 x ^ 2 + x ^ 2 + 8 x + 4 - 4 x = 6
- x ^ 2 + x ^ 2 + 8 x - 4 x + 4 = 6
4 x + 4 = 6 Subtract 4 to both sides
4 x + 4 - 4 = 6 - 4
4 x = 2 Divide both sides by 4
4 x / 4 = 2 / 4
x = 2 / ( 2 * 2 )
x = 1 / 2
y = 2 - x
y = 2 - 1 / 2
y = 4 / 2 - 1 / 2
y = 3 / 2
The solutions are :
x = 1 / 2 , y = 3 / 2