Algebra – Study Guide

Graphing Equations


One of the most common methods of graphing an equation is by using an x-y T-chart. T-charts use a few points to plot and connect a
line. Normally we choose between 3 and 5 x-values to “plug in” and solve for y (the x-values should be both positive and negative and should stay close to 0). A typical set that can be used for most problems are x={-2, -1, 0, 1, 2}.

Example: Given that y=2x-1. Substituting the x={-2, -1, 0, 1, 2} to find the corresponding y values, the chart will correspond as follows:



-2 -5
-1 -3
0 -1
1 1
2 3


For example, with x=-2, y=2x-1=2(-2)-1=-5. Follow this same procedure for all of the x-values that you have selected. Plotting and connecting them gives the finished graph:

Graphing Inequalities

To graph an inequality you first look to see if it has an “or equal to” sign, =, a less than or equal to sign \leq or a greater than or equal to sign, \geq. If there is a \geq or \leq symbol, you will use a solid line for the graph. If it is only a less than or greater than sign, < or >, respectively, you will use a dashed line instead.

When plotting an inequality, plot the line like normal using the appropriate dashed or solid lien and then shade the appropriate side of the line that describes which side of the graph is greater or less than. To determine which part of the line to shade, you choose a test point [(x,y) coordinate] that is not on the line and put the x and y-values into the equation. If the statement this creates is true, you hade towards that point. If the statement this creates is incorrect, you shade away from the point. If possible, choosing the origin is a good choice of test point (remember it cannot fall on the line in order to use it).

Example: Graph the following inequality

y\leq 2x +1

First make a T-chart as plugging x={-2,-1,0,1,2} to find the corresponding y values



-2 -3
-1 -1
0 1
1 3
2 -5


Since the inequality is less than or equal to, draw a solid line for y=2x+1 Then, to decide which direction to shade, pick a point (let’s use (0,0)) and see if that point makes the inequality true or false. 0\leq2(0)+1 gives us 0\leq1, which is true. Therefore, shade toward the point (0,0)


Example 2: Graph the following inequality


First make a T-chart with x={-4, -2, 0, 2, 4} substitutions and the corresponding y-values



-4 -4
-2 -3
0 -2
2 -1
4 0

Since, the inequality is greater than, draw a dashed line for y=\frac{1}{2}\cdotX -2 Then, to decide which direction to shade, pick a point (let’s use (0)) and see if that point makes the inequality true or false. 0> \frac{1}{2}(0) -2 gives us 0>-2, which his true. Therefore, shade toward the point (0,0).

Distribution and the FOIL Method

The Distributive Property a(x+y)=ax+ay

Multiplying a number across a set of parentheses is the same as multiplying each of the terms inside the parentheses by that number and adding or subtracting them.


3(x+1)=3 \cdot x +3 \cdot 1 = 3x+3


4x(2x+8)=4x\cdot 2x + 4x \cdot 8 =8x^2+32x

The FOIL Method

FOIL (first-outer-inner-last) is a method used to multiply two binomial expressions. Multiply each pair of terms and add them together.
Example: Expand the following expression (x=3)(x+2)

The foil method says to expand it into simpler form:

“first” : (x)(x) = x^2
“outer” : (x)(2) = 2x
“inner” : (3)(x) = 3x
“last” : (3)(2) = 6

Then, you can add these terms together. The foil method says to expand into a simpler form:

(x + 3)(x + 2) = x^2 +2x +3x +6

Combining like terms gives the final answer of


Example: Expand the following expression

(x\cdot x)+(x \cdot (-1))+(4 \cdot x )+(4 \cdot (-1))

Combining like terms gives the final answer of


Factoring Trinomials

When factoring trinomials, there are a few different techniques that can be used. The simplest is to try and find a common factor in each of the terms that can be factored out of the expression. Then, you factor the resulting expression like normal.

Example: Factor x^3-x^2-2x

First, factor an x out of each of the terms:


Next, factor the binomial in the parentheses:


So, the factors are x, x+1, and x-2. The zeros are 0, -1, and 2.

Special Trinomials

There are a few types of special trinomials that are easy to factor once you know the pattern. Perfect square trinomials can be factored by following these rules:


Example: Factor x^2+2x+1
First, recognize that this is a perfect square trinomial with a=x, 2ab=2x, and b=1
Second, apply the appropriate rul from above the factor.

Differences of Squares

Another useful factoring technique involves a difference of squares. It follows this pattern:

This means that if you have 2 squares subtracted from each other (whether they are variables, numbers, or a combination of the two doesn’t matter), the expression can be factored.

Example: Factor x^2-9

Recognize that this is a difference of squares because x\cdot x = x^2 and 3^2=9

You then know that a=x and b=3. So, x^2-9=(x+3)(x-3)

Quadratic Formula:

The general form of at quadratic equation is

ax^2+bx+c=0 where a\neq0
The Quadratic Formula to find the root(s) for any quadratic equation is

x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}
The various possibilities of roots are determined by what is called the discriminate. THe discriminate is the expression contained underthe square root symbol: b^2-4ac


if b^2-4ac>0 \rightarrowthere are two real roots
if b^2-4ac=0 \rightarrow there is a real repeated root
if b^2-4ac<0 \rightarrow there are two complex roots

Example: 2x^2-4x-3=0
There are factors of (2)(-3)=-6 that add up to -4, so this quadratic cannot be factored; Instead, use the Quadratic Formula. In this case, a-2, b-4, and c-3.

x=\frac{-(-4)\pm \sqrt{(-4)^2-4(2)(-3)}}{2(2)}
= \frac{4\pm \sqrt{16+24}}{4} = \frac{4\pm \sqrt{40}}{4} = \frac{4\pm \sqrt{4} \sqrt{10}}{4}
= \frac{4\pm 2\sqrt{10}}{4}=\frac{2(2\pm \sqrt{10})}{2(2)}=\frac{2\pm \sqrt{10}}{2}
=\frac{2-\sqrt{10}}{2}, \frac{2+\sqrt{10}}{2}
\approx -0.58113883, 2.5811388=x
Then the answers are x=-0.58 and x=2.58, rounded to two decimal places.

Cross Multiplication

Cross multiplication is a way to solve proportions. Examples of equations where cross multiplication can be used are shown below:

\frac{4}{16}=\frac{x}{2} or \frac{6}{y} = \frac{3}{5}


To solve problems with cross multiplication, first draw an ‘X’, of where the multiplication is going to take place, connecting the diagonal terms as shown below:

The numbers of variables in the denominator should be multiplied with the diagonal term in the numerator and set equal to the other set of the diagonal terms multiplied together.

(2 \cdot 4)=(16 \cdot x) or (6 \cdot 5) = (y \cdot 3)
18=16x or 30=3y
The equation can then be solved by dividing to isolate the variable.

x=\frac{1}{2} or y=10


Cross multiplication can still be used even with binomial in the fraction.

The only difference is that we now have to distribute that 6 to both of the terms in the binomial.

(1 \cdot 6) =((3 \cdot x)-(3 \cdot 3))
After the 3 has been distributed, the problem can be solved as above.

6 = 3x -9
3x = 15
x = 5

Systems of Linear Equations

Systems of linear equations can be used to model and solve lots of interesting, real-world problems. A system of linear equations is made up of two or more equations. In order to completely solve the system, the number of variables must equal the number of equations. IN order to completely solve the system, the number of variables must equal the number of equations. An example of a system with two equations and two unknowns is shown below:


There are there ways to solve systems of linear equations: graphing, substitution, and elimination. We will go through an example of each method.

Method 1: Graphing

The first step in the graphing method is to convert the equations to slope-intercept form. This is done by isolating the y in the given equations.

Now that we have the equations in point-slope form we can easily graph them. If the two lines intersect at one point (x,y), the system has no solution. If the two lines are exactly the same, the system has an infinite number of solutions. The plots below show all of the possible types of solutions.

There is one solution at the point (-1,2). The solution to the system is x=-1 and y=2

The two lines are parallel. There are no solutions to this system.

The two lines are exactly the same. Since they are right on top of each other there are an infinite number of solutions to the system.

Method 2: Substitution

We are going to use a similar technique in the substitution method. But in this case, we are only going to solve one of the equations.

y=\frac{-1}{2}x+\frac{3}{2} \rightarrow -3x+4y=11

The next step in the substitution process is to take the first equation and plug it into the second equation. The arrow above shows where to make the substitution (plug in -\frac{1}{2}x+\frac{3}{2} for y).


Using the distributive property and combining like terms we can now solve the equation from x.


The last step in the substitution method is plug our newly solved x back into one of the equations and solve for y. It doesn’t matter which equation, the both give the same answer.

2x+4y=6 -3x+4y=11
2(-1)+4y=6 -3(-1)+4y=11
-2+4y=6 3+4y=11
4y=8 4y=8
y=2 y=2

The substitution method has now given us that x=-1 and y=2. Notice that this is exactly the same answer we found in the graphing method.

Method 3: Elimination

The final method for solving systems of linear equations is called elimination. In elimination, we either add or subtract the two equations to or from each other. Doing this eliminates either the x or y variable and allows us to solve for the other. We are going to start off with a slightly different set of equations for this example.

x+2y=3 (Equation #1)
-3x+4y=11 (Equation #2)

The first step in elimination is to get either the x or y coefficients the same in both equations. In this example we have ‘2y’ is the first equation and ‘4y’ in the second equation. In order to get the same coefficient in both equations, we have to multiply the entire first equation by 2.


Now we have ‘4y’ in both equations. If your equations already have a common coefficient then the first step can be skipped. Now the second equation is subtracted from the first equation. This eliminates the y variable and allows us to solve for x


The last step, as with the substitution method, is to plug the x back into either one of the original equations and solve.

x+2y=3 -3x+4y=11
1(-1)+2y=3 -3(-1)+4y=11
-1+2y=3 3+4y=11
2y=4 4y=8
y=2 y=2

Just like in the first two methods we get the solution to the system of x=-1 and y=2.


The important thing to remember with percent problems is to pay attention to the wording.

Example: 19 is what percent of 30?

To go from a sentence to algebra you should look for key words. “Is” is the first key word in this problem. This is the same as saying “is equal to” and can be replaced by a “=”. So now we have 18=”what percent” of 30. What percent is the thing you are solving for and can be replaced by a variable (x). “Of” in this case means to multiply. So now we have 18=x \cdot 30 for our equation. Solving gives x=0.6 which is a decimal answer and needs to be changed into a percent. To go from a decimal to a percent you multiply 100\% and you get x=60\% as the final answer. This problem can also be set up as a proportion, knowing that a percent is always going to be out of a total 100\%. In this problem the proportion can be set up as follows: \frac{18}{30}=\frac{x}{100}. Then you can cross-multiply 18 \cdot 100=30x, solve for x to get 60\%.

Example 2: What is 20\% of 80?

Again replace “what” with “x”, replace “is” with “=”, and replace “of” with “\cdot” and you get x=20\% \cdot 80. Convert the % to a decimal and you get x=.2 \cdot 80 which means x=16. Proportions can also one used in this set up: \frac{x}{80}=\frac{20}{100}. Then you cross-multiply 20 \cdot 80=100x, solve for x to get x=16.

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