How do you solve this problem?

x^2 + y^2 = 63 ----- A
x^2 - 3y^2 = 27 ----- B

Subtract B from A,

x^2 + y^2 - x^2 + 3y^2 = 63 - 27
4y^2 = 36
y^2 = 9
or y = 3 or -3

Substituting y = +/- 3 in A,
x^2 + 9 = 63
x^2 = 63 - 9
x^2 = 54
x = 3sqrt(6) or -3sqrt(6)

Solution set = { (3sqrt(6), 3), (3sqrt(6), -3), (-3sqrt(6), 3), (-3sqrt(6), -3) }

HTH!

Value of x:
3(63 - x²) = x² - 27
189 - 3x² = x² - 27
4x² = 216
x² = 54
x = +/- 7.3484693

Values of y:
= √(63 - 54)
= √9
= +/- 3

Answer: x = 7.3484693, - 7.3484693; y = 3, - 3

Proof (2nd equation):
(- 7.3484693)² - 3(- 3²) = 27
54 - 3(9) = 27
54 - 27 = 27

x^2 + y^2 = 63
x^2 - 3y^2 = 27

First find an equation for x^2
x^2 = 63 - y^2

now sub x^2 into the second equation: x^2 - 3y^2 = 27
(63 - y^2) - 3y^2 = 27

-4y^2 = -36

y^2 = 9
y = ±3

Now sub y^2 into the equation for x^2 : x^2 = 63 - y^2

x^2 = 63 - (9)

x^2 = 54

x = ±√(54)

x = ±√(9 * 6)

x = ±3√(6)
y = ±3

x^2 = 63 - y^2

63 - y^2 - 3y^2 = 27
63 - 4y^2 = 27
-4y^2 = 27 - 63
y^2 = 36/4 = 9
y = 3 or - 3

x^2 = 63 - 9 = 54
x = + /- sqrt(54)

3x^2 + 3y^2 =189
x^2 - 3y^2 = 27
+_________________
4x^2=216
x^2=54 y^2=9

x=3sqrt(6) 0r x=-3sqrt(6) y=3 or y=-3

4 y ² = 36

y ² = 9

y = ± 3

x ² + 9 = 63

x ² = 54

x = ± 3 √6

x = ± 3 √6 , y = ± 3

Solve by using substitution.

x^2 + y^2 = 63
x^2 - 3y^2 = 27

x^2 + y^2 = 63
x^2 = (63 - y^2) ...... (1)

x^2 - 3y^2 = 27

Substitute (1) for x^2.
(63 - y^2) - 3y^2 = 27
63 - y^2 - 3y^2 = 27
-4y^2 = 27 - 63
-4y^2 = -36
y^2 = -36/(-4)
y = ±√9
y = ±3

x^2 + y^2 = 63

Substitute 3 for y.
x^2 + 3^2 = 63
x^2 = 63 - 9
x = ±√54
x = ±√(3^2 * 6)
x = ±3√6

∴ x = ±3√6, y = ±3

set both equations equal to x. then make the equations equal to each other.
y^2= 63 - x^2 subtract the x^2
y= sqrt(63 - x^2) then take the root of both sides
simplify this to get your first equation

-3y^2 = 27-x^2 (subtract the x^2)
y^2= (27-x^2)/-3 divide by -3.
square root both sides
simplify this to get your second equation

now you have two values equal to y ( i'm not going to simplify those for you )
take the first equation and second equation (both equal to y) equal to each other like so..
y= sqrt(63 - x^2)
y= sqrt((27-x^2)/-3))

sqrt(63 - x^2) = sqrt((27-x^2)/-3))
simplify the above equation to solve for X.

for the finals step, plug your value for X back into one of the previous equations.

then you're done!!

ps sorry i didn't have paper to work this out properly.

3x^2 + 3y^2 = 189
x^2 - 3 y^2 = 27 implying: 4x^2 = 216. That is x^2 = 54, and so x = +/- 3sqrt(6). Plug that into either equation to get: y^2 = 9 or y = +/- 3. Those are your answers.

Number 1.

To get y:

x^2 + y^2 = 63
x^2 - 3y^2 = 27

x^2 + y^2 = 63
-1(x^2 - 3y^2 = 27)

x^2 + y^2 = 63
-x^2 + 3y^2 = -27

Cancel x^2 and -x^2
y^2 = 63
3y^2 = -27

4y^2 = 36 (Divide both sides by 4)
y^2 = 9 (Square root)
y = 3

To get x:

x^2 + 3^2 = 63
x^2 + 9 = 63
x^2 = 63 - 9
x^2 = 54 (Square root)

If a term isn't a perfect square, break the term down into a square and a co-factor if possible. In this case, it's possible.
x = √9 √6
x = 3√6

Checking:
(3√6)^2 + 3^2 = 63
9*6 + 9 = 63
54 + 9 = 63

Solution set is (3√6, 3)

Hope that helps.