Evaluate :(1^0+2^0+3^0) = (x^0+y^0)
Answers
Answer:
Often it is necessary to look at several functions of the same independent variable. Consider the prior example where x, the number of items produced and sold, was the independent variable in three functions, the cost function, the revenue function, and the profit function.
In general there may be:
n equations
v variables
Solving systems of equations
There are four methods for solving systems of linear equations:
a. graphical solution
b. algebraic solution
c. elimination method
d. substitution method
Graphical solution
Example 1
given are the two following linear equations:
f(x) = y = 1 + .5x
f(x) = y = 11 - 2x
Graph the first equation by finding two data points. By setting first x and then y equal to zero it is possible to find the y intercept on the vertical axis and the x intercept on the horizontal axis.
If x = 0, then f(0) = 1 + .5(0) = 1
If y = 0, then f(x) = 0 = 1 + .5x
-.5x = 1
x = -2
The resulting data points are (0,1) and (-2,0)
Graph the second equation by finding two data points. By setting first x and then y equal to zero it is possible to find the y
intercept on the vertical axis and the x intercept on the horizontal axis.
If x = 0, then f(0) = 1 + .5(0) = 1
If y = 0, then f(x) = 0 = 1 + .5x
-.5x = 1
x = -2
The resulting data points are (0,1) and (-2,0)
Graph the second equation by finding two data points. By setting first x and then y equal to zero it is possible to find the y intercept on the vertical axis and the x intercept on the horizontal axis.
If x = 0, then f(0) = 11 - 2(0) = 11
If y = 0, then f(x) = 0 = 11 - 2x
2x = 11
x = 5.5
The resulting data points are (0,11) and (5.5,0)
At the point of intersection of the two equations x and y have the same values. From the graph these values can be read as x = 4 and y = 3.
Example 2
given are the two following linear equations:
f(x) = y = 15 - 5x
f(x) = y = 25 - 5x
Graph the first equation by finding two data points. By setting first x and then y equal to zero it is possible to find the y intercept on the vertical axis and the x intercept on the horizontal axis.
If x = 0, then f(0) = 15 - 5(0) = 15
If y = 0, then f(x) = 0 = 15 - 5x
5x = 15
x = 3
The resulting data points are (0,15) and (3,0)
Graph the second equation by finding two data points. By setting first x and then y equal to zero it is possible to find the y intercept on the vertical axis and the x intercept on the horizontal axis.
If x = 0, then f(0) = 25 - 5(0) = 25
If y = 0, then f(x) = 0 = 25 - 5x
5x = 25
x = 5
The resulting data points are (0,25) and (5,0)
From the graph it can be seen that these lines do not intersect. They are parallel. They have the same slope. There is no unique solution.

Example 3
given are the two following linear equations:
21x - 7y = 14
-15x + 5y = -10
Rewrite the equations by putting them into slope intercept form.
The first equation becomes
7y = -14 + 21x
y = -2 + 3x
The second equation becomes
5y = -10 + 15x
y = -2 + 3x
Graph either equation by finding two data points. By setting first x and then y equal to zero it is possible to find the y intercept on the vertical axis and the x intercept on the horizontal axis.
If x = 0, then f(0) = -2 +3(0) = -2
If y = 0, then f(x) = 0 = -2 + 3x
3x = 2
x = 2/3
The resulting data points are (0,-2) and (2/3,0)
From the graph it can be seen that these equations are equivalent. There are an infinite number of solutions.

Algebraic solution
This method will be illustrated using supply and demand analysis. This type of analysis is derived from the work of the great English economist Alfred Marshall.
Q = quantity and P = price
P (s)= the supply function and P (d) = the demand function
When graphing price is placed on the vertical axis. Thus price is the dependent variable. It might be more logical to think of quantity as the dependent variable and this was the approach used by the great French economist, Leon Walras. However by convention economists continue to graph using Marshall’s analysis which is referred to as the Marshallian cross.
The objective is to find an equilibrium price and quantity, i.e. a solution where price and quantity will have the same values in both the supply function and the price function.
QE = the equilibrium quantity PE = the equilibrium price
For equilibrium
supply = demand
or P (s) = P (d)
Given the following functions
P (s) = 3Q + 10 and P (d) = -1/2Q + 80
Set the equations equal to each other and solve for Q.
P (s) = 3Q + 10 = -1/2Q + 80 = P (d)
3.5Q = 70
Q = 20 The equilibrium quantity is 20.
Substitute this value for Q in either equation and solve for P.
P (s) = 3(20) + 10
P (s) = 70
P (d) = -1/2(20) + 80
P (d) = 70 The equilibrium price is 70.