x^+y^=89; x^-y^=-39;How do I solve the system of equations?

a = x²
b = y²

a+b = 89
a-b = -39
---------------------
2a = 50
a = 25
b = 64

x² = 25
x = ±5

y² = 64
y = ±8

x+y = 13 x-y = 5 fixing for y: -one million(x+y) = (-one million)13 :: Multiply the two aspects (of the precise equation) by utilising -one million -x - y = -13 x - y = 5 -2y = -8 :: The X's cancel out and -13 + 5 = -8 y = 4 :: divide the two aspects by utilising -2 -------------------- fixing for X: x + y = 13 x - y = 5 2x = 18 :: The Y's cancel out and 13 + 5 = 18 x = 9 :: Divide the two aspects by utilising 2 ------------------- Checking your answer 9 + 4 = 13 9 - 4 = 5 the two equations are maximum superb.

x^2 + y^2 = 89 (solve by using substitution)
x^2 - y^2 = -39

x^2 + y^2 = 89
x^2 = 89 - y^2

x^2 - y^2 = -39
(89 - y^2) - y^2 = -39
89 - y^2 - y^2 = -39
-2y^2 = -39 - 89
-2y^2 = -128
y^2 = -128/-2
y^2 = 64
y = ±√64
y = ±8

x^2 + y^2 = 89
x^2 + 8^2 = 89
x^2 = 89 - 64
x^2 = 25
x = ±√25
x = ±5

∴ x = ±5 , y = ±8

Q. x^2 + y^2 = 89 --- A
Q. x^2 - y^2 = -39 --- B

A + B:
=> (x^2 + y^2) + (x^2 - y^2) = 89 + -39
=> 2x^2 = 50
=> x^2 = 25
=> x = +/- 5

A - B:
=> (x^2 + y^2) - (x^2 - y^2) = 89 - -39
=> 2y^2 = 128
=> y^2 = 64
=> y = +/- 8

Answer: x = +/- 5; y = +/- 8

however because there are two equations and two unknows it is possible. because there just x and y if you look at their signs you will see that one of the ys are positive and the other is negative. this is very good because you don't have to do work yourself in find what to cancel with with what.
ADD the two equations together.
x^+y^=89
+x^-y^=-39

you get 2x^=50 and then it is just a simple algebraic equation.

x^=25 and then what ever the value for the ^ is you take it to the root of 25

this means x=+/-5 you take this value and you plug it back in to one of the equations to solve for the y value.

so (5)^2+y^=89
25+y^=89
y^=64

y^2=64
y=(64)^1/2
y=+/-8

so your final answers are x=5 y=8 plug these values into both your original equations to make sure it matches both of them in this case it dose.

x^2+y^2=89 and x^2-y^2=-39 is so x=+/-5 and y=+/-8

good luck i hope this helps

(1) x^2+y^2 = 89
(2) x^2-y^2 = -39
Subtracting (2) from (1), we get:
(3) x^2+y^2-x^2--y^2 = 89--39
Which is the same as:
2(y^2) = 128
So y^2 = 128/2 = 64
So y = root(64) = +/- 8

Substituting into (1):

x^2+8^2 = 89 = x^2+64
So x^2 = 89-64 = 25
So x = root(25) = +/- 5

x = +/- 5
y = +/- 8

2x^2 = 50; x^2 =25; x = 5 or -5

2y^2 = 128; y^2 = 64; y = 8 or -8.

Sum is 2x^2 =50
x^2 =25
x=5 or -5
25 +y^2 =89
y^2 =64
y=8 or -8
Answers: (5,8), (-5,8), (5, -8). (-5, -8)

Solve each one for a variable (ex. make them both equal x) and then set them equal to each other and solve to find y.

add them to find x

substract the second equation from the 1st equation to find y

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x^2 + y^2 = 89
x^2 - y^2 = -39

Answer:
a) 2x^2 =50 ; x = +/- 5
b) 2y^2 = 128 , y = +/- 8