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MATH: BASIC ALGEBRA Section II DePaul Math Placement Test

Equation Solving There are a number of algebraic terms you should know in order to be able to talk and think about algebra. Variable: A variable unknown quantity, written as a letter. X and Yare the most commonly used letters for variables, but can be represented by any letter in the English or any Greek alphabets. Constant: A quantity that does not change is termed as a constant. It can also be a variable’s coefficient. In an algebraic equation, you’ll find that addition and subtraction signs often separate terms from one another. For example, in the equation: 3x 3 +2x 2 -7x+4 = x-1 Expression: An expression can be as simple as a single constant term, like 5. It can also be as complicated as the sum or difference of of constants and variables, such as {(x 2 + 2) 3 – 6x} ⁄ 7x 5 . Expressions don’t include an “equals” sign, which is what differentiates expressions from equations. Expressions therefore cannot be solved; they can only be simplified. Equation: Two expressions linked by an equal sign.

2 Manipulating Equations Now that you know how to set up an equation (previous section), the next thing you need to do is solve for the value that the question asks for. The most important thing to remember when manipulating equations is that each side of the equation must be manipulated in the same way. By treating the two sides of the equation in the same way , you won’t violate the equality of the equation. You will have to change the form of the equation—that’s the point of manipulating it. But the equation will always remain true as long as you do the same thing to both sides. What happens when you manipulate the equation 3x + 2 = 5 , with x= 1 . Subtract 2 from both sides: (3x + 2)-2 = 5-2 2 x+0=3 3(1)= 3 3 =3 This simple addition operation shows that you can tamper with the equation as long as you tamper the same way on both sides. If you follow this rule, you can manipulate the question without affecting the value of the variables .

Solving an Equation with One Variable To solve an equation with one variable, you must manipulate the equation to isolate that variable on one side of the equation. remember PEMDAS. Here you will go in reverse order ( SADMEP ! ). The idea is to “undo” everything that is being done to the variable so that it will be isolated in the end. Let’s look at an example: Solve for x in the equation [(3x 2 +5)X3]/4 +1 = 61 First, subtract 1 from both sides of the equation: [(3x 2 +5) X 3] /4+1 -1 = 61-1 => [(3x 2 +5) X 3]/4 = 60 Then, multiply both sides of the equation by 4: (3x 2 +5) X 3 = 60 X 4 => (3x 2 +5) X 3 = 240 Next, divide both sides of the equation by 3: (3x 2 +5) X 3 ÷3 = 240÷3 =>(3x 2 +5) = 80 Now, subtract 5 from both sides of the equation: 3x 2 +5-5 = 80 -5 => 3x 2 = 75 Again, divide both sides of the equation by 3 : 3x 2 ÷ 3 = 75÷3 => x 2 = 25 Finally, take the square root of each side of the equation: x = ±5 , We have isolated x to show that x = ±5 .

Solving an Equation with One Variable Sometimes the variable that needs to be isolated is not located conveniently. For example, it might be in a denominator, or an exponent. Equations like these are solved the same way as any other equation, except that you may need different techniques to isolate the variable. Let’s look at a couple of examples: Solve for x in the equation 1 /x +2 =4 1 /x = 2 1 = 2x x =1/2 The key step is to multiply both sides by x to extract the variable from the denominator. It is not at all uncommon to have to move the variable from one side to the other in order to isolate it. Having just given you a very basic introduction to solving equations, we’ll reemphasize two things: 1.Do the same thing to both sides. 2.Work backward (with respect to the order of operations).

Distributing and Factoring Distributing and factoring are two of the most important techniques in algebra. They give you ways of manipulating expressions without changing the expression’s value. In other words, distributing and factoring are tools of reorganization. Since they don’t affect the value of the expression, you can factor or distribute one side of the equation without doing the same for the other side of the equation. The basis for both techniques is the following property, called the distributive property: a X (b+c+…) = a X b +a X c ‘ a’ can be any kind of term, from a variable to a constant to a combination of the two .

Factoring Factoring an expression is essentially the opposite of distributing. Consider the expression 4x 3 – 8x 2 + 4x , for example. You can factor out the greatest common factor of the terms, which is 4x : 4x 3 – 8x +4x = 4x(x 2 -2+1) The expression simplifies further: 4x –( x 2 -2 +1) = 4x(x-1) 2 See how useful these techniques are? You can group or ungroup quantities in an equation to make your calculations easier. In this example from the previous section on manipulating equations, we distributed and factored to solve an equation. Distributing eliminates parentheses, and factoring creates them.

Combining Like Terms There are other steps you can take to simplify expressions or equations. Combining like terms is one of the simpler techniques you can use, and it involves adding or subtracting the coefficients of variables that are raised to the same power. For example, by combining like terms, the expression x 2 - x 3 +4x 2 +4 x 2+ 3x 2 can be simplified to (-1+3)x 3 + (1+4) x 2 = 2 x 3 + 5x 2 by adding the coefficients of the variable x 3 together and the coefficients of x 2 together. The point is, you’d rather have one term, 7x 2 , instead of x 2 , 3x 2 , –3x 2 , 2x 2 , and 4x 2 all floating around in your expression. A general formula for combining like pairs looks like this: ax k + bx k + cx k = x k (a+b+c)

Zero Product Rule When the product of any number of terms is zero, you know that at least one of the terms is equal to zero. For example, if xy = 0 , you know that either: x = 0 and y ≠ 0, y = 0 and x ≠ 0, or x = y = 0 This is useful in a situation like the following: (x+4)(x-3) = 0 (x+4) =0 or (x-3) = 0 By the zero product rule, you know that (x + 4) = 0 or (x – 3) = 0. In this equation, either x = –4 or x = 3 , since one of the expressions in parentheses must be equal to 0. Consider this equation: 3x 2 (x+2) =0 Again, since 3x 2 or (x + 2) must equal 0, x = 0 or x= –2 Keep your eye out for a zero product. I t’s a big time-saver, especially when you have multiple answers to choose from.

Systems of Equations Sometimes a question will have a lone equation containing two variables, and using the methods we’ve discussed thus far will not be enough to solve for the variables. Additional information is needed, and it must come in the form of another equation. For example, that a single equation uses the two variables x and y . Try as you might, you won’t be able to solve for x or y . But given another equation with the same two variables x and y , the values of both variables can be found. These multiple equations containing the same variables are called systems of equations. There are essentially two types of systems of equations that you will need to be able to solve. 1. First step involves easier type involves s ubstitution 2. The second involves manipulating equations simultaneously .

Substitution Simply put, the substitution method involves finding the value of one variable in one equation and then substituting that value into the other equation to solve for the other variable. Here’s a straightforward example: If x – 4 = y – 3 and 2y = 6 , what is x ? In this case, we have two equations. The first equation contains x and y . The second contains only y. To solve for x , you must solve for y in the second equation and then substitute that value for y in the first equation, eliminating the second variable from that equation. If 2y = 6 , then y = 3 , and substituting that into the first equation: x = y – 3+4 = 3-3+4 = 4

Substitution Here is a slightly more complicated example: Suppose 3x = y + 5 and 2y – 2= 12k . Solve for x in terms of k . Again, you cannot solve for x in terms of k using just the first equation. Instead, you must solve for y in terms of k in the second equation and then substitute that value in the first equation to solve for x . 2y – 2= 12k 2y = 12k+2 y= 6k + 1 Then substitute y = 6k + 1 into the equation 3 x = y + 5. 3x = y + 5 3x = (6k+1)+5 3x = 6k + 6 x = 2k + 2

Linear Equations Linear Equations refer to equations that can be added or subtracted from each other in order to find a solution. Consider the following example: Q. Suppose 2x + 3y = 5 and –1x – 3y = –7 . What is x ? A. In this particular problem, you can find the value of x by adding the two equations together: 2x +3y =5 +(-1x) -3y =-7 x =-2 And Plug the value of x in either equation to find the value of y . Taking the first equation , 2x+3y =5, x =-2. 2(-2) +3y =5. -4 +3y =5 3y =5+4 3y = 9 y = 9/3 y = 3

Linear Equations Here’s another example: Q. 2x + 3y = –6 and –4x + 16y = 13 . What is the value of y ? A. The question asks you to solve for y , which means that you should find a way to eliminate one of the variables by adding or subtracting the two equations. 4x is simply twice 2x , so by multiplying the first equation by 2, you can then add the equations together to find y. 2 X (2x + 3y = –6) = 4x + 6y = –12 Now add the equations and solve for y. 4x +6y =-12 +(-4x) +16y = 13 22y = 1 y= 1/22

Polynomials and Binomials Polynomial is an expression that contains one or more algebraic terms, each consisting of a constant multiplied by a variable raised to a power greater than or equal to zero. For example , x 2 + 2x + 4 is a polynomial with three terms . The interesting thing to note is that the third term is 4x 0 = 4 . In such situations where M (x 0 ) is equal to M . Where is a M is a constant. Here’s a different situation : 2 x –1 , on the other hand, is not a polynomial because x is raised to a negative power. Binomial is a polynomial with exactly two terms: x + 5 and x 2 – 6 are both binomials.

Multiplying Binomials There is a simple acronym that is useful in helping to keep track of the terms as you multiply binomials: FOIL . It stands for F irst, O uter, I nner, L ast . This is the order in which you multiply the terms of two binomials to get the right product. For example, if asked to multiply the binomials (x + 1)(x – 3), you first multiply the first terms of each binomial: x X x = x 2 Next, multiply the outer terms of the binomials: x X 3 =3x Then, multiply the inner terms: 1 X x = x Finally, multiply the last terms: 1 X 3 =3 Combine like terms and you have your product: 2x 2 + -2x +6x + -6 = 2x 2 +4x-6 Here’s a few more examples: (y+3)(y-7) = y 2 -7y+3y -21 = y 2 -4y -21 (-x+2)(4x+6) = -4x 2 -6x +8x +12 = -4x 2 + 2x +12

Multiplying Polynomials It may seem like a daunting task. But when the process is broken down, multiplying polynomials requires nothing more than distribution and combining like terms—and some attention to detail. To find a product of a polynomial, just distribute the terms of the first polynomial into the second polynomial individually, and combine like terms to formulate your final answer. For Example: (x 2 + x+4)(2x+5x 2 -6x-3)= x 2 (2x 3 +5x 2 -6x -3)+ x(2x 3 +5x 2 -6x-3) = 2x 5 +5x 4 -6x 3 -3x 3 +2x 4 +5x 3 -6x 2 -3x = 2x 5 +7x 4 +7x 3 +11x 2 -27x -12 If we look at the FOIL method a little more closely, we can see how each of these terms is constructed: (x+a)(x+b)=x 2 +(a+b) x +ab You can see that the constant term is the product of the two constants in the original binomials and the x coefficient is simply the sum of those two constants. In order to factor x 2 + 10 x + 21 into two binomials ( x + a )( x + b ), you must find two numbers whose sum is 10 and whose product is 21 .

Slope of Straight Line- Slope-Intercept Form Slope The slope of a line is a measurement of how steeply the line climbs or falls as it moves from left to right. More technically, it is a line’s vertical change divided by its horizontal change, informally known as “the rise over run.” Given two points on a line, call them ( x 1 , y 1 ) and ( x 2 , y 2 ) , the slope of that line can be calculated using the following formula of a slope – intercept form : Slope = y 2 –y 1 x 2 - x 1 The variable most often used to represent slope is m. Here’s an example, the slope of a line that contains the points (–2, –4) an d (6, 1) is: m= 1-(-4) = 5 6-(-2) 8

Point-Slope Form The point-slope form of the equation of a line is: y-y 1 = m(x-x 1 ) where m is the slope of the line and ( x 1 , y 1 ) is a point on the line. The point-slope form and slope-intercept form are just alternative ways of expressing the same equation. In fact, the slope-intercept form is the point-slope form taken at the y-intercept, or the point ( 0, y 1 ): y - y 1 = m(x-0) y - y 1 = mx y = mx- y 1 Since, y 1 =b (the y -intercept is simply the y -coordinate of the point at which x=0), the two forms are equal. The slope-intercept form of the line equation is the more common of the two, but the point-slope form is extremely useful when all the information you have is the slope and a point (hence “point-slope”).

Point-Slope Form Q. What is the slope-intercept equation of the line that contains the point (3,4 ) and is perpendicular to the line y = x – 6 ? A. To answer this question, you need to first find the slope of the line whose equation you are trying to write. Fortunately, the question gives you the slope of a perpendicular line, and we know that the slope of a line is the opposite reciprocal of the slope of the line to which it is perpendicular. Thus, the slope is –1⁄ (1/3) = –3 . If the line contains the point ( 3, 4 ), its point-slope equation is y – 4 = –3(x – 3) . To convert this to slope-intercept form, use algebra: y- 4 =-3(x-3) y- 4 =-3x+9 y =-3x+13 Here try it yourself: What is the slope-intercept form of the equation of the line that contains the points ( 5, 3 ) and (–1, 8)?

Positive and Negative Slopes You can easily determine whether the slope of a line is positive or negative just by looking at the line. If a line slopes uphill as you trace it from left to right, the slope is positive. If a line slopes downhill as you trace it from left to right, the slope is negative. You can get a sense of the magnitude of the slope of a line by looking at the line’s steepness. The steeper the line , the more extreme the slope will be; The flatter the line , the smaller the slope will be. Note that an extremely positive slope is bigger than a moderately positive slope while an extremely negative slope is smaller than a moderately negative slope. Look at the lines in the figure on the left hand side and try to determine whether the slope of each line is negative or positive and which has a greater slope: Lines a and b have positive slopes, and lines c and d have negative slopes. In terms of slope magnitude, line ma > mb > mc > md.

Special Slopes Here are a few slopes you should recognize by sight. If you can simply see one of these lines and identify its slope without having to do any calculations you will save yourself a lot of time. A line that is horizontal has a slope of zero . In this case there is no “rise,” y 2 – y 1 = 0 , and thus, m = ( y 2 – y 1 )/( x 2 – x 1 ) = 0/( x 2 – x 1 ) = 0 . A line that is vertical has an undefined slope . In this case, there is no “run,” and x 2 – x 1 = 0. Thus , m = ( y 2 – y 1 )/( x 2 – x 1 ) = ( y 2 – y 1 )/0 and any fraction with 0 in its denominator is, by definition, undefined. A line that makes a 45º angle with a horizontal has a slope of 1 or –1 . This makes sense because the “rise” equals the “run,” and y 2 – y 1 = x 2 – x 1 or y 2 – y 1 = –( x 2 – x 1 ). Now, take the four lines on the left hand size. Let’s decide which is which. Line a has slope 0 because it is horizontal. Line b has slope - 1 because it slopes downward at 45 0 as you move from left to right. Line c has slope 1 because it slopes upward at 45 0 as you move from left to right. Line d has undefined slope because it is vertical.

Parallel and Perpendicular Lines Parallel lines are lines that don’t intersect. In coordinate geometry, they can also be described as lines with the same slope. Perpendicular lines are lines that intersect at a right angle. In coordinate geometry, perpendicular lines have opposite, reciprocal slopes. That is, a line with slope m is perpendicular to a line with a slope of –(1 / m ). In the figure, lines q and r both have a slope of 2, so they are parallel. Line s is perpendicular to both lines q and r and thus has a slope of –(1/ 2). Now, test your knowledge of the topics discussed by clicking on the sample test links given below.

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