Chapter 1 - Equations, Inequalities, Problem Solving

33
Assessment 1

1
- Assessment 1
Chapter 2 - Linear Inequalities and Absolute Value

24
Assessment 2

1
- Assessment 2
Chapter 3 - Linear Equations with Two Variables

28
Assessment 3

1
- Assessment 3
Chapter 4 - Systems of Linear Equations

23
Assessment 4

1
- Assemssment 4
Chapter 5 - Exponents, Radicals and Polynomials

27
Assessment 5

1
- Assessment 5
Chapter 6 - Exponential and Logarithmic Functions

30
Assessment 6

1
- Assessment 6
Final Exam

2
- Final Exam
- DRC Final Exam

Introductory Algebra

1.6 – Formulas

Formulas

To begin our discussion on formulas, we need to identify parts of formulas. A constant is an unchanging number, while a variable is a number that can change. For example, in the expression $3x+2y$ , the numbers 3 and 2 are unchanging, thus, they are constants. The letters $x$ and $y$ , however, can represent any number. $x$ and $y$ then, are variables.

Perimeter is the length of the sides of a 2-dimensional object.

Perimeter of a Rectangle
is the sum of all of the sides of a rectangle: two lengths added to two widths, or: $P=l+l+w+w$ Which we can simplify to be $P=2l+2w$

Perimeter of a Rectangle

is the sum of all of the sides of a rectangle: two lengths added to two widths, or:

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$P=l+l+w+w$
Which we can simplify to be
$P=2l+2w$

A rectangle has a length of 12 inches and a width of 5. What is the perimeter?
We can simply substitute the length and width values into the formula, and simplify: $P=2l+2w$ $P=2(12)+2(5)$ $P=24+10$ $P=34$

Perimeter of a Triangle
with sides $a$ , $b$ , and $c$ is given by the sum of these three sides, that is: $P=a+b+c$

Perimeter of a Triangle

with sides $a$ , $b$ , and $c$ is given by the sum of these three sides, that is:

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$P=a+b+c$

A Triangle has lengths of 3, 4 and 5. Find its perimeter.
The perimeter is the sum of these sides: 3+4+5=12

Area is a measure of two-dimensional space.

Area of a Rectangle
The area of a rectangle is $A=l\times w$ . Where $l$ is the length of the rectangle and $w$ is the width. We can then use this formula to calculate the area for a rectangle with known length and width.

A rectangle has a length of 5, and a width of 8. Find its area.
Utilizing our formula, we have: $A=l\times w$ $A=5\times 8$ $A=40$

Some more difficult shapes can be “cut” into smaller pieces. The area of such a shape is the sum of the areas of the smaller pieces.

Find the area of the shape below
Because the object is not a rectangle, we can not use the rectangle formula to find its area. But we can create rectangles by cutting the object. Let’s draw such a cut: In drawing this cut, we have created two rectangles. The rectangle on the left has a side length of 6, while the width is not immediately known. This can be gathered from the fact that the botton of the original object was 6, and the top line is 4. the width of the smaller, left rectangle is the difference: 6-4=2. The larger, right rectangle has length of 8 and width of 4: Thus, the area of the left rectangle is $6\times2=12$ while the area of the right rectangle is $8\times4=32$ : Thus, the total area of the object is $12+32=44$

Find the area of the shape below

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Because the object is not a rectangle, we can not use the rectangle formula to find its area. But we can create rectangles by cutting the object. Let’s draw such a cut:

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In drawing this cut, we have created two rectangles. The rectangle on the left has a side length of 6, while the width is not immediately known. This can be gathered from the fact that the botton of the original object was 6, and the top line is 4. the width of the smaller, left rectangle is the difference: 6-4=2.

The larger, right rectangle has length of 8 and width of 4:

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Thus, the area of the left rectangle is $6\times2=12$ while the area of the right rectangle is $8\times4=32$ :

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Thus, the total area of the object is $12+32=44$

Distance, Rate and Time
A formula that we will encounter periodically is the distance formula. The distance ( $d$ ) traveled by an object moving at a rate ( $r$ ) for a time period ( $t$ ) is given by the formula: $d=r\cdot t$

A runner travels 10 miles per hour for two hours, what is his distance traveled?
plugging in these values into the formula, we have: $d=rt$ $d=10\cdot2$ $d=20$

A car drives 50 miles for 4 hours. What is the distance traveled?
$d=rt$ $d=50\cdot4$ $d=200$

Variation

Often, two amounts are related to each other either directly, or indirectly (such as inversely). We say that an amount is directly related, or directly proportionate to another amount if the first amount is a multiple of the second. That is:

Direct Variation
An amount $y$ varies directly to an amount $x$ if, for some constant value $k$ we have the equation $y=kx$

The questions involving direct variation generally describe one set of values for the variation, and then ask you to find another set of values. To do this, we must use the first set of values to solve for $k$ , the variation constant, and then use that constant to find the second set of values.

$y$ varies directly as $x$ . If $y$ is 15 when $x$ is 5, find $y$ when $x$ is 8.
We are given the first set of values that solve the variation equation. “y is 15 when x is 5” – This tells us that in the variation equation: $y=k\cdot x$ We can substitute the $x$ and $y$ value in for the $x$ and $y$ in the equation: $15=k\cdot 5$ And we can solve this equation for the $k$ value: ${\color{red}\frac{1}{5}}\cdot15=k\cdot 5\cdot{\color{red}\frac{1}{5}}$ $3=k$ We now know the constant $k$ for this equation. We can then use this to find the $y$ value for when $x$ is 8: $y=3\cdot 8$ So $y$ is 24 when $x$ is 8

$y$ varies directly as $x$ . If $y$ is 15 when $x$ is 5, find $y$ when $x$ is 8.

We are given the first set of values that solve the variation equation. “y is 15 when x is 5” – This tells us that in the variation equation:
$y=k\cdot x$
We can substitute the $x$ and $y$ value in for the $x$ and $y$ in the equation:
$15=k\cdot 5$
And we can solve this equation for the $k$ value:
${\color{red}\frac{1}{5}}\cdot15=k\cdot 5\cdot{\color{red}\frac{1}{5}}$
$3=k$ We now know the constant $k$ for this equation. We can then use this to find the $y$ value for when $x$ is 8: $y=3\cdot 8$
So $y$ is 24 when $x$ is 8

$y$ varies directly as $x$ . If $y$ is 4 when $x$ is 3, find $y$ when $x$ is 21.
We insert our known $x$ and $y$ values for the first known set of values into the direct variation equation: $y=kx$ $4=k\cdot 3$ And we solve to find that $k=\frac{4}{3}$ . Then we can insert the now known value of $k$ and the next value for $x$ to find its corresponding $y$ value. $y=\frac{4}{3}\cdot 21$ $y=28$

$y$ varies directly with $x$ . If $y$ is 24 when $x$ is 6, find the value of $y$ when $x$ is 5.
$y=k\cdot x$ $24=k\cdot 6$ $k=4$ So when $x$ is 5, $y=4\cdot 5$ $y=20$

Another way that amounts can relate to each other is called indirect variation, or inverse variation. If a value, $y$ , inversely varies with another value, $x$ , then we say that y is equivalent to some constant $k$ value divided by the $x$ value. That is:

Indirect (Inverse) Variation
An amount $y$ varies inversely with $x$ if for some constant $k$ we have the equation $y=\frac{k}{x}$

Inverse variation problems can be solved in a similar fashion as direct variation problems.

$y$ varies inversely as $x$ . If $y$ is 8 when $x$ is 2, find the value of $y$ when $x$ is 8.
Similar to direct variation, we can insert the known values of $x$ and $y$ into the inverse variation equation to find the value of $x$ . $y=\frac{k}{x}$ $8=\frac{k}{2}$ $16=k$ And now that we know our constant $k$ value, we can find the $y$ value that corresponds to $x=8$ : $y=\frac{16}{8}$ $y=2$

$y$ varies inversely with $x$ . If $y$ is 20 when $x$ is 3, find the value of $y$ when $x$ is 5.
Inserting the known solution into the equation, we have: $y=\frac{k}{x}$ $20=\frac{k}{3}$ $60=k$ And we find the second solution: $y=\frac{60}{5}$ $y=12$

We might also see other variations. It is important to carefully read the questions when dealing with these types of problems.

$y$ varies inversely with the square of $x$ . If $y$ is 2 when $x$ is 3, find the value of $y$ when $x$ is 2.
In this particular problem, $y$ is inversely variant with the square of $x$ . This means we have the equation: $y=\frac{k}{x^2}$ We can again input the first solution of the equation to find $k$ : $y=\frac{k}{x^2}$ $2=\frac{k}{3^2}$ $2=\frac{k}{9}$ $18=k$ We now know $k$ , so we can input our next $x$ value and find the corresponding $y$ value: $y=\frac{18}{2^2}$ $y=\frac{18}{4}$ $y=\frac{9}{2}$ So the corresponding $y$ value is $\dfrac{9}{2}$ , or equivalently, 4.5

$y$ varies inversely with the square of $x$ . If $y$ is 2 when $x$ is 3, find the value of $y$ when $x$ is 2.

In this particular problem, $y$ is inversely variant with the square of $x$ . This means we have the equation:
$y=\frac{k}{x^2}$
We can again input the first solution of the equation to find $k$ :
$y=\frac{k}{x^2}$
$2=\frac{k}{3^2}$
$2=\frac{k}{9}$
$18=k$ We now know $k$ , so we can input our next $x$ value and find the corresponding $y$ value: $y=\frac{18}{2^2}$ $y=\frac{18}{4}$ $y=\frac{9}{2}$
So the corresponding $y$ value is $\dfrac{9}{2}$ , or equivalently, 4.5

Solving for a Random Variable

In many situations, it is required to solve an equation containing many variables for one of them. To do so, we can employ the Addition and Multiplication principles of equalities to isolate the desired variable.

Solve the equation for $r$ : $ar-b=g$
We do not have any numbers in this equation, but that is okay. We can still use the principles to isolate $r$ . We will start by adding $b$ to each side of the equation: We can then completely isolate $r$ by dividing each side of the equation by $a$ : $\frac{ar}{\color{red}a}=\frac{g+b}{\color{red}a}$ We can simplify the left side, but there is nothing we can do to simplify the right, so we leave it as it is, and have completely solved the equation for $r$ : $r=\frac{g+b}{a}$

Solve the equation for $r$ : $ar-b=g$

We do not have any numbers in this equation, but that is okay. We can still use the principles to isolate $r$ . We will start by adding $b$ to each side of the equation:

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We can then completely isolate $r$ by dividing each side of the equation by $a$ :

$\frac{ar}{\color{red}a}=\frac{g+b}{\color{red}a}$
We can simplify the left side, but there is nothing we can do to simplify the right, so we leave it as it is, and have completely solved the equation for $r$ :

$r=\frac{g+b}{a}$

Solve the given equation for $y$ : $ax+by=c$
Again, we will start by adding or subtracting any terms away from the $y$ term. Here, we subtract $ax$ from each side. And we can then completely isolate $y$ by dividing $b$ from each side: $\frac{by}{\color{red}b}=\frac{c-ax}{\color{red}b}$ And simplify the left side: $y=\frac{c-ax}{b}$