Answer:
f(x)=x^2+3
f(x+y)=(x+y)^2+3
this is done by replacing x with (x+y)
f(x+y)=x^2+2xy+y^2+3
therefore : f(x+y)= x^2+y^2+2xy+3
Answer:
second one
Step-by-step explanation:
am pretty sure
You can see that the term
appears in both equations. In this cases, we can leverage this peculiarity and subtract the two equations to get rid of the repeated term. So, if we subtract the first equation from the second, we have
![(-4x -6y) - (-10x-6y) = 54-0 \iff 6x = 54 \iff x = 9](https://tex.z-dn.net/?f=%20%28-4x%20-6y%29%20-%20%28-10x-6y%29%20%3D%2054-0%20%5Ciff%206x%20%3D%2054%20%5Ciff%20x%20%3D%209%20)
Now that we know the value of
, we can substitute in any of the equation to deduce the value of
: if we use the first equation, for example, we have
![-10x-6y=0 \iff -10\cdot 9 - 6y = 0 \iff -90-6y=0 \iff 6y = -90 \iff y = -15](https://tex.z-dn.net/?f=%20-10x-6y%3D0%20%5Ciff%20-10%5Ccdot%209%20-%206y%20%3D%200%20%5Ciff%20-90-6y%3D0%20%5Ciff%206y%20%3D%20-90%20%5Ciff%20y%20%3D%20-15%20)
It would take Olive 30 minutes to walk home from the news station.