Pre Algebra Study Hall

Basic operations

Addition, subtraction, multiplication, and division should already have already been mastered. And besides, a calculator is allowed on the SAT and therefore these operations shouldn’t be a factor. Just take the time to double check calculations, both by hand and calculator—calculations by hand can easily mean silly goof-ups and calculations by calculator can mean typo errors.

Substituting values for variables

Most equations have at least one variable. A variable is the letter in the equation. It's called a "variable" because it's value varies, it can be different at different times. For instance, a symbol that looks like this: "5" we understand to always mean "five". A symbol that looks like this: $x$ might stand for "five" or "twenty-seven" or "243.287" or whatever.

It’s necessary to be able to substitute values (actual numbers) for variables (usually shown as letters such as $x$ or $y$). For instance:

If $x=4$ and $y=5$, then $x^2+2y=$

To solve this equation, substitute “4” in the x spot and “5” in the y spot to get:


The answer is $(4 \times 4) + (2 \times 5)=26$. Or, $16 + 10 = 26$.

Order of operations

The order of operations defines the order in which an algebraic equation is solved. Without a set order, answers would vary.

For instance, take the problem:


If you simply move from left to right, you’ll get 14 minus 2 equals 12, times 10 equals 120. That’s incorrect. The order of operations tells us that multiplication comes before addition. So, the correct procedure is 2 times 10 is 20, and 14 minus 20 is -6. -6 is the correct answer.

The correct order of operations is usually shown with the acronym: PEMDAS. Each letter stands for an operation, as follows:

  1. Parentheses
  2. Exponent
  3. Multiply
  4. Divide
  5. Add
  6. Subtract

For instance, take the problem:

$18+3(4-2^3)=6x + {4^3 \over 2}$

To attack this problem and solve for x you’ll need to use the Order of Operations or PEMDAS correctly. First handle the parentheses—use PEMDAS within the parentheses. Notice that inside the parentheses is an exponent, take care of it next so that $2^3$ becomes 8. $4 - 8$ then becomes -4. Looking at the whole equation again, there’s another exponent. $4^3$ becomes 64.

Now the problem reads: $18+3(-4)=6x+{64\over2}$

Multiply next. $3(-4) = -12$.

Divide next. ${64\over2} = 32$.

Now the problem reads: $18-12=6x+32$. This becomes $6=6x+32$ when becomes $-26=6x$.

Finally, the solution is: ${-26\over6}=x$ which reduces to ${-13\over3}=x$. This is your answer.

The importance here is to remember that not sticking to the Order of Operations would yield a wrong answer. Use PEMDAS!

Solving linear equations with a single variable

A “linear equation” is a fancy way to refer to a “regular” algebra problem. The term “linear equation” implies that the equation could be graphed on a 2D graph (an graph with x and y coordinates).

The two main rules in solving a linear or algebraic equation are to:

  • Isolate the variable on one side of the “=” sign
  • Do the same thing to both sides of the “=” sign

For example:

$10x-28=-8$ Solve for x.

To solve this equation, first add 28 to both sides to get:


Now divide both sides by 10 to get $x=2$. This isolates the variable and thus, 2 is the answer.


Fractions abound on the SAT and ACT. Manipulating them successfully is crucial. Make sure you’re comfortable adding, subtracting, multiplying, and dividing them. Also, it’s critical to reduce fractions all the way.

  • Adding fractions: find the LCD (least common denominator), add numerators, reduce
  • Subtracting fractions: find the LCD, subtract numerators, reduce
  • Multiplying fractions: simply multiply numerators and denominators, then reduce
  • Dividing fractions: “flip” or invert the number you’re using to divide, next multiply straight across, then reduce. For example:

You may need to increase the font size to read these fractions. (PC users: CTRL + and MAC users: APPLE +)

${{1\over2}\over{3\over4}} =x$

“Flip” the 3 and 4 so that the equations now appears as:

${{1\over2}\over{4\over3}} =x$

Bring the denominator up even with the numerator, then multiply across to yield:


Now reduce down to

${2\over3}$ and that’s your answer.
Just to double check, use your calculator and type in .5 divided by .75 and you’ll get .666666667, which of course, is $2/3$.

Fractions as decimals as percentages

Fractions, decimals, and percentages can be turned into one another and vice-versa. They’re interchangeable. This fact is helpful for double checking math by use of a calculator. This is true also on a basic calculator that cannot handle fractions. For example, a fraction problem can be easily converted to decimals and then tabulated on even the simplest of calculators.

An example of fraction/decimal/percent interchangeability might be: ${2\over5}$ is equal to .40 which is 40%. This number is achieved by dividing the denominator into the numerator, or 5 into 2. Also, $3 {2\over5}$ is equal to 3.4 which is 340%. This number is achieved by changing an “improper fraction” (one with numbers and fractions like ${17\over5}$) to a “proper fraction” (one with just a numerator and denominator like ). This change is done by multiplying the denominator by the root number, then adding the numerator, or in this case 5 times 3 then plus 2, to yield ${17\over5}$. 17 is then divided by 5 to yield 3.4 and 340%.

Converting word problems to algebraic equations

Word problems abound on standardized tests. Don’t let them intimidate you. Word problems are simply math problems written in “English” rather than in “Math.” Just as a phrase in Spanish such as, “Donde esta el bano?” can be converted to English as, “Where is the bathroom?”, mathematical word problems can usually be easily converted from “English” to “Math”.

Consider this word problem: 39% of what number is 186?

This may seem intimidating with its odd numbers, but it’s very easy to solve. The word of refers to multiplication. The phrase “what number” can be our variable x. The word is refers to the equals sign. Therefore, our algebraic equality now reads: $.39*x=186$ This is easily solved by dividing both sides of the equation by .39. 186 divided by .39 yields 476.9. That is to say, 39% of 476.9 is 186.

In all honesty, all word problems will not be as simple as the one above. They usually involve more words and a longer “story.” In fact, they’re sometimes called “story problems.” Don’t let the story (a) distract you or (b) somehow intimidate you. In essence, they’re usually rather simple algebra problems—you just have to convert the words to numbers, symbols, and variables and thus set up the equation.

Here are a few tips.

  • Don’t let the wording fool you or intimidate you! You can solve it!
  • Figure out what needs to be known and let that be x.
  • Take the time to define your variables. This means, go ahead and write down something like: x=final price of the shirt. Or maybe, x=average speed during the trip.
  • Prepositions like the words “of” or “at” refer to multiplication. For instance, “18% of 52” is the same as $.18*52=9.36$; or “6 at $2.95 each” is the same as $6*2.95=17.70$
  • The word “is” refers to equality
  • The word “less” refers to subtraction
  • The word “more” refers to addition
  • Percentages and percent discounts are frequent, such as 15% off a shirt priced at $39.95. Remember, the quickest way to achieve the discount price is to simply multiply .85 times $39.95 to get $33.96, the answer.

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