Math Prob 35?

Question

Solve the system:

x^2 + y^2 = 25

x^2 - y^2 = 7

Answer 1

x^2 + y^2 = 25

x^2 - y^2 = 7

2x^2 = 32
x^2 =16
x = +/-4
y^2 = 9
y = +/-3

Solution set (x=4,y=3) , (x= -4, y= -3)

Answer 2

x + y = 5
x = 5-y

(5-y)^2 - y^2 = 7 (5-x)^2 x^2 - =7
25 -y^2 - y^2 =7 x^2 -25 -x^2 =7
25 -7 - 2y^2 - 7=7-7 -2x^2 -25 +25 =7 +25
18 - 2y^2 =0 -2x^2 =32
-2y^2 =18 x^2 =-16
-2y^2/-2 = 18/-2 x = -4 0r 4
y^2 = -9
y = 3 or -3



x^2 -25 -x^2 =7
-2x^2 -25 +25 =7 +25
-2x^2 =32
x^2 =-16
- x = -4 0r 4




I hate imaginary numbers by the way
if x equals 4
16 +9 = 25
so the real numbers numbers x =4 and y=3 fit for the first problem
and 16 -9 =7 so theres your solution no imaginary numbers

Answer 3

x^2 + y^2 = 25

x^2 - y^2 = 7 add
2x^2=32
x^2=16
x=+/-4

Answer 4

x^2 + y^2 = 25............This looks very much like our old friend the 3,4,5 right angled triangle

3^2 + 4^2 = 5^2.......you know one of them is 3 and the other is 4 but you don't yet know which is which

Let's see what happens when you sutract 3^2 from 4^2

is 16 - 9 = 7.....rather as expected, it confirms the first calculation

you tell by inspection, (looking at it) that

x = 4 and y = 3

Answer 5

x^2 + y^2 = 25
x^2 - y^2 = 7
-------------------
2x^2 = 32 (Divide both sides by 2)
x^2 = 16 (Take square root from both sides)
x = +4 or -4

4^2 - y^2 = 7
16 - y^2 = 7 (Add y^2 and subtract 7 from both sides)
y^2 = 9 (Take square root from both sides)
y = +3 or -3
QED

Answer 6

step one: x^2= 25 - y^2
step two: (25 - y^2) - y^2 =7
step three: 18 - 2y^2=0 and y = 3
finaly: x^2 = 25-9 and x=4

=)

Answer 7

Addition to first answer:
x = 4, y = 3
x = - 4, y = 3
x = 4, y = - 3
x = - 4, y = - 3

Answer 8

adding 2x^2=32
x^2=16
x=+/-4
subs
16+y^2=25
y^2=9
y=+/-3
so x=+/-4 and y=+/-3