Answer:
<h2>PART A: x = 15 and y = -8</h2><h2>PART B: x = -5 and y = 3</h2>
Step-by-step explanation:
PART A:
The first equation is 
The second equation is 
Putting the value of y from (a) in the above equation, we get 
From (a) we get y = 7 - x = 7 - 15 = -8
PART B:
Here the system of equations are 
From (a) we get, 
Putting the above value of x in (b) we get, 
Hence, 
To solve for this, we need to find for the value of x
when the 1st derivative of the equation is equal to zero (or at the
extrema point).
So what we have to do first is to derive the given
equation:
f (x) = x^2 + 4 x – 31
Taking the first derivative f’ (x):
f’ (x) = 2 x + 4
Setting f’ (x) = 0 and find for x:
2 x + 4 = 0
x = - 2
Therefore the value of a is:
a = f (-2)
a = (-2)^2 + 4 (-2) – 31
a = 4 – 8 – 31
a = - 35
Answer: The 1 goes in the hundreds place and the 4 goes in the tens place and the 5 goes in the ones place
Step-by-step explanation:
We can add or subtract integer. We can also subtract like terms.
