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How to solve for #x,y# this system #{(x = y-2),(4y+x^2=16):}# ?

Answer 1

Julia Adams

See below.

Calling #x, y# the two numbers such that #x < y# we have:

#{(x = y-2),(4y+x^2=16):}#

The first equation being substituted into the second

#4(x+2)+x^2=16# now solving for #x# we obtain

#x = -2(1pm sqrt3)# so

#y = 2-2(1pm sqrt3)#

When we rule out negative options, then

#x = 2(sqrt3-1)# and

#y = 2 sqrt3#

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Answer 2

Genesis Allen

To solve the system ((x = y - 2)) and ((4y + x^2 = 16)):

Substitute (x = y - 2) into the second equation: (4y + (y - 2)^2 = 16).
Expand and simplify the equation: (4y + y^2 - 4y + 4 = 16).
Combine like terms: (y^2 + 4 = 16).
Subtract 4 from both sides: (y^2 = 12).
Take the square root of both sides: (y = \pm \sqrt{12} = \pm 2\sqrt{3}).

Now, for each value of (y), find the corresponding value of (x) using (x = y - 2):

For (y = 2\sqrt{3}): (x = 2\sqrt{3} - 2).
For (y = -2\sqrt{3}): (x = -2\sqrt{3} - 2).

Therefore, the solutions for (x) and (y) are: (x = 2\sqrt{3} - 2) and (y = 2\sqrt{3}), or (x = -2\sqrt{3} - 2) and (y = -2\sqrt{3}).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.