- Draw Pictures
- Apply Logic
- Answer the Question That They Ask You
- Check the Choices
- Pick Numbers for the Variables
- Skip Around
- Read the Questions Carefully
- Look for "New Operations"
- Multiple Choice Versus Student-Produced Response
- Practice Questions
- Answer Explanations
This book assumes a basic understanding of algebra. Our focus will be on reviewing some general concepts and applying those concepts to questions that might appear on the PSAT math sections.
Understanding Linear Equations with One Variable
In a linear equation with one variable, the variable cannot have an exponent or be in the denominator of a fraction. An example of a linear equation is 2x + 13 = 43. The PSAT will most likely require you to solve for x in that equation. Do this by isolating x on the left side of the equation, as follows:
2x + 13 = 43
2x = 43 13
2x = 30
x = , or 15
Understanding Systems of Equations
A system of equations is a set of equations that is satisfied by the same set of values of the variables. On the PSAT, a system of equations will generally take the form of two linear equations with two variables. Following is an example of a system of equations, and the steps necessary to solve it:
4x + 5y = 21
5x + 10y = 30
The first step will be to make either the x-values or the y-values cancel each other out. In the above system, it will be easier to work with the y-values, since 5 is a factor of 10. Multiply each element in the first equation by -2:
-2(4x) + -2(5y) = -2(21)
-8x – 10y = -42
Now, you can add the like terms in each equation:
(-8x + 5x) = -3x
(-10y + 10y) = 0
-42 + 30 = -12
-3x = -12
Therefore, notice that the two y-terms cancel each other out. Solving for x, you get x = 4. Now, choose one of the original two equations, plug 4 in for x, and solve for y:
4(4) + 5y = 21
16 + 5y =21
5y = 5
y = 1
The solutions for the system of equations are x = 4 and y = 1.
Understanding Polynomial Operations and Factoring Simple Quadratic Expressions
A polynomial is the sum or difference of expressions like 2x2 and 14x. The most common polynomial takes the form of a simple quadratic expression, such as: 2x2 + 14x + 8, with the terms in decreasing order. The standard form of a simple quadratic expression is ax2 + bx + c, where a, b, and c are whole numbers. When the terms include both a number and a variable, such as x, the number is called the coefficient. For example, in the expression 2x, 2 is the coefficient of x.
The PSAT will often require you to evaluate or solve a polynomial by substituting a given value for the variable.
For example: If x = -2, what is the value of 2x2 + 14x + 8 = ?
2(-2)2 + 14(-2) + 8 =
2(4) + (-28) + 8 =
8 – 28 + 8 = -12
You will also be required to add, subtract, multiply, and divide polynomials. To add or subtract polynomials, simply combine like terms, as in the following examples:
(2x2 + 14x + 8) + (3x2 + 5x + 32) = 5x2 + 19x + 40
(8x2 + 11x + 23) – (7x2 + 3x + 13) = x2 + 8x + 10
To multiply polynomials, use the Distributive Property to multiply each term of one polynomial by each term of the other polynomial. Following are some examples:
(3x)(x2 + 4x – 2) =(3x3 + 12x2 – 6x)
Remember the FOIL Method to help solve some polynomial problems: multiply the First terms, then the Outside terms, then the Inside terms, then the Last terms.
For example: If (2x + 5)(x – 3) = ax2 + bx + c for all values of x, what is the value of b?
First terms: (2x)(x) = 2x2
Outside terms: (2x)(-3) = -6x
Inside terms: (5)(x) = 5x
Last terms: (5)(-3) = -15
Now put the terms in decreasing order:
2x2 + (-6x) + 5x + (-15) =
2x – x – 15
Therefore, the value of b is 1, because the coefficient of x is 1.
You might also be asked to find the factors or solution sets of certain simple quadratic expressions. A factor or solution set takes the form, (x ± some number). Simple quadratic expressions will usually have 2 of these factors or solution sets. The PSAT might require you to calculate the values of the solution sets.
For example: If (2x + 5)(x – 3) = 0, what are all the possible values of x?
Set both elements of the equation equal to 0.
(2x + 5) = 0
(x – 3) = 0
Now solve for x:
2x + 5 =
2x = -5
x 3 =
x = 3
The possible values of x are and 3.
Following are some general factoring rules that might prove useful for answering PSAT math questions:
Finding the difference between two squares: a2 – b2 = (a + b)(a – b)
Finding common factors, such as: x2 – 2x = x(x + 2)
Factoring quadratic equations, such as: x2 +2x –8 = (x + 4)(x – 2)
Understanding Linear Inequalities with One Variable
Linear inequalities with one variable are solved in almost the same manner as linear equations with one variable: by isolating the variable on one side of the inequality. Remember, though, that when multiplying both sides of an inequality by a negative number, the direction of the sign must be reversed.
For example: If 3x + 4 > 5x + 1, then x =
First, isolate x on one side of the inequality.
3x – 5x > 1 – 4
-2x > -3
Now, since you have to divide both sides by -2, remember to reverse the inequality sign.
Understanding Inequalities and Absolute Value Equations
An inequality with an absolute value will be in the form of |ax + b| > c, or | ax + b|< c. To solve |ax + b| > c, first drop the absolute value and create two separate inequalities with the word OR between them. To solve |ax + b|< c, first drop the absolute value and create two separate inequalities with the word AND between them. The first inequality will look just like the original inequality without the absolute value. For the second inequality, you must switch the inequality sign and change the sign of c. For example,
To solve |x + 3| > 5, first drop the absolute value sign and create two separate inequalities with the word OR between them.
x + 3 > 5 OR x + 3 < -5
Solve for x:
x > 2 OR x < -8
To solve |x + 3| < 5, first drop the absolute value sign and create two separate inequalities with the word AND between them:
x + 3 < 5 AND x + 3 > -5
Solve for x:
x < 2 AND x > -8
A function is a set of ordered pairs in which no two of the ordered pairs has the same x-value. In a function, each input (x-value) has exactly one output (y-value). An example of this relationship would be y = x2. Here, y is a function of x because for any value of x, there is exactly one value of y. However, x is not a function of y because for certain values of y, there is more than one value of x. The domain of a function refers to the x-values, whereas the range of a function refers to the y-values. If the values in the domain correspond to more than one value in the range, the relation is not a function.
For example: Let the function f be defined by f(x) = 2(3x). What is the value of f(5)?
Solve this problem by substituting 5 for x wherever x appears in the function:
f(x) = x2 – 3x
f(5) = (5) 2 – (3)(5)
f(5) = 25 – 15
f(5) = 10