Question

X^2-y^2=48 and x-y=2. The solution to the given system of equations is (x,y). What is the value of x + y?

Answer 1

Answer:

x= 13

y=11

x^2 - y^2 = 48        and       x - y = 2

                                           x = 2 + y

(2 + y)^2 - y^2 = 48

(2 + y) (2 + y) - y^2 = 4

(4 + 2y +2y + y^2) -y^2 = 48

4 + 4y + y^2 - y^2 = 48

4 + 4y = 48

4y = 48 -4

4y = 44

y= 44/4

y= 11

Plug y back into the original equation:

x^2 - y^2 = 48

x^2 - 11^2 = 48

x^2 - 121 = 48

x^2 = 121 + 48

x^2 = 169

The square root of 169 is 13

So x= 13

x + y

13 + 11 = 24

Answer 2

Answer:

24

Step-by-step explanation:

We have a system of equations, so we can start with the second equation:

$x-y=2$

And move y to the other side to isolate x:

$x=2+y$

Now we can replace x into the first equation:

$(2+y)^{2} -y^{2} =48\\ (2+y)(2+y)-y^{2} =48\\ 4+4y+y^{2} -y^{2} =48$

Notice that we can cancel out the y squared terms:

$4+4y=48\\ 4y=44\\ y=11$

Now we just need to find x by replacing y into the second equation:

$x-11=2\\ x=13$

And finally, add them:

$x+y=11+13\\ x+y=24$

Hope this helps.