Here is a collection of some of the most commonly-made algebra mistakes I've seen. If you can learn to avoid these, you'll also avoid a lot of frustration and wrong turns as you learn to solve more complicated problems. It's worth it to pay attention in early algebra classes and practice, practice, practice!
All of the tools in your growing toolkit will be necessary from time to time. Try not to forget them.
There are many examples of elementary algebra operations in the exercises and written solutions in some of the sections on the left. Check those out, too. The most important thing to remember about algebra, in my view, is this: More things in algebra are the same than different. Most of algebra, and therefore much of mathematics, is repeating the same rules and operations, just in different circumstances. The more you can learn to see things as similar, the better you'll be at solving problems you've never seen before.
I've put this one first because it's one of the biggest misconceptions I've noticed, and it's at the root of many mistakes made doing algebra. I'm afraid that we (educators) may have done you a disservice a while back by assigning you exercises like this:
Because of problems like these, you might have got the impression the the equal sign means "the answer goes here →". It does not mean that.
The equal sign means that what's on the left side of the equation has the same value as what's on the right — always. It implies balance.
Please go to the Algebra section to learn more about how to use the equal sign.
The equal sign ( = ) cannot be violated. If what you do on one side has any effect, you must do it to the other side, too.
OK, I have to contradict myself right away: Actually, there's only one operation, because multiplication is just repeated addition ... but let's keep it simple. Focusing on addition and multiplication will help you navigate a lot of more complicated arithmetic and algebra.
If you know how to multiply, then you know how to divide: Division is just multiplication by the reciprocal.
Thinking of division in this way especially makes division by a fraction easier to cope with. The second example in the box below illustrates this. A duanting double-fraction expression is easily reduced to a multiplication if you remember that division by a fraction is just multiplication by the reciprocal of that fraction.
Remember that the reciprocal of an integer, like 2, is "1 over 2," or ½. Likewise, the reciprocal of ½ is 2/1 = 2.
Here's an example, 2 divided by two-fifths:
$$ \require{cancel} \frac{2}{\frac{2}{5}} = \cancel{2} \cdot \frac{5}{\cancel{2}} = 5 $$
If you know how to add, then you also know how to subtract: Subtraction is just addition of the negative.
Using this definition in reverse can sometimes save you a lot of trouble. Very often you will encounter an expression like:
That "dangling negative sign" can get lost in further steps, especially when the expressions get more complicated. Do yourself a favor and rewrite it as
There's one fewer sign to get lost or create confustion there.
This is a big one. I've seen students who absolutely refused to work with fractions. While they could do a lot with a calculator (some calculators even do symbolic arithmetic with fractions), there was always a point where they just had to miss something because of a fear of working with fractions. If you just learn a few simple rules, no fraction will ever be daunting again.
Even numbers that don't have a denominator have a denominator. It's one (1), and we just have a convention of not writing denominators of 1 or multipliers of 1 because they're unnecessary.
It will help you to do arithmetic and algebra if, when you need to multiply a fraction by a number or a variable, that that number of variable can be written as 3/1 or x/1.
Try to get used to thinking about fractions.
· 10% is 10/100, which is 1/10.
· 5/8 of an inch is 1/8th more than half an inch.
· Which is smaller, 3/75 or 1/24 ?
You can download the fraction table on the right by clicking on it. Stare at it once in a while to see relationships between diferent kinds of fractions - fractions with different denominators. The fraction table was inspired by a neat book: Strength in Numbers; Discovering the Joy and Power of Mathematics in Everyday Life, by Sherman K. Stein.
* Strictly speaking, irrational numbers like π cannot be represented by fractions, but we could always express π as a fraction: π/1
Narrow screens: scroll right and left to view wide equations
Multiplication of fractions is easy. Just multiply numerators and multiply denominators – "multiply across".
But ... make sure to pay attention to the other rules of algebra and properties of operations (like the distributive property of multiplication). Look at the examples on the left.
When you work with application problems (word problems) the word "of" almost always means "multiply". For example 1/2 of 1/4 is (1/2)*(1/4) = 1/8. That makes sense: There are two eighths in a fourth, so half of a fourth is an eighth.
Multiply these fractions, then check your answers.
$$\frac{1}{2} \cdot \frac{1}{3} = \; \rule{2cm}{0.2mm}$$
$$\frac{1}{2} \cdot \frac{1}{3} = \frac{1 \cdot 1}{2 \cdot 3} = \frac{1}{6}$$
$$\frac{1}{2} \cdot \frac{2}{3} = \; \rule{2cm}{0.2mm}$$
$$\frac{1}{2} \cdot \frac{2}{3} = \frac{1 \cdot \cancel{2}}{\cancel{2} \cdot 3} = \frac{1}{3}$$
$$\frac{9}{7} \cdot \frac{8}{11} = \; \rule{2cm}{0.2mm}$$
$$\frac{9}{7} \cdot \frac{8}{11} = \frac{9 \cdot 8}{7 \cdot 11} = \frac{72}{77}$$
$$\frac{4}{3} \cdot \frac{1}{2} = \; \rule{2cm}{0.2mm}$$
$$\frac{4}{3} \cdot \frac{1}{2} = \frac{4 \cdot 1}{3 \cdot 2} = \frac{4}{6} = \frac{2}{3}$$
$$\frac{2}{3} \cdot \frac{x+1}{3} = \; \rule{2cm}{0.2mm}$$
$$ \begin{align} \frac{2}{3} \cdot \frac{x+1}{3} &= \frac{2(x+1)}{3 \cdot 3} \\[5pt] &= \frac{2x + 2}{9} \; \text{ OR } \; \frac{2(x+1)}{9} \end{align}$$
$$\frac{x - 1}{2} \cdot \frac{2}{x + 1} = \; \rule{2cm}{0.2mm}$$
$$ \begin{align} \frac{x - 1}{2} \cdot \frac{2}{x + 1} &= \frac{\cancel{2}(x-1)}{\cancel{2} (x+1)} \\[5pt] &= \frac{x-1}{x+1} \end{align}$$
$$\frac{x}{3} \cdot \frac{2}{x + 1} = \; \rule{2cm}{0.2mm}$$
$$\frac{x}{3} \cdot \frac{2}{x + 1} = \frac{3x - 2}{4x}$$
$$\frac{3x - 2}{4} \cdot \frac{1}{x} = \; \rule{2cm}{0.2mm}$$
$$\frac{3x - 2}{4} \cdot \frac{1}{x} = \frac{3x - 2}{4x}$$
Addition of fractions is a little more tricky. We can't add two fractions that do not have the same denominator; it's like adding cows to seatcushions-- what's the result?
When adding fractions that have the same denominator, just add the numerators and keep the same denominator. It's like this: One pie plus another pie is two pies; one hat plus another hat is two hats; and one third plus another third is two thirds. Done.
If you don't have a common denominator, try to find a common multiple of both and use that as your target common denominator. There's an example on the right.
If you can't find a common multiple, you'll have to "punt" and make the common denominator the product of the two you have.
Multiplying by one is always OK
Example:
Narrow screens: scroll right and left to view wide equations
A trick for adding fractions
If we look a little more carefully at our example of finding a common denominator above, we can turn it into a nifty trick for adding two fractions. Here it is again:
$$\begin{align} \frac{a}{b} + \frac{c}{d} &= \frac{a}{b} \left( \frac{d}{d} \right) + \frac{c}{d} \left( \frac{b}{b} \right) \\[5pt] &= \frac{ad + cd}{bd} \end{align}$$
Now notice the pattern. In the numerator of the result, we have products that are funny "cross products" of numerators and denominators, ad (in green below) and cd (in magenta). And the denominator of the result is just the product of the denominators. It looks like this:
Here's an example with numbers, adding two fractions that would otherwise be a real pain to add:
And here's another example. Provided you know and understand where it comes from, this can be a real time-saving trick.
$$ \begin{align} \frac{9}{7} + \frac{3}{5} &= \frac{9 \cdot 5 + 3 \cdot 7}{7 \cdot 5} \\[5pt] &= \frac{66}{35} \end{align}$$
Add these fractions, then check your answers.
$$\frac{1}{2} + \frac{1}{3} = \; \rule{2cm}{0.2mm}$$
$$\frac{1}{2} + \frac{1}{3} = \frac{3(1)+2(1)}{2(3)} = \frac{5}{6}$$
$$\frac{1}{2} + \frac{4}{5} = \; \rule{2cm}{0.2mm}$$
$$\frac{1}{2} + \frac{4}{5} = \frac{5(1)+2(4)}{2(5)} = \frac{13}{10}$$
$$\frac{9}{7} + \frac{8}{11} = \; \rule{2cm}{0.2mm}$$
$$ \begin{align} \frac{9}{7} + \frac{8}{11} &= \frac{9(11)+8(7)}{7(11)} \\[5pt] &= \frac{99+56}{77} \\[5pt] &= \frac{155}{77} \end{align}$$
$$\frac{4}{3} + \frac{1}{2} = \; \rule{2cm}{0.2mm}$$
$$\frac{4}{3} + \frac{1}{2} = \frac{4(2)+3(1)}{3(2)} = \frac{11}{6}$$
$$\frac{2x}{3} - \frac{x}{3} = \; \rule{2cm}{0.2mm}$$
$$\frac{2x}{3} - \frac{x}{3} = \frac{2x - x}{3} = \frac{x}{3}$$
Don't forget to look for a common denominator!
$$\frac{2x}{3} - \frac{x}{4} = \; \rule{2cm}{0.2mm}$$
$$ \begin{align} \frac{2x}{3} - \frac{x}{4} &= \frac{4(2x)-3(x)}{4(3)} \\[5pt] &= \frac{8x - 3x}{12} \\[5pt] &= \frac{5x}{12} \end{align}$$
$$\frac{x-1}{3} - \frac{(x-2)}{2} = \; \rule{2cm}{0.2mm}$$
$$ \begin{align} \frac{x-1}{3} - \frac{(x-2)}{2} &= \frac{2(x-1)-3(x-2)}{3(2)} \\[5pt] &= \frac{2x - 2 - 3x + 6}{6} \\[5pt] &= \frac{-x + 4}{6} = \frac{4-x}{6} \end{align}$$
$$\frac{2x - 3}{4x} + \frac{1}{x} = \; \rule{2cm}{0.2mm}$$
$$ \begin{align} \frac{2x - 3}{4x} + \frac{1}{x} &= \frac{x(2x-3) + 4x(1)}{4x(x)} \\[5pt] &= \frac{2x^2 - 3x + 4x}{4x^2} \\[5pt] &= \frac{2x^2 + x}{4x^2)} \\[5pt] &= \frac{\cancel{x}(2x + 1)}{3x^\cancel{2}} \\[5pt] &= \frac{2x + 1}{3x} \end{align}$$
This is really multiplying by x/x, not x/1. Remember, every number is a fraction, so 3 is 3/1, and so on.
A college math professor acquaintance tells me that so many college freshman come to his school (an Ivy League school) with this misunderstanding that he and his colleagues call it "the second fundamental theorem of algebra". it is: $(a + b)^2 = a^2 + b^2
When expressions like $(x + y)^n$ are multiplied correctly (unless n = 1), there are always "cross terms," terms containing both components of the binomial, $x$ and $y$. That's the "$2xy$" in the green box above.
This is a good point to review the algebra of multiplying binomials, things that look like (a + b). There are two ways to look at it, and you should be very familiar with each.
First, the acronym F.O.I.L. stands for "
Another way to look at binomial multiplication is just using the distributive property of multiplication. You can see that the outcome is the same:
The pattern (x - a)(x + a) is a special one. The product is x^{2} - a^{2}, a difference of perfect squares. You should memorize this pattern because it will pop up a lot, and knowing it will make your life easier. Below are a few examples of binomial multiplication. Try them and check your answers if you need the practice
Multiply the binomials (#8 contains a trinomial) and simplify the resulting expression as much as possible. Make sure to check your solutions.
$(x + 7)(x - 3)$
$$ \begin{align} &= x(x-3) + 7(x-3) \\[5pt] &= x^2 - 3x + 7x -21 \\[5pt] &= x^2 + 4x - 21 \end{align}$$
$(x - 3)(x - 3)$
$$ \begin{align} &= x(x-3) - 3(x-3) \\[5pt] &= x^2 - 3x - 3x + 9 \\[5pt] &= x^2 - 6x + 9 \end{align}$$
$(x - 3)(x + 3)$
$$ \begin{align} &= x(x+3) - 3(x+3) \\[5pt] &= x^2 + \cancel{3x} - \cancel{3x} - 9 \\[5pt] &= x^2 - 9 \end{align}$$
$(2x + 4)(5x - 9)$
$$ \begin{align} &= 2x(5x - 9) + 4(5x - 9) \\[5pt] &= 10x^2 - 18x + 20x - 36 \\[5pt] &= 10x^2 + 2x - 36 \end{align}$$
$(-x-1)(x - 1)$
$$ \begin{align} &= -x(x - 1) + -(x-1) \\[5pt] &= -x^2 + x - x + 1 \\[5pt] &= -x^2 + 1 = 1 - x^2 \end{align}$$
$(2x - 1)^2$
$$ \begin{align} &=(2x - 1)(2x - 1) \\[5pt] &= 2x(2x-1) - (2x - 1) \\[5pt] &= 2x^2 - 2x - 2x + 1 \\[5pt] &= 2x^2 -4x + 1 \end{align}$$
$\left( \frac{1}{2} x - 1 \right) \left( x - \frac{1}{2} \right)$
$$ \begin{align} &= \frac{1}{2}x\left(x - \frac{1}{2}\right) - \left(x - \frac{1}{2}\right) \\[5pt] &= \frac{1}{2} x^2 - \frac{1}{4}x - x + \frac{1}{2} \\[5pt] &= \frac{1}{2} x^2 - \frac{5}{4}x + \frac{1}{2} \end{align}$$
$(x - 1)(x^2 + 2x - 1)$
$$ \begin{align} &= x(x^2 + 2x - 1) - (x^2 + 2x - 1) \\[5pt] &= x^3 + 2x^2 - x - x^2 - 2x + 1 \\[5pt] &= x^3 + x^2 - 3x + 1 \end{align}$$
This is a very intuitive look at why (x + a)(x + b) = x^{2} + (a+b)x + ab.
We know that the area of a rectangle is its width times its height. The outer rectangle shown here has height (x + a) and width (x + b). Clearly the area of the outer rectangle is the sum of the areas of the four smaller rectangles that make it up: xx, xb, ax and ab.
Calculating the area in that way is identical to FOIL-ing. Take a look.
Here are a few mistakes made frequently when canceling numbers or variables from the numerator and denominator of a fraction:
Be very careful when you cancel if there is an addition or subtraction operation on either side of the fraction bar. That complicates things. You can't just cancel variables or numbers where it's convenient. What you cancel must be present in every part of the addition or subtraction expression. It might be helpful to remember that
In this case $x$ is present in all terms of the fraction, so we can divide each term by $x$. But be careful, $\frac{x}{x} = 1$, so
IN this case, we can divide both terms in the numerator by $x$, but we have to remember that $\frac{x^2}{x} = x$.
Remember that "cancelling" numbers or variables on either side of an equation or across a fraction bar is really dividing something by itself, which results in 1, which we can omit if it's multiplied or divided, but must keep if its added or subtracted.
The answer is: As long as there's only one negative sign, it just doesn't matter. If either the numerator or denominator are negative, then the quotient (the value of the fraction) is negative. And if we hold the negative sign out front, it's the same thing. Put it wherever it's most convenient for the problem on which you're working.
but remember that
$$\frac{-a}{-b} = -\left( \frac{-a}{b} \right) = -\left( \frac{a}{-b} \right) = \frac{a}{b}$$
So much confusion, especially when applying algebra to solving problems involving units (physics, chemistry, engineering ...) can be avoided by not writing fractions with slash ( / ), but with a horizontal line.
It's just too easy to forget about a denominator if it doesn't dangle down below the rest of the expression(s), and it's tough to correctly cancel units when they're all on the same line.
Many algebra errors are made by forgetting or improperly using the laws of exponents. Here are the laws and some examples of common mistakes:
Operation | Rule | Example |
---|---|---|
product |
$$a^m \cdot a^n = a^{m+n}$$ Add exponents |
$$2^2 \cdot 2^3 = (2 \cdot 2)(2 \cdot 2 \cdot 2) = 2^5$$ |
quotient |
$$\frac{a^m}{a^n} = a^{m-n}$$ Subtract exponents |
$$\frac{2^3}{2^2} = \frac{ 2\cdot \cancel{2} \cdot \cancel{2}}{\cancel{2} \cdot \cancel{2}} = 2^{3-2} = 2$$ |
power |
$$(a^m)^n = a^{m\cdot n}$$ Multiply exponents |
$$(2^2)^3 = (2 \cdot 2 \cdot 2)(2 \cdot 2 \cdot 2)(2 \cdot 2 \cdot 2) = 2^9$$ |
inverse |
$$a^{-1} = \frac{1}{a}$$ Take the reciprocal |
$$2^{-1} = \frac{1}{2}$$ |
zero exponent |
$$a^0 = 1$$ | $$2^0 = 1$$ |
If you have not learned about the concept of functions in math yet, don't worry about this one. But as soon as you do, come back to it. It's a very commonly-made mistake.
Let's begin with the function $f(x) = x^2$, and we'll find $f(a + b)$.
Now $f(a + b) = (a + b)^2$, which is not $a^2 + b^2$, as we saw in pitfall 4 above. So $f(a + b) \ne f(a) + f(b)$. The function does not distribute to the $a$ and the $b$. It "operates" on them at the same time.
There are many examples of this common error that crop up all of the time. Here are a few:
$$ \begin{align} log(x + y &\ne log(x) + log(y) \\[5pt] ln(x + y) &\ne ln(x) + ln(y) \\[5pt] sin(x + y) &\ne sin(x) + sin(y) \\[5pt] cos^{-1}(x + y) &\ne cos^{-1}(x) + cos^{-1}(y) \end{align}$$
Just remember:
Functions don't distribute.
If you have a favorite pitfall you'd like me to incorporate into this page, please send it along and I'll try to get to it.
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