Math Study Guide for the SAT Exam
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More Algebraic Concepts
Absolute Value Expressions, Equations, and Inequalities
The absolute value of a number or expression is its distance from 0. Absolute values are always positive because they represent a magnitude. Expressions and equations containing absolute values (represented by an expression surrounded by two vertical lines) can be manipulated in much the same way expressions and equations lacking absolute values can be manipulated.
Consider the following:
\[3 2x + 3 = 12\]The goal, in this case, is to isolate the absolute value; this is accomplished by dividing both sides by \(3\):
\[2x + 3 = 4\]To solve an absolute value that has been isolated, rewrite the problem as two equations, one equal to positive \(4\) and one equal to \(4\). Solve each equation for \(x\):
\(\vert{2x+3}\vert = 4\) is rewritten as:
\(2x + 3 = 4\) and \(2x + 3 = 4\)
So:
\[x = \frac{1}{2}\] \[x = \frac{7}{2}\]These answers can be verified by substituting them back into \(x\) and evaluating.
Inequalities can also contain an absolute value. In the case of inequalities containing less than or less than and equal to signs (\(<, \le\)), the original inequality must be rewritten without the absolute value to indicate that the solution set includes all values between (and sometimes including) the positive and negative integer expressions in the original inequality. For example:
\[6x 8 \le 52\]is rewritten as:
\[52 \le 6x  8 \le 52\]Which can be broken apart to form two inequalities that can be solved individually:
\(52 \le 6x  8\) and \(6x  8 \le 52\)
So,
\(\frac{44}{6} \le x\) and \(x \le 10\)
This solution set would be graphed as two closed (shaded) circles at \(x = \frac{44}{6}\) and \(x = 10\), with the values between them being shaded as well. The closed circles indicate that the values at the beginning and end of the shaded line are also part of the solution.
Greater than or greater than or equal to inequalities containing an absolute value are solved in a similar fashion, for example:
\[23x2 \ge 18\]In this case, division by \(2\) results in:
\[3x  2 \ge 9\]Which can now be rewritten as:
\[3x  2 \ge 9\]or
\[3x  2 \le 9\]So,
\[x \ge \frac{11}{3}\]or
\[x \le \frac{7}{3}\]This solution set would be graphed with one closed circle at \(x = \frac{11}{3}\) and an arrow pointing to the right indicating that all values larger than \(\frac{11}{3}\) are also true. Another closed circle would be at \(x = \frac{7}{3}\) with an arrow pointing to the left of this value indicating that all numbers smaller than \(\frac{7}{3}\) are also true.
Specific Algebraic Skills to Practice
Besides knowing what mathematical terms mean, you need to be able to do things with them: perform operations, manipulate them, and generally use them to solve problems. Be sure you understand these procedures and know when to use them.
Create a Linear Equation from a Given Situation
Consider the following situation:
A company sells products that cost \(\$5.00\). Excusing any of their production costs, generate a linear equation modeling the company’s revenue if they sell \(13\) products.
In this case, there is a linear relationship between the number of products sold and the total revenue generated. If \(1\) product is sold, the revenue is \(\$5.00\); if \(3\) products are sold, the revenue is \(3 \cdot 5.00 = 15.00\). We can model this relationship as a function of the number of products sold, with each product netting \(\$5.00\):
Revenue = \(\$5.00 \times\) number of products sold
Let the variable \(p\) represent the unknown number of products sold, and \(R(p)\) represent the revenue generated:
\(R(p) = 5.00 \cdot p\) is the linear equation modeling this situation.
Always remember to consider what the independent variable is (in this case number of products sold), and what the dependent variable is (in this case the revenue generated), and create an equation that relates them appropriately.
Solve or Interpret a Linear Expression or Equation in One Variable
Consider the following singlevariable equation:
\[2  3(x + 2)  4x = 3 + 4x\]To solve this equation, manipulate the equation such that a single \(x\) is on one side of the equation, and a single value is on the other side. In this case, we will begin by distributing the \(3\) through the parentheses:
\[2 3x  6  4x = 3 + 4x\]Now, combine like terms on the left side of the equation:
\[4  7x = 3 + 4x\]Now, either add \(7x\) to both sides of the equation, subtract \(4x\) from both sides of the equation, add \(4\) to both sides of the equation, or subtract \(3\) from both sides of the equation. To avoid a negative sign in the variable’s coefficient, we will add \(7x\) to both sides:
\[4 = 3 + 11x\]Because we want all of the numbers on one side and only \(1 \,x\) on the other side, now subtract \(3\) from both sides. Remember that any operation performed on one side of the equation must also be performed on the other side of the equation to maintain equality:
\[7 = 11x\]Notice that the variable is currently multiplied by \(11\). To undo this operation, and to isolate just \(1\) \(x\), it is necessary to divide the right side by \(11\); and, to keep the equation equal, we must also divide the left side by \(11\):
\[\frac{7}{11} = \frac{11x}{11}\] \[\frac{7}{11} = x\]Our answer can be verified by substituting \(\frac{7}{11}\) into our original \(x\) and checking to see if the equation is true:
\[2  3(\frac{7}{11} + 2)  4 \cdot \frac{7}{11} = 3 + 4 \cdot \frac{7}{11}\] \[2  3(\frac{15}{11}) + \frac{28}{11} = 3  \frac{28}{11}\] \[2  \frac{45}{11} + \frac{28}{11} = 3  \frac{28}{11}\] \[\frac{22}{11}  \frac{45}{11} + \frac{28}{11} = \frac{33}{11}  \frac{28}{11}\] \[\frac{5}{11} = \frac{5}{11}\]The solution is correct.
On exams, presuming enough time is available, always verify your answer using the method above.
Work with Linear Inequalities in One Variable
Consider the following linear inequality in one variable:
\[3 + 2(4  x) \le 4x  1\]As with linear equations, linear inequalities are solved by manipulating the inequality such that the variable is on one side of the inequality and a value is on the other side. One important thing to remember when working with inequalities is that any time the inequality is multiplied or divided by a negative value, the direction of the inequality switches. For example, \(4x \ge 8\) becomes \(x \le 2\).
Let’s solve the inequality from above:
Distribute the \(2\): \(3 + 8  2x \le 4x  1\)
Combine like terms: \(11  2x \le 4x  1\)
Rearrange: \(12 \le 6x\)
Divide both sides by \(6\): \(2 \le x\)
As before, this solution can be verified by selecting a value for \(x\) that is larger than or equal to \(2\) and testing if it yields a true statement. We will leave the verification to you.
You will also be tested on your understanding of graphs of linear inequalities. The solution determined above would graph as a closed circle at \(x = 2\) with an arrow pointing to the right to indicate that all \(x\) values larger than \(2\) are also part of the solution set. The closed circle indicates that \(2\) is also part of the solution set. An open circle, contrastingly, indicates that the value where the circle is located is not part of the solution set.
Create a Linear Function Showing the Relationship Between Two Quantities
Consider the following situation:
In a chemical reaction, the concentration of reactant \(A\) decreases as the concentration of reactant \(B\) increases. It is determined that there is a linear relationship between the concentrations of the reactants as the reaction progresses. At one point, there are found to be \(10\) parts of reactant \(A\) for every \(1\) part of reactant \(B\). At another point, there are found to be \(4\) parts of \(A\) for every \(4\) parts of \(B\). Express the relationship between the two reactants using a linear function.
From the information provided, it can be seen that the concentrations of the reactants exhibit an inverse relationship. We can rewrite the relative concentrations as points on an \(AB\)coordinate plane, in which the \(A\)axis represents the concentration of \(A\), and the \(B\)axis represents the concentration of \(B\).
We are given two points then: \((10, 1)\) and \((4, 4)\). From our knowledge of linear equations, we can determine the slope of the line that relates these two variables:
Slope = \(\frac{y_2  y_1}{x_2  x_1} = \frac{4  1}{4  10} = \frac{1}{2}\)
Now that we have the slope, we can use the pointslope formula to determine the linear function describing the relationship:
\[(y  4) = \frac{1}{2} (x  4)\] \[y  4 = \frac{1}{2}x  2\] \[y = \frac{1}{2}x + 2\]where \(y\) is the concentration of \(B\), and \(x\) is the concentration of \(A\).
Work with Linear Equations in Two Variables
Consider the following problem:
\[3xy + 4x  2y + 3(x  y) = 4y + 2x\]Solve for \(y\) in terms of \(x\).
Because we have only one equation, but two variables, we can only express one variable in terms of the other. To solve for \(y\) in terms of \(x\) means to end up with an equation with \(y\) on one side and everything else on the other.
The first step, in this case, entails distributing the \(3\) through the parentheses so that every term is clarified:
\[3xy + 4x  2y + 3x  3y = 4y + 2x\]Now, move every term containing a \(y\) to one side of the equation and everything else to the other side of the equation:
\[3xy  2y  3y  4y = 2x  4x  3x\]Combine like terms:
\[3xy  9y =  5x\]Notice that \(y\) can be factored out of the expression on the left:
\[y(3x  9) = 5x\]We can now divide both sides by the parentheses to solve for \(y\) in terms of \(x\):
\[y = \frac{5x}{3x9}\]Work with Linear Inequalities in Two Variables
Consider the following linear inequality:
\[3x  4(y + 3) > 2y + 2(3  x)\]As with a linear equation with two variables, we will begin by manipulating the inequality to solve for \(y\) in terms of \(x\):
Distribute the \(4\):
\[3x  4y  12 > 2y + 6  2x\]Combine like terms through rearrangement:
\[4y  2y > 2x + 3x + 6 + 12\] \[6y > x + 18\]Recall, when multiplying or dividing both sides of an inequality by a negative value, the inequality sign switches directions:
\[\frac{6y}{6} < \frac{x + 18}{6}\] \[y < \frac{1}{6}x  3\]To graph this solution, we first find two points that lie along the line. Because this is already in slopeintercept form, we know that \((0,3)\) is one point on the line, and using the slope of \(\frac{1}{6}\), we can subtract \(1\) from the \(y\) value of the \(y\)intercept and add \(6\) to the \(x\) value of the \(y\) intercept:
\[(0 + 6, 3 + 1) = (6, 4)\]Plot the points \((0, 3)\) and \((6, 4)\), and because our inequality has a less than symbol, draw a dotted line extending through both points. Because \(y\) is less than the graphed line, the solution set is every point below the dotted line. The solution set is indicated by a shaded region.
Work with Systems of Two Linear Equations in Two Variables
Consider the following system of equations:
\[4x + y = 6\] \[3x  2y = 8\]Three methods of solving this system should be available to you: elimination, substitution, and graphing.
Elimination
The goal in solving a system by elimination is to multiply one of the equations by a value such that combining both equations will result in the elimination of one of the variables. In this case, by multiplying the top equation by \(2\), the coefficients of the \(y\) variable will cancel and the \(x\) variable can be found:
\(2(4x + y = 6)\) becomes \(8x + 2y = 12\)
Adding the first equation with the second yields:
\[5x = 20\] \[x = 4\]Now that \(x\) has been solved, its value can be substituted into either equation to solve for \(y\):
\[4(4) + y = 6\] \[16 + y = 6\] \[y = 10\]The solution set is \((4, 10)\).
Substitution
Solving by elimination is not always the most efficient method. Sometimes it is best to solve a system using substitution. The goal of solving by substitution is to express one of the variables in terms of the other. The expression can then be substituted into the other equation, and the value for the variable can be found:
\[4x + y = 6\] \[3x  2y = 8\]Solve the first equation for \(y\):
\[y = 4x + 6\]Now substitute this expression for \(y\) into the other equation:
\[3x  2(4x + 6) = 8\] \[3x + 8x  12 = 8\] \[5x = 20\] \[x = 4\]Substituting this value of \(x\) into either equation will yield the value of \(y\), \(10\).
Graphing
Solving a system of equations using the method of graphing entails graphing both lines and finding their point of intersection. There are three cases of solutions:

The two lines do not intersect. This happens when the two lines are parallel but have different \(y\) or \(x\) intercepts. In this case, there is no solution.

The two lines intersect at one point. This happens when the two lines are distinct but do not share the same slope. In this case, there is one solution.

The two lines intersect at every point. This happens when the two lines are the same. In this case, there are infinite solutions.
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