We are given the functions:
<span>S (p) = 40 + 0.008 p^3 --->
1</span>
<span>D (p) = 200 – 0.16 p^2 --->
2</span>
T o find for the price in which the price of supply equals
demand, all we have to do is to equate the two equations, equation 1 and 2, and
calculate for the value of p, therefore:
S (p) = D (p)
40 + 0.008 p^3 = 200 – 0.16 p^2
0.008 p^3 + 0.16 p^2 = 160
p^3 + 20 p^2 = 20,000
p^3 + 20 p^2 – 20,000 = 0
Calculating for the roots using the calculator gives us:
p = 21.86, -20.93±21.84i
Since price cannot be imaginary therefore:
p = 21.86
Step-by-step explanation:
Difference per month = 28
=> January = February - 28 = 66-28 = 38
System 1: The solution is (x, y) = (-4, 5)
System 2: The solution is 
<em><u>Solution:</u></em>
<em><u>Given system of equations are:</u></em>
2x + 3y = 7 ------ eqn 1
-3x - 5y = -13 --------- eqn 2
We can solve by elimination method
Multiply eqn 1 by 3
6x + 9y = 21 ------ eqn 3
Multiply eqn 2 by 2
-6x - 10y = -26 ------- eqn 4
Add eqn 3 and eqn 4
6x + 9y -6x - 10y = 21 - 26
-y = -5
y = 5
Substitute y = 5 in eqn 1
2x + 3(5) = 7
2x + 15 = 7
2x = -8
x = -4
Thus the solution is (x, y) = (-4, 5)
<h3><em><u>
Second system of equation is:</u></em></h3>
8 - y = 3x ------ eqn 1
2y + 3x = 5 ----- eqn 2
We can solve by susbtitution method
From given,
y = 8 - 3x ----- eqn 3
Substitute eqn 3 in eqn 2
2(8 - 3x) + 3x = 5
16 - 6x + 3x = 5
3x = 16 - 5
3x = 11

Substitute the above value of x in eqn 3
y = 8 - 3x

Thus the solution is 