x+y=12....and x-y=8, what is the value of x^2+y^2?

Question

yes how do you solve this...
as in the method...

Answer 1

x^2+y^2 = [(x+y)^2 + (x-y)^2]/2
= [(12)^2 + (8)^2]/2
= [144 + 64]/2
= 208/2
= 104

Answer 2

First, solve the two as a system of equations.
Let: (i) x + y = 12, and (ii) x - y = 8. Let's try elimination: (i) - (ii) gives 2y = 4 => y = 2.

Next, substitute y = 2 into one of the equations, I'm going to go with (ii), so x - 2 = 8 => x = 10

Now, plug this information into the expression x^2 + y^2:

10^2 + 2^2 = 100 + 4 = 104.

Therefore 104 is the answer.

Answer 3

You can use the addition method to solve.
x + y = 12
x - y = 8
-------------
2x = 20
x = 10

Now you have the value of x we can substitute that back into either equation to get the value of y

10 + y = 12
y = 2

So y = 2 and x = 10 we can now solve:
10^2 + 2^2 = 100 + 4 = 104

Answer 4

First add the two equations to eliminate y and solve for x.

x+y=12
+(x-y = 8)
2x = 20
x = 10

Put your value for x into one of the original equations to solve for y.

10 + y = 12
y = 2

Now you know that x is 10 and y is 2, just square both of them and add them together.

10^2 + 2^2 = 100+4=104

Answer 5

x+y = 12
x-y = 8
------------
2x = 20 thus x = 20/2 or 10 and y = 2

so x^2+y^2= 10^2 +2^2
= 100 +4
=104

Answer 6

2x = 20 (upon adding equations)
x = 10
y = 2
x² + y² = 100 + 4 = 104

Answer 7

first solve for x and y

x + y = 12
x - y = 8
------------
2x = 20
x = 10

10 + y = 12
y = 2

so, 10^2 + 2^2 = 104

Answer 8

x+y=12 . . . . . . . . . . . . . . .(1)
x-y=8 . . . . . . . . . . . . . . . . (2)
Squaring(1)
x²+2xy+y²=144 . . . . . . . . .(3)
Squaring(2)
x²-2xy+y²=64 . . . . . . . . . . (4)
Adding (3) &(4)
2x²+2y²=208
2(x²+y²)=208
(x²+y²)=208/2=104............Ans.
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Answer 9

You solve the two linear equations for x and y, then substitute the x and y in the quadratic.

Answer 10

x=10
y=2