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Algebra Posts

Top 5 Tested Intermediate Algebra Concepts on the ACT

By posted August 22, 2011

1. Quadratics – Quadratic equations have three terms and are in the form ax² + bx + c. An example of a quadratic is x² – 5x + 6. To find the factors of this equation, we must set up our set of two parentheses: (        )(        )

The first term in both parentheses must be x, since x multiplied by x is the only way to get x². Then we look at the coefficient of the second term, -5. It’s important to include the sign in front of the integer as part of the coefficient. One of the rules of quadratic equations is that the second terms in the two factors must add together to equal the middle term’s coefficient. So we need to think of two numbers that add together to give us -5.

Already, we can think of many combinations: -6 and 1, -2 and -3, -200 and 105. So which pair is it? Now we have to look at the integer that’s the third term of the quadratic. Here it’s  + 6. Another rule of quadratic equations is that the third term of the quadratic equation will equal the product of the second terms in the two factors. So not only do we need the two numbers to add together to equal -5, but we need them to multiply together to equal + 6. Therefore the factors must be: (x – 2) (x – 3). The “roots” or the “solutions” for this quadratic would be 2 and 3.

2.  Systems of Equations – The ACT will often present you with two or more equations with multiple variables. Remember the “n equations with n variables rule.” If there are 2 variables in an equation (for example, x and y), then there must be 2 equations that each contain those variables in order to solve. The two common ways to solve are Substitution and Combination. Jordan Schonig reviews each method in detail here.

3. Functions –It’s helpful to think of (x, f(x)) as another way of writing (x, y). For many function questions, you can Pick Numbers for the variables to solve! Read more »

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Elementary Algebra Pt. 1

By posted February 24, 2010

Algebra is a branch of mathematics that describes equations and arithmetic “symbolically.” Whereas in normal arithmetic an expression like “2+2” has a definite answer, in algebra unidentified numbers can be symbolically represented by variables; “a,b,c,x,y, or z”. A simple addition in algebra could look like this “a+b=c”. With these concepts, we can look at an algebraic equation.

“x+6 = 16”

When we are presented with an algebraic equation like this, we can actually “solve the equation” which just means we can find the numeric value of the ‘x‘ variable in the equation. The numeric value of x that satisfies the equation, keeps the equation true, is called the solution to the equation.

So how do we start solving any equation? Well, there are a number of operations we can do on any equation and these operations are our main tools in Algebra. In Algebra, the number one concept to understand is that on equality. Any equation, or equality, can be modified by doing the same arithmetical operation to BOTH SIDES of the equation. For example, let’s take another look at the equation “x+6 = 16”. Let’s subtract a constant from both sides of the equation, in this case 6.

(x+6) – 6 = (16) – 6 On the left side of the equation, 6-6 = 0, so those two terms cancel out leaving only x. And on the right side of the equation, 16-6 = 10, so 10 remains on the right side. So after we have subtracted 6 from both sides, we are left with this equation: x = 10. Notice the form of this final equation. What is left is actually the solution to the equation! If we now know that x is equal to 6, then if we substitute 6 for x in the original equation, we’ll find that the equality is still true.

X + 6 = 16 : (10) + 6 = 16 Notice that 10 + 6 does equal 16, so we know that x = 10 is a correct solution to the equation. This was a very simple example of solving an algebraic equation. Once again, the most important concept to understand about solving algebraic equations is how to do operations on BOTH sides of the equation. The main method of solving a simple equation like this is called “isolating the variable” where on one side of the equation is only the variable term and on the other side is everything else. With this, we have the basics of elementary algebra.

Let’s try a harder problem and go through all the concepts so far:

3x + 24 = 8 + 5x : notice that there are 4 terms in the equation; 3x and 24 in the left expression, 8 and 5x in the right expression. So if we want to solve the equation, we need to isolate the variable, ‘x‘. In order to do that, we have to do arithmetical operations to each side of the equation. Since there a term with x on both sides of equation, we can do an operation to manipulate the equation so that x is only on one side of the equation. Since 3x is less than 5x, we can subtract 3x from both sides of the equation thereby leaving the equation with x on only one side of the equation.

(3x + 24) – 3x = (8 + 5x) – 3x :

3x – 3x + 24 = 8 + 5x – 3x :

24 = 8 + 2x
So now that we have only one x term on one side of the equation, we can take the next step in isolating x by subtracting 8 from both sides of the equation.

(24) – 8 = (8 + 2x) – 8 :

16 = 2x
One more step and we can completely isolate x thereby finding the solution for the equation. Let’s divide both sides of the equation by 2.

(16)/2 = (2x)/2 :

8 = x
Now we have solved the equation for x. If we go back to the original equation and substitute 8 for x, we will hopefully find that the equality still holds true.

3*(8) + 24 = 8 + 5*(8) :

24 + 24 = 8 + 40 :

48 = 48

This just about wraps up the basics of elementary algebra. In the next part I will go more in depth and develop the basic concepts discussed in this first part. For now, there are a few definitions to understand that are important for every stage of algebra from here on out.

Definitions:

term – Any number or variable, coefficient or not, in an equation or expression. Example (terms are in bold):

3x + 4 = y/2 + 10

variable – A symbol used to represent either a single number or a set of possible numbers. Example: “a,b,c,x,y,z”

expression – A set of numbers, variables, and various arithmetical operations connecting each term. Example “4x²+ 2x + 9”

equation – An equation says that two “things”, either terms or expressions, are equal to each other.

Example: “7 – c = 1 + 2c”

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How to Factor “Factoring”

By posted January 11, 2010

To factor in mathematics is to express a quantity as the product of two or more quantities. Take the integer 8 for example. Two factors of 8 are 2 and 4. We can express this as an equation: 2 x 4 = 8. There are two types of expressions you will need to factor on the SAT: polynomials and quadratics.

A polynomial is a numerical expression with more than one term. For example, 4x + 2x. If we were asked to factor this expression, we would want to see what is common to both terms and then divide it from each term. Right away, we see that both terms contain x. We also notice that one of the terms contains 2 and that 2 is a factor of the coefficient of the first term, 4. So we know we can factor out 2x from this equation. Once we’ve found one of the factors, we put it in parentheses and leave what is left behind in another set of parentheses. Here’s how it would look:

4x + 2x = (2x)(2 + 1)

We must have a 1 left in the second parentheses to act as a place-holder for the 2x. A quick way to check your work is to distribute the 2x across the terms and make sure you get 4x + 2x. It’s important to make sure you have the correct factors and this extra step is a quick way to make sure you don’t make any simple mistakes.

The quadratic equation is a special polynomial. It has three terms and is in the form ax² + bx + c. The first term will always be squared, the middle term will always have the same variable, and the last term will contain only an integer.

An example of a quadratic is x² – 5x + 6. To find the factors of this equation, we must set up our set of two parentheses:

(        )(        )

The first term in both parentheses must be x, since x multiplied by x is the only way to get x². Then we look at the coefficient of the second term, -5. It’s important to include the sign in front of the integer as part of the coefficient. One of the rules of quadratic equations is that the second terms in the two factors must add together to equal the middle term’s coefficient. So we need to think of two numbers that add together to give us -5.

Already, we can think of many combinations: -6 and 1, -2 and -3, -200 and 105. So which pair is it? Now we have to look at the integer that’s the third term of the quadratic. Here it’s  + 6. Another rule of quadratic equations is that the third term of the quadratic equation will equal the product of the second terms in the two factors. So not only do we need the two numbers to add together to equal -5, but we need them to multiply together to equal + 6. Therefore the factors must be:

(x – 2) (x – 3)

Those are the two “factors” of our original equation: x² – 5x + 6. If an SAT question asks for the roots or solutions of a quadratic equation, you would simply set the two factors each equal to zero and solve for x.

(x – 2) = 0                   (x – 3) = 0

+2   +2                        +3    +3

x  = 2                           x  = 3

Therefore the “roots” or “solutions” to this quadratic are 2 and 3. For more practice with factoring, create your own game on Grockit and choose to focus on Algebra questions!

To factor in mathematics is to express a quantity as the product of two or more quantities. Take the integer 8 for example. Two factors of 8 are 2 and 4. We can express this as an equation: 2 x 4 = 8. There are two types of expressions you will need to factor on the SAT: polynomials and quadratics.

A polynomial is a numerical expression with more than one term. For example, 4x + 2x. If we were asked to factor this expression, we would want to see what is common to both terms and then divide it from each term. Right away, we see that both terms contain x. We also notice that one of the terms contains 2 and that 2 is a factor of the coefficient of the first term, 4. So we know we can factor out 2x from this equation. Once we’ve found one of the factors, we put it in parentheses and leave what is left behind in another set of parentheses. Here’s how it would look:

4x + 2x = (2x)(2 + 1)

We must have a 1 left in the second parentheses to act as a place-holder for the 2x. A quick way to check your work is to distribute the 2x across the terms and make sure you get 4x + 2x. It’s important to make sure you have the correct factors and this extra step is a quick way to make sure you don’t make any simple mistakes.

The quadratic equation is a special polynomial. It has three terms and is in the form ax² + bx + c. The first term will always be squared, the middle term will always have the same variable, and the last term will contain only an integer.

An example of a quadratic is x² – 5x + 6. To find the factors of this equation, we must set up our set of two parentheses:

(        )(        )

The first term in both parentheses must be x, since x multiplied by x is the only way to get x². Then we look at the coefficient of the second term, -5. It’s important to include the sign in front of the integer as part of the coefficient. One of the rules of quadratic equations is that the second terms in the two factors must add together to equal the middle term’s coefficient. So we need to think of two numbers that add together to give us -5.

Already, we can think of many combinations: -6 and 1, -2 and -3, -200 and 105. So which pair is it? Now we have to look at the integer that’s the third term of the quadratic. Here it’s  + 6. Another rule of quadratic equations is that the third term of the quadratic equation will equal the product of the second terms in the two factors. So not only do we need the two numbers to add together to equal -5, but we need them to multiply together to equal + 6. Therefore the factors must be:

(x – 2) (x – 3)

Those are the two “factors” of our original equation: x² – 5x + 6. If an SAT question asks for the roots or solutions of a quadratic equation, you would simply set the two factors each equal to zero and solve for x.

(x – 2) = 0                   (x – 3) = 0

+2   +2                        +3    +3

x  = 2                           x  = 3

Therefore the “roots” or “solutions” to this quadratic are 2 and 3. For more practice with factoring, create your own game on Grockit and choose to focus on Algebra questions!

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