College Algebra

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Previous Lessons
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 nonlinear systems of equations.
- Use nonlinear systems of equations to solve applied problems.
- Graph nonlinear systems of inequalities.


We can label our first equation, 1, and our second equation, 2, and solve by elimination.

First, we'll rearrange our second equation so we have `-2x^2+y^2 = 10`.

And then we can go ahead and subtract that from our first equation. So we have `x^2+2x^2 = 3x^2`, `y^2-y^2 = 0`, and `25-10 = 15`. So if we divide both sides by 3, we have `x^2 = 5`. So `x = +-sqrt(5)`.

If we backsubstitute this into equation 1, we have `(+-sqrt(5))^2+y^2 = 25`, or `5+y^2 = 25`. And if we subtract 5 on both sides, we have `y = +-sqrt(20)`. And since 20 is `5*4`, we can pull out a 2, and leave the 5 inside of our radical.

So our solutions are the `(sqrt(5), 2sqrt(5))`, `(-sqrt(5), 2sqrt(5))`, `(sqrt(5), -2sqrt(5))`, and `(-sqrt(5), -2sqrt(5))`.



So let's check our solutions in our first equation.

So for `(sqrt(5), 2sqrt(5))`, we have `sqrt(5)^2+(2sqrt(5))^2 = 25`, or `5+20 = 25`, which is a true statement.

Then for `(-sqrt(5), 2sqrt(5))`, we have `(-sqrt(5))^2+(2sqrt(5))^2 = 25`. That's `5+20 = 25` again.

Then the `(sqrt(5), -2sqrt(5))` is the `sqrt(5)^2+(-2sqrt(5))^2 = 25`, which gives us `5+20 = 25` again.

And then `(-sqrt(5))^2+(-2sqrt(5))^2 = 25`, which is another true statement.

So it checks with our first equation.



Now let's check all of our solutions in our second equation.

So `y^2-2x^2 = 10`, or `(2sqrt(5))^2-2(sqrt(5))^2 = 10`. So we have `20-10 = 10`, which is a true statement.

And then if we plug in our second point, `(2sqrt(5))^2-2(-sqrt(5))^2 = 10`, which is 20-10, again, equals 10, a true statement.

And then for our third point, `(-2sqrt(5))^2-2(sqrt(5))^2 = 10`; that's another `20-10 = 10`. True statement.

And then for our last point, `(-2sqrt(5))^2-2(-sqrt(5))^2 = 10`, or `20-10 = 10`, a true statement.

So all of our solutions check.



Eddie Kim, the owner of The Graphics Studio, is designing a rectangular advertisement with an area of 28 square inches and perimeter of 22 inches. Find the dimensions of the advertisement.

So here's our rectangular advertisement; let's let the length be represented by big `L`, and the width be represented by big `W`. Then the area, the 28 square inches, is equal to the length times the width, and the perimeter, 22, is equal to 2 times the length + 2 times the width.

We can solve our first equation for L, and get `L = 28/W`, and plug that into equation 2. So we have `22 = 2(28/W)+2W`. Now let's divide everything by 2. So we have `11 = 28/W + W`. And then let's multiply both sides by `W`. So we have `11W = 28+W^2`. So if we bring everything over to the right, we have `W^2-11W+28 = 0`, which we can factor and get `0 = (W-7)(W-4)`. Since -7 and -4 gives us positive 28, but if we check our middle terms, we have -7W and -4W, which do add up to -11W.

So if we solve this equation for W, we have `W=7" or "4`.

If we plug this into equation 1, we have `L = 7 or 4`. So the dimensions of our advertisement are 7 by 4 inches.



Graph the system of inequalities. And then find the coordinates of the points of intersection of the graphs of the related equations.

`y >= x^2-5` and `y <= 4x`. So let's call our first inequality, 1, and our second inequality, 2.

Our first related equation is `y = x^2-5`. And our second related equation is `y = 4x`.

This is the graph of the parabola, `y = x^2`, shifted down 5 units. So the vertex is at the point (0, -5). And since the `x^2` term is positive, it opens up.

Now let's choose a test point of (0, 0), and plug this into our inequality to see where we need to shade. So if we use the test point (0, 0), we have `0 >= 0^2-5`, or `0 >= -5`. This is a true statement, so our shading will include our test value of (0, 0). So we're going to shade everything on and above our equation, `y = x^2-5`. Notice that we have a solid line, because our inequality includes the equal sign.

Now let's graph our related equation, `y = 4x`. The y-intercept is at the point (0, 0), and we have a slope of 4. And once again, we have a solid line, because the inequality includes the equal sign. Now let's pick a test value. We'll choose the point (1, 0). So we need to plug this into our second inequality. So we have `0 <= 4*1`, or `0 <= 4`. Since this is a true statement, our shading is going to include our test value.

So the solution to our system of inequalities is the overlapping region. And we have two points of intersection. It's going to intersect once here, and once somewhere up there. We can determine these points of intersection, by setting our related equations equal to each other and solving. So we have `x^2-5 = 4x`. And if we bring 4x to the left, we have `x^2-4x-5 = 0`. And we can factor this, so we have `(x-5)(x+1) = 0`. Since `x*x` gives us x^2, and `-5*1 = -5`, and our middle terms add up to -4x.

So we have `x = 5" and "-1`. And if we plug this into equation 2, when `x = 5`, we have `y = 20`. And when `x = -1`, we have `y = -4`. So our points of intersection are (5, 20) and (-1, -4).