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Mathematics
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If x^2+4y^2=40 and xy=6, what will be the value of x+2y
First, we must solve the system of equations formed for x^2+4y^2=40 and xy=6
We choose substitution method to do so:
Go to one equation and solve x: x= 6/y
And now substitute in the other equation: (6/y)^2+4y^2=40
Solving the equation:
36/y^2+4y^2=40
Here we have a biquadratic equation, and we have to reduce it to a
quadratic by substituting: t=y^2
36+4t^2=40t
4t^2-40t+36=0
Dividing by 2:
2t^2-20t+18=0
Solving the second grade equation applying the quadratic
formula: t= (-b±(b^2-4*a*c)^1/2) /2*a
While a=2, b=-20 and c=18
The roots for t are: 9, 1
While t=y^2, then y=t^1/2
So, y = {±1,±3}
Substituting this values in x=6/y then the solutions are the pair of (x,y) values: (6,1), (-
6,-1), (2,3) and (-2,-3)
Substituting those pair of values in x+2y:
(6,1) → x+2y=6+2*1=8
(-6,-1) → x+2y=-6+2*-1=-8
(2,3) → x+2y=2+2*3=8
(-2,-3) → x+2y=-2+2*-3=-8
Finally, If x^2+4y^2=40 and xy=6, the value of x+2y will be: ±8

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If x^2+4y^2=40 and xy=6, what will be the value of x+2y First, we must solve the system of equations formed for x^2+4y^2=40 and xy=6 We choose substitution method to do so: Go to one equation and solve x: x= 6/y And now substitute in the other equation: (6/y)^2+4y^2=40 Solving the equation: 36/y^2+4 ...
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