Download Math Analysis

Math Analysis
Chapter 7
Monday 24 May: Section 7-1 Solving Systems of Equations with Two Variables;
Sections 7-1: Solving Systems of Equations with Two Variables
Solving Systems of equations with two variables by Substitution:
1. Solve either of the equations for one variable in terms of the other, if necessary. (If possible, solve
a variable whose coefficient is 1 or −1 to avoid working with fractions).
2. Substitute the expression found in step 1 into the other equation. This will result in an equation
with one variable.
3. Solve the equation containing one variable.
4. Back-substitute the value found in step 3 into one of the original equations. Simplify and find the
value of the remaining variable.
5. Check the proposed solution in both of the system’s given equations.
Practice: 1. Solve by substitution method:
3x + 2y = 4
2x + y = 1
Solving Systems of equations with two variables by Elimination:
1. If necessary, rewrite both equation in the form Ax + By = C.
2. If necessary, multiply either equation or both equations by appropriate nonzero numbers so that the
sum of the x-coefficients or the sum of the y-coefficients is 0.
3. Add the equations in step 2. The sum is an equation in one variable.
4. Solve the equation in one variable.
5. Back-substitute the value obtained in step 4 into either of the given equation and solve for the other
variable.
6. Check the proposed solution in both of the system’s given equations.
Practice 2: Solve by the elimination method (aka the addition method):
4x + 5y = 3
2x – 3y = 7
Math Analysis Notes Prepared by Mr. Hopp
1
Remember: If solving the system by either method you get a true statement without a variable in the
equation the system has infinitely many solutions which we will now be defining that solution.
Practice 3: Solve the system using any method:
x – 4y = −8
5x – 20y = −40
Applications
Strategy for Problem Solving Using Systems of Equations
Step 1: Read the problem carefully. Attempt to state the problem in your own words and state what the
problem is looking for. Use variables to represent unknown quantities.
Step 2: Write a system of equations that models the problem’s conditions.
Step 3: Solve the system and answer the problem’s question.
Practice 4: A chemist needs to mix 18% acid solution with 45% acid solution to obtain 12 liters of 36%
acid solution. How many liters of each of the acid solutions must be used?
Math Analysis Notes Prepared by Mr. Hopp
2
Revenue and Cost Functions
A company produces and sells x units of a product.
Revenue Function:
R(x) = (price per unit sold)x
Cost Function:
C(x) = fixed cost + (cost per unit produced)x
The break-even point is the intersection of the graphs of the revenue and cost functions (aka solution of
the system). The x-coordinate of the point reveals the number of units that a company must produce and
sell so that money coming in, the revenue, is equal to the money going out, the cost. The y-coordinate of
the break-even point gives the amount of money coming in and going out.
Practice 5: A company that manufactures running shoes has a fixed cost of $300,000. Additionally, it
costs $30 to produce each pair of shoes. They are sold at $80 per pair.
a) Write the cost function, C, of producing x pairs of running shoes.
b) Write the revenue function, R, from the sale of x pairs of running shoes.
c) Determine the break-even point. Describe what this means.
Have students work in groups and complete the practice problems page 723
(7,11,15,18,24,29,33,36,39,47,48)
Math Analysis Notes Prepared by Mr. Hopp
3
Tuesday 25 May Section 7-2 Solving Systems of Equations with Three Variables.
Section 7-2: Systems of Linear Equations in Three Variables.
Solving Linear Systems in Three Variables by Eliminating Variables.
1. Reduce the system to two equations in two variables. This is usually accomplished by taking two
different pairs of equations and using the elimination method to eliminate the same variable from
both pairs.
2. Solve the resulting system of two equations in two variables using elimination or substitution. The
result is an equation in one variable that gives the value of that variable.
3. Back-substitute the value of the variable found in step 2 into either of the equations in two
variables to find the value of the second variable.
4. Use the values of the two variables from steps 2 and 3 to find the value of the third variable by
back-substitution into one of the original equations.
Practice 6: Solve the system:
x + 4y – z = 20
3x + 2y + z = 8
2x – 3y + 2z = −16
Application: Writing the equation of a quadratic equation knowing three points that the equation passes
through. General form of a quadratic equation is: y = ax2 + bx + c
In order to find the quadratic equation passing through 3 given points write the system of equations that
models to given information and solve for the coefficients a, b, and c.
Practice 7: Find the quadratic function whose graph passes through the points (1, 4), (2, 1) and (3, 4).
Have students work in groups and complete the practice problems page 734 (5,8,11,14,16,17,21,25)
Math Analysis Notes Prepared by Mr. Hopp
4
Wednesday 26 May Section 7-3 Partial Fractions
After completing today’s notes you should be able to do the following:
P
1. Decompose , where Q has only distinct linear factors
Q
2.
Decompose
P
, where Q has repeated linear factors
Q
3. Decompose
P
, where Q has a non-repeated prime quadratic factor
Q
4.
P
, where Q has a non-repeated prime quadratic factor
Q
Decompose
The reason we want to be able to find the partial fraction of decomposition of a rational expression is to be
able to study rates of change. Calculus is the study of rates of change and partial fraction of
decomposition is an algebraic technique used to find a function if its rate of change is known.
Steps in Partial Fraction Decomposition
1. Set up the partial fraction decomposition with the unknown constants A, B, C, etc., in the
numerator of the decomposition.
2. Multiply both sides of the resulting equation by the least common denominator.
3. Simplify the right-side of the equation.
4. Write both sides in descending powers, equate coefficients of like powers of x, and equate constant
terms.
5. Solve the resulting linear system for A, B, C, etc.
6. Substitute the values for A, B, C, etc., into the equation in step 1 and write the partial fraction
decomposition.
P( x)
The Partial Fraction Decomposition of
: Q(x) Has Distinct Linear Factors.
Q( x)
The form of the partial fraction decomposition for a rational expression with distinct linear factors in
the denominator is:
P( x)
A
B


 ... (until number of linear factors are matched)
(a1 x  b1 )(a2 x  b2 )(a3 x  b3 )...(an x  bn ) a1 x  b1 a2 x  b2
Practice 1: Write the form of the partial fraction decomposition of the rational expression. It is not necessary to solve for
the constants.
(a)
11x  10
( x  2)( x  1)
(b)
Practice 2: Write the partial fraction decomposition of:
1
x(2 x  3)
3x  50
( x  9)( x  2)
Math Analysis Notes Prepared by Mr. Hopp
5
P( x)
: Q(x) Has Repeated Linear Factors.
Q( x)
The form of the partial fraction decomposition for a rational expression containing the linear factor (ax
+ b) occurring n times as its denominator is:
P( x)
A
B
?
(keep adding constant terms until you match the power of the linear factor)


 ... 
n
2
(ax  b)
ax  b (ax  b)
(ax  b) n
The Partial Fraction Decomposition of
Practice 3: Write the form of the partial fraction decomposition of the rational expression. It is not necessary to solve for
the constants.
(a)
x 2  3x  5
( x  2)( x  3)3
(b)
x2
( x  1)2 ( x  1)2
x2  2 x  7
Practice 4: Write the partial fraction decomposition of:
x( x  1) 2
Math Analysis Notes Prepared by Mr. Hopp
6
P( x)
: Q(x) Has Non-Repeated, Prime Quadratic Factor.
Q( x)
If ax2 + bx + c is prime quadratic factor of Q(x), the partial fraction decomposition will contain a term
of the form
Ax  B
2
ax  bx  c
The Partial Fraction Decomposition of
Practice 5: Write the form of the partial fraction decomposition of the rational expression. It is not necessary to solve for
the constants.
(a)
x2  x  8
( x  1)( x 2  2 x  2)
(b)
Practice 6: Write the partial fraction decomposition of:
5 x3  2 x 2  x  4
( x 2  1)( x 2  3x  5)
5x2  6 x  3
( x  1)  x 2  1
P( x)
: Q(x) Has a Prime, Repeated Quadratic Factor.
Q( x)
The form of the partial fraction decomposition for a rational expression containing the prime factor ax2
+ bx + c occurring n times as its denominator is:
P( x)
Ax  B
Dx  C
?
(keep writing linear equations as numerator)
 2

 ... 
2
n
2
2
2
(ax  bx  c)
ax  bx  c (ax  bx  c)
(ax  bx  c)n
The Partial Fraction Decomposition of
Practice 7: Write the form of the partial fraction decomposition of the rational expression. It is not necessary to solve for
the constants.
(a)
2 x3  x  3
( x  1)( x 2  1)3
(b)
2
x( x  x  4)3
2
Math Analysis Notes Prepared by Mr. Hopp
7
Practice 8: Write the partial fraction decomposition of:
x3  4 x 2  9 x  5
( x 2  2 x  3) 2
4 x 2  3x  14
Practice 9: Write the partial fraction decomposition of:
x3  8
Thursday 27 May Section 7-4 Systems of Nonlinear Equations in 2 Variables; Section 7-5 Systems of
Inequalitites
After completing today’s notes you should be able to do the following:
1. Solve nonlinear systems by substitution or elimination methods
2. Graph a system of inequalities.
3. Solve applied problems involving systems of inequalities
Remember that the same steps for solving a nonlinear system is the same as if it was linear. See page 5 for steps to solve
systems of equations:
Section 7-4 Systems of Nonlinear Equations
Practice 1: Solve by substitution method:
x2  y  1
4 x  y  1
Math Analysis Notes Prepared by Mr. Hopp
8
x  2y  0
Practice 2: Solve by substitution method:
Practice 3: Solve:
( x  1) 2  ( y  1) 2  5
3x 2  2 y 2  35
4 x 2  3 y 2  48
Math Analysis Notes Prepared by Mr. Hopp
9
Practice 4: Find the length and width of a rectangle whose perimeter is 20 feet and whose area is 21 square feet.
Section 7-5 Systems of Inequalities
x y 2
Practice 5: Graph the solution set of the system: 2  x  1
y  3
Remember graphing equations of circles: (x – h)2 + (y – k)2 = r2 where (h, k) = center and r = radius. We are now going to
graph inequalities that are circles.
Practice 5: Graph the solution set of the system:
( x  1)2  ( y  1)2  25
( x  1)2  ( y  1)2  16
Math Analysis Notes Prepared by Mr. Hopp
10
Practice 6: Graph the solution set of the system:
x2  y 2  1
y  x2  0
Math Analysis Notes Prepared by Mr. Hopp
11