Next Generation Advanced Algebra and Functions Study Guide for the ACCUPLACER Test
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General Information
This is the test taken by entering college students who are majoring in a STEMrelated field or in a nonSTEM field that requires a great deal of math expertise. It is assumed that you are already comfortable with most of the concepts and procedures reviewed in our practice materials for the other two Accuplacer Next Generation Math tests: Arithmetic and Quantitative Reasoning, Algebra, and Statistics. If you have any doubts about this, please look at the study guides, practice questions, and flashcards for those tests on our site. Most of the concepts covered here are more advanced and require prior knowledge of many prerequisite skills that are covered on those tests.
The 20 questions on the Accuplacer Next Generation Advanced Algebra and Functions test address the following skill areas in a fairly equal manner, with perhaps slightly more emphasis on the areas noted in italics.

Linear equations

Linear applications and graphs

Factoring

Quadratics

Functions

Radical and rational equations

Polynomial equations

Exponential and logarithmic equations

Geometry concepts for Algebra 1

Geometry concepts for Algebra 2

Trigonometry
You also need to be comfortable working with slope, radians and degrees, and “must be true” statements. Facility with relevant labeled graphs and skill in determining their meaning is also required.
Note: It would be best to check with your testing center for confirmation, but our information indicates that no handheld calculators are allowed during any Accuplacer tests. Apparently, there are some questions that allow calculator use, and an onscreen calculator will appear on your screen for those questions, only.
Linear Equations
A linear equation relates a variable to a constant, or to the product of a constant and another variable. As expected, linear equations produce straight lines when graphed.
These are all examples of linear equations:
\(y = 3x + 2\)
\(y = 4\)
\(x = 0\)
It is worthwhile to become familiar with the different linear equation forms and the graphs they produce.
The equations \(y = c\) and \(x = c\), where c is a constant, produce horizontal and vertical lines, respectively.
It is also useful to become familiar with the terms intercept and slope. The xintercept of a line is the point at which a line crosses the \(x\)axis. The yintercept of a line is the point at which a line crosses the \(y\)axis.
The slope of a line is the ratio of the difference in \(y\)values of two points to the difference in \(x\)values of those same points. It is defined algebraically as:
\[m = \frac{y_2  y_1}{x_2  x_1}\]Consider the line passing through the points \((1, 2)\) and \((4, 6)\). To determine the slope, plug the respective \(x\) and \(y\) values into the equation for the slope of the line:
\[m = \frac{6  (2)}{4  (1)} = \frac{8}{5}\]When given two points, it is possible to calculate the slope, as we’ve just done, and it is also possible to derive an overall equation for the line, from which we can then find both the \(x\) and \(y\) intercepts.
Using a general equation, called the pointslope form of a line, we can derive the equation of the line. The pointslope form of a line is defined algebraically as:
\((y  y_1) = m(x  x_1)\), where \((x_1, y_1)\) is a point on the line and \(m\) is the slope of the line
It does not matter which point you select, but it is wise to choose a point that could potentially reduce the overall complexity of the calculations. In this case, selecting the point containing only positive values may simplify the calculation:
\((y  6) = \frac{8}{5}(x  4)\), which can be solved for \(y\) as follows:
\(y  6 = \frac{8}{5}x  \frac{32}{5}\) and adding \(6\) (in the form of \(\frac{30}{5}\)) to both sides:
\(y = \frac{8}{5}x  \frac{32}{5} + \frac{30}{5}\), which reduces to:
\[y = \frac{8}{5}x  \frac{2}{5}\]This final form is known as the slopeintercept form of a line because the equation provides both the slope of the line and the \(y\)intercept of the line. Stated generally, the slopeintercept form of a line is:
\(y = mx + b\), where \(m\) is the slope and \(b\) is the \(y\)intercept.
In the case of the example above, we’ve already solved for the slope, but the \(y\)intercept is \(\frac{2}{5}\).
The \(x\)intercept can be found by substituting \(0\) for the \(y\) and solving for \(x\):
\(0 = \frac{8}{5}x  \frac{2}{5}\), and adding the fraction to both sides:
\(\frac{2}{5} = \frac{8}{5}x\), and multiplying both sides by the reciprocal of the \(x\) coefficient:
\(\frac{5}{8} \cdot \frac{2}{5} = x\), which simplifies to:
\[\frac{1}{4} = x\]The \(x\)intercept is then \((\frac{1}{4}, 0)\).
To summarize:
The slopeintercept form of a line is \(y = mx + b\), which readily provides the slope and \(y\)intercept of a line.
The pointslope form of a line is \((y  y_1) = m(x  x_1)\), which enables derivation of the line’s equation when the slope of the line and one of its points are known.
Lastly, there is the general equation of a line: \(Ax + By = C\), where A, B, and C are constants.
Each of these line forms can be manipulated to generate any of the others, and for any line, they will all produce the same graphical representation.
Creating Linear Equations
Creating equations involves translating a word problem or diagram into a relationship between algebraic expressions. Initially, this translation process will prove difficult, but as with everything, consistent practice develops the skill.
Consider the following one variable example:
“Ryan is half as old as Dominique. If Dominique is \(38\), how old is Ryan?”
Ryan’s unknown age can be represented as the variable \(r\) and related to Dominique’s age, \(38\). It is given that Ryan is half as old as Dominique; written algebraically:
\(r = \frac{38}{2}\), so \(r = 19\) and Ryan is 19 years old.
Consider the following two variable example:
Nancy is hired as a baker for a math club meeting. She earns a flat rate of \(\$10\) for the three hours she spends at the meeting. In addition to her flat rate, she earns \(\$0.25\) for every baked good sold. Her total pay is only a function of her flat rate and the number of goods sold.
We can use a linear equation to represent Nancy’s pay. By generating a linear equation that relates her total pay to the number of baked goods sold, we can predict how much money she will make if she sells a particular number of goods, and we may also be able to solve for how many goods were sold if we know how much money she earned.
We will use variables to represent the unknown quantities we wish to relate: \(P\) to represent her pay, and \(g\) to represent the number of goods sold and her flat rate.
Generating a relationship from the information provided thus far may produce an equation that reads something like:
“Nancy’s total pay, \(P\), is equal to the number of goods she sells, plus the flat rate she is paid.”
However, there is an inconsistency in units if we directly translate this expression into an algebraic equation. Adding the number of goods (which has units of “number of goods”) will not combine with her flat rate of $10 to give a payment that is in dollars. To resolve this inconsistency we must first multiply the number of goods sold by the price per good. Doing so eliminates the “number” unit and leaves a “dollar” unit, which can then be combined with her flat rate to find her total pay. Stated algebraically:
\(P = 0.25g + 10\), where \(P\) is total pay, and \(g\) is number of goods sold.
It is very important to maintain awareness of units in any equation so that an answer can be verified for reasonableness, and so that any error can be easily traced to its source.
Say Nancy sells \(4\) goods, her total pay would then be:
\[P = 0.25(4) + 10 = 1 + 10 = 11\]It is also worthwhile to notice that the equation can be solved for \(g\):
\[P  10 = 0.25g\]Dividing both sides by \(0.25\) (which is the same as multiplying both sides by \(4\)) gives us:
\[4(P  10) = g\]Now, if we have a value for \(P\), \(g\) can be directly computed.
Solving Linear Equations
Linear equations in one variable are solved by isolating the variable in question. Solving for a variable entails performing the order of operations in reverse so that the operations performed on the variable in question can be undone, and the variable itself can be found. This process is better explained through an example.
Consider this equation:
\[\frac{5(x + 3)}{2} = 10\]In this case, our goal is to end with \(x = C\), where C is some constant. In order to arrive at this final equation, it is necessary to undo all of the operations that are performed on the variable \(x\), starting with the most “recent” operation.
Looking at the expression on the left side, it can be seen that the first operation performed on \(x\) is the addition of \(3\). This entire expression, \((x + 3)\), is then multiplied by \(5\) and divided by \(2\). So, when isolating the variable \(x\), it is first necessary to undo the multiplication and division operations. To undo multiplication, it is necessary to divide and to undo division, it is necessary to multiply:
\[\frac{5(x + 3)}{2} = 10\]The first step is to multiply both sides by \(2\):
\[2 \cdot \frac{5(x + 3)}{2} = 10 \cdot 2\]Notice that this cancels out the division by \(2\) on the left side and ultimately brings us closer to isolating \(x\). The next step is to divide both sides by \(5\) to eliminate the multiplication by \(5\) on the left side:
\[\frac{5(x + 3)}{5} = \frac{20}{5}\]which simplifies to:
\[x + 3 = 4\]The variable \(x\) is being increased by \(3\), so to solve for just one \(x\), it is necessary to subtract \(3\) from both sides:
\[x + 3  3 = 4  3\]which simplifies to:
\[x = 1\]We began with a somewhat complicated expression, but we have discovered that it is equivalent to the equation \(x = 1\). It is wise to verify our answer when possible. Because we are claiming that \(x\) is equal to a particular value, and we have the original expression, we can determine whether \(x\) is actually equal to 1 by substituting this value back into the original expression and determining whether the substitution yields a true statement.
We began with \(\frac{5(x + 3)}{2} = 10\), and we are claiming that \(x = 1\), so:
\(\frac{5((1) + 3)}{2} = 10\), and following the order of operations for the left hand side:
\(\frac{20}{2} = 10\), or \(10 = 10\), which is a true statement, and our claim is verified.
For this equation, only \(x = 1\) will give a true statement.
In cases where you are given only one equation that contains multiple variables, it is only possible to solve for any one variable in terms of the other variables. In order to assign a value to any variable, it is necessary to have as many equations as you have variables. Consider, for example, this equation:
\[3(x + y) = 2z\]We can solve for \(z\) by dividing both sides by \(2\):
\[z = \frac{3(x + y)}{2}\]We can solve for either \(x\) or \(y\) by dividing both sides by \(3\) and subtracting by \(y\) if solving for \(x\), and subtracting by \(x\) if solving for \(y\):
\(x + y = \frac{2z}{3}\), which can then give:
\(x = \frac{2z}{3}  y\), or \(y = \frac{2z}{3}  x\)
Simplifying
Simplifying is the process of rearranging and manipulating equations and expressions to produce an equation or expression that is “simpler” in form. You may recall the many days spent learning how to reduce (or simplify) fractions, or you may have learned the concept of factoring; both are methods for simplifying expressions and will be elaborated upon in the following sections.
Linear Equations
Consider this linear equation:
\[3x + \frac{2(3x)}{4} = \frac{5  (3x + 2)}{6}\]Simplifying the equation entails combining like terms and solving for the variable. It is not always possible to immediately combine like terms, but through performing arithmetic operations (being sure to follow the order of operations), like terms may be found which can then be combined. The equation simplifies as follows:
\[3x + \frac{2(3x)}{4} = \frac{5  (3x + 2)}{6}\]Begin by distributing terms through the parentheses:
\[3x + \frac{6 + 2x}{4} = \frac{5 + 3x  2}{6}\]Now multiply both sides by \(12\) to eliminate the denominator from each side:
\(36x + 3(6 + 2x) = 2(5 + 3x  2)\), and distributing terms again:
\(36x  18 + 6x = 10 + 6x  4\), and combining like terms:
\(42x  18 = 6x + 6\), and moving terms to the opposite side to isolate the variable:
\(36x = 24\), and dividing both sides by \(36\) and reducing:
\[x = \frac{36}{24} = \frac{3}{2}\]Linear Inequalities
Aside from one subtle difference, linear inequalities are simplified in the same way that linear equations are simplified. The difference between the two is that any time a linear inequality is multiplied or divided by a negative value, the direction of the inequality sign switches (a greater than becomes a less than and vice versa).
Starting with the inequality: \(3(2x+6) \leq 18\)
Distribute the \(3\): \(6x  18 \leq 18\)
Add \(18\) to both sides: \(6x \leq 36\)
Divide both sides by \(6\) (remembering to flip the direction of the inequality since we’re dividing by a negative number): \(x \geq 6\)
So the simplified form of the inequality is \(x \geq 6\).
Solving Systems of Two Linear Equations
The solution of a system of two linear equations is defined as the point at which the two lines intersect. Consequently, the possible solution falls into three distinct cases: there is no solution, there is one solution, and there are infinite solutions.
There is no solution to a system of two linear equations that share the same slope, but are not multiples of one another. For example:
The system \(y = 2x\) and \(y = 2x  4\) has no solution because these lines never intersect and neither line can be formed by multiplying the other by a constant.
The system \(y = 2x\) and \(y = \frac{1}{2}x + 6\) has one solution because the lines intersect only once.
The system \(2y = 4x + 4\) and \(y = 2x + 2\) has infinite solutions, because the two lines intersect at every point (they are the same line).
Systems of equations can be solved in many ways, but the methods of substitution, elimination, and graphing are the three most common methods that you will encounter prior to advanced math courses.
As an example, consider the system of equations, which we will solve using substitution and elimination:
\[2x + y = 4\]and
\[2x + y = 2\]The method of substitution entails solving one of the equations for a variable, then substituting the expression that the variable is equal to into the same variable of the other equation. Begin by solving the first equation for \(y\):
\[y = 2x + 4\]The y variable in the second equation can be replaced with the expression \(2x + 4\):
\[2x + (2x + 4) = 2\]Notice that the equation now only contains one variable, which can be solved exactly by combining like terms and isolating the variable:
\(4x + 4 = 2\), which becomes
\(4x = 6\), which becomes
\[x = \frac{3}{2}\]Now, the value of \(x\) can be substituted into either equation to solve for the variable \(y\):
\(2(\frac{3}{2}) + y = 4\), which becomes
\(3 + y = 4\), so \(y = 1\)
The claim is then that the solution set is: \((\frac{3}{2}, 1)\). To verify the solution set, substitute the values of \(x\) and \(y\) into either equation:
\(2\frac{3}{2} + 1 = 4\), and evaluating the left side
\(3 + 1 = 4\), which is a true statement, so the solution set of the system of equations is \((\frac{3}{2}, 1)\).
Now we will solve the same system using the method of elimination. Elimination entails combining two equations such that one of the variables is eliminated in the process. In the case where two equations do not directly combine to eliminate one of the variables, one of the equations can be multiplied by an integer value so that elimination is possible through combination.
Given the original system:
\[2x + y = 4\]and
\[2x + y = 2\]it can be seen that adding the two equations together eliminates the \(x\) variable:
\(2x + (2x) + y + y = 4 + (2)\), which becomes
\(2y = 2\), which simplifies to \(y = 1\)
And, as before, this value of \(y\) would then be substituted into either original equation to determine the \(x\) value.
The method of graphing will be dealt with in the coming sections.
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