Solve the following system of equations -x^2+y^2=16
-2x^2+y^2=16
Provide your answer below
(x, y) (x, y)

The solutions are (0, 4) and (0, -4)

Step-by-step explanation:

Solve this system using substitution:  

Take the negative of the first equation and add the result to the second one:

-x^2 + y^2 = 16 becomes x^2 - y^2 = -16

Then:

x^2    - y^2 = -16

-2x^2 + y^2 = 16

-x^2 = 0, so x must be 0.

By the first equation, if x = 0, -(0)^2 + y^2 = 16, or

y^2 = 16, or y = ±4

The solutions are (0, 4) and (0, -4)

answer: 48

just remember to use pemdas:

parentheses

expressions

multiplication

division

addition

subtraction

let's add {f(x)=x+1}f(x)=x+1 and {g(x)=2x}g(x)=2x together to make a new function.

f(x)+g(x) =(x+1)+(2x)=x+1+2x=3x+1

let's call this new function hh. so we have:

{h(x)}={f(x)}+{g(x)}{=3x+1}h(x)=f(x)+g(x)=3x+1

we can also evaluate combined functions for particular inputs. let's evaluate function hh above for x=2x=2. below are two ways of doing this.

method 1: substitute x=2x=2 into the combined function hh.

h(x)

h(2)

​    

=3x+1

=3(2)+1

=7

​ since h(x)=f(x)+g(x)h(x)=f(x)+g(x), we can also find h(2)h(2) by finding f(2) +g(2)f(2)+g(2).

first, let's find f(2)f(2):

f(x)

f(2)

​    

=x+1

=2+1

=3

​  

now, let's find g(2)g(2):

g(x)

g(2)

​    

=2x

=2⋅2

=4

​  

so f(2)+g(2)=3+4=\greend7f(2)+g(2)=3+4=7.

notice that substituting x =2x=2 directly into function hh and finding f(2) + g(2)f(2)+g(2) gave us the same answer!