College Algebra

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Open Chapter Ch. 1: Graphs, Functions, and Models
Lesson #1 Graphing
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Lesson #2 Graphing Functions
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Lesson #3 Slope of Linear Functions
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Lesson #4 Linear Equations and Modeling
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Lesson #5 Linear Equations, Functions, Zeros and Applications
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Lesson #6 Linear Inequalities
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Open Chapter Ch. 2: More on Functions
Lesson #7 Analyzing Functions
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Exam Exam 1
Lesson #8 Algebra for Functions
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Lesson #9 The Composition of Functions
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Lesson #10 Symmetry
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Lesson #11 Transformations
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Lesson #12 Equations of Variation
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Open Chapter Ch. 3: Quadratic Functions and Equations; Inequalities
Lesson #13 Complex Numbers
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Lesson #14 Quadratic Equations, Functions, Zeros and Models
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Lesson #15 Analyzing Graphs of Quadratic Functions
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Lesson #16 Rational and Radical Equations
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Exam Midterm Exam
Lesson #17 Absolute Value Equations and Inequalities
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Open Chapter Ch. 4: Polynomial Functions and Rational Functions
Lesson #18 Polynomial Functions and Models
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Lesson #19 Graphing Polynomial Functions
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Lesson #20 Polynomial Division
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Lesson #21 The Zeros of Polynomial Functions
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Lesson #22 Rational Functions
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Lesson #23 Polynomial and Rational Inequalities
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Open Chapter Ch. 5: Exponential Functions and Logarithmic Functions
Lesson #24 Inverse Functions
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Lesson #25 Exponential Functions and Graphs
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Lesson #26 Logarithmic Functions and Graphs
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Exam Exam 3
Lesson #27 Properties of Logarithmic Functions
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Lesson #28 Solving Exponential Equations and Logarithmic Equations
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Lesson #29 Applications and Models: Growth and Decay; Compound Interest
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Open Chapter Ch. 6: Systems of Equations
Lesson #30 Systems of Equations in Two Variables
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Lesson #31 Systems of Inequalities and Linear Programming
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Lesson #32 Nonlinear Systems of Equations and Inequalities
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Lesson #33 Sequences and Series
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Exam Final Exam

Assignments:

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Lesson Objectives:

- Solve by graphing.
- Solve by substitution or elimination method.
- Use a system of two linear equations to solve an applied problem.


A system of equations consists of two or more equations considered simultaneously. The corresponding unknowns have the same values. In order to find all the unknown numbers in a system of equations, we must have as many equations as there are unknown numbers.

Solving a system of equations graphically:

Each point at which the graphs intersect is a solution to both equations.



Solve graphically:

`x+y = 3` and `2x+y = 0`.

Start by solving for y in each equation. So in the first equation, we have `y = 3-x`, and in the second equation, we have `y = -2x`.

Now you can plug these equations into your calculator, under `y=`. And use your calculator to determine the point of intersection. The graphs intersect at a single point, (-3, 6). So (-3, 6) is the solution of the system of equations.

Now let's check our solution. Plug (-3, 6) into our first equation, `x+y = 3`. So we get `(-3)+(6) = 3`; `3 = 3`, a true statement. Now let's check it with the second equation. So we have `2x+y = 0`, or `2(-3)+(6) = 0`. This gives us `-6+6`, or `0=0`, another true statement. So (-3, 6) is the solution.



Solving a system of equations algebraically:

Systems of equations are solved by combining the equations so as to obtain a single equation with one unknown number.

And you can accomplish this by substitution or elimination. In substitution, we solve for a variable in one of the equations, and substitute the value of that variable into the second equation.

In elimination, we use addition or subtraction to eliminate one of the variables. And then we solve for the remaining variable and substitute the value for that variable into either of the original equations.



Solve using the substitution method.

`x+y = 10` and `2x-3y = -2`. We'll call the first equation, 1, and the second equation, 2.

Now let's solve 1 for x. So we get `x = 10-y`. Then we substitute `10-y` in for x in equation 2. So we get `2(10-y)-3y = -2`.

So now we have an equation in one variable, which we know how to solve. Let's start by distributing 2 into the parentheses. So we get `20-2y-3y = -2`. And then go ahead and combine the -2y and the -3y; so we get -5y. And now if we subtract 20 from both sides, we get `-5y = -22`. And then divide both sides by -5. So we get `y = 22/5`.

So now we substitute `22/5` in for y in either of our equations. This is called back substitution. So if we plug it into equation 1, we have `x+22/5 = 10`. Now subtract `22/5` from both sides, and we have `x = 10-22/5`. And we can make common denominators by multiplying the top and bottom of `10/1` by 5. So we have `50/5 - 22/5`. Now we can just subtract the numerators, 50-22, and get `28/5`. So `x = 28/5`.

So the solution for the system of equations is the ordered pair `(28/5, 22/5)`.



And now we can check our solution by plugging into both of our equations.

So if we plug into 1, we have `28/5+22/5 = 10`. This gives us `50/5 = 10`, or `10 = 10`, a true statement. Now if we plug into equation 2, we have `2(28/5)-3(22/5) = -2`. Let's go ahead and divide everything by 2. So we get `28/5-3/2(22/5) = -1`. Now 2 goes into 22, 11 times, so we have `28/5` minus `3(11/5)`, which is `33/5`, is equal to -1.

So now we can just subtract the numerators; `28-33 = -5`. So we have `-5/5 = -1`, or `-1 = -1`, another true statement. So this verifies that `(28/5, 22/5)` is the solution to our system of equations.



Solve the following system of equations using the elimination method.

`x+3y = 8` and `x-3y = -5`. Since the coefficients of y in both of our equations is 3 and -3, we can eliminate y by adding our equations. So if we go ahead and add them, we're left with `2x = 3`, and we get `x = 3/2`.

Now backsubstitute `3/2` for x in either of our equations. Let's go ahead and plug into our first equation. So we have `3/2+3y = 8`. And now if we subtract `3/2` from both sides, we have `3y = 8-3/2`. Now `8/1` can be multiplied by 2 in the numerator and denominator so that we get `16/2`. So we have `16/2-3/2 = 3y`. Now we can subtract the numerators, `16-3 = 13`, so we have `13/2 = 3y`. And if we multiply both sides by 1/3, the 3's cancel on the left, and we have `y = 13/6`.

So the solution to our system of equations is the ordered pair, `(3/2, 13/6)`.



There are three types of systems.

The last question we did was called a Consistent Independent System, because there was exactly one solution, or one common point between our two equations. If you were to graph the equations, you would see that we have two perpendicular lines.

Another type of systems is called Inconsistent Independent. Now this is a graph of two parallel lines. There are no common points and no solutions.

And finally, there's a system called Consistem Dependent. These are lines that are identical. There are infinitely many common points, and infinitely many solutions.



A riverboat takes 2 hours to travel downstream 55 kilometers. It takes 3 hours to travel 60 kilometers upstream. What is the speed of the boat and the speed of the current?

We'll need to translate our situation using mathematical language by creating more than one equation with more than one variable. So notice that we're given a time and a distance. Time and distance are related by the following equation: `"distance" = "rate"*"time"`.

Now, downstream, we have that the riverboat takes 2 hours to travel 55 kilometers. And upstream, we have that the boat takes 3 hours to travel 60 kilometers. Now let's go ahead and let the speed of the boat be represented by b, and the speed of the current be represented by c.

Since the boat is going to be traveling with the current, downstream, the rate is b+c. And then since the boat will be traveling against the current, upstream, the rate will be b-c.

So our system of equations is `55 = (b+c)*2` and `60 = (b-c)*3`. Now let's go ahead and simplify both of our equations, by dividing by 2 on both sides in our first equation, and dividing by 3 on both sides in our second equation. So we have `27.5 = b+c`, and then we have `20 = b-c`.

Let's use the elimination method. Notice that we have a positive c, and a -c here. So if we go ahead and add our equations, the c's will cancel, and we're left with `47.5 = 2b`, or `23.75 = b`, if we divide both sides by 2.

Now let's backsubstitute 23.75 into either one of our equations. I'll use our first equation, `27.5 = b+c`. So we have `27.5 = (23.75)+c`. And then if we subtract both sides by 23.75, then we're left with `c = 3.75`.

So the speed of the boat is 23.75 kilometers per hour, and the speed of the current is 3.75 kilometers per hour.