Algebra

Quadratic formula and examples

A quadratic equation is any equation that can be written as $ax^2+bx+c=0$, for some numbers $a$, $b$, and $c$, where $a$ is nonzero. The quadratic formula is one method of solving this type of question. Below, we will look at several examples of how to use this formula and also see how to work with it when there are complex solutions.

Table of Contents

Basic examples of applying the formula
When there are complex solutions (involving $i$)
Summary

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Examples of applying the quadratic formula

Looking at the formula below, you can see that $a$, $b$, and $c$ are the numbers straight from your equation. Applying this formula is really just about determining the values of $a$, $b$, and $c$ and then simplifying the results.

image showing the quadratic formula and the values used in the formula

But, it is important to note the form of the equation given above. If your equation is not in that form, you will need to take care of that as a first step. Let’s take a look at a couple of examples.

Example

Solve: $x^2-x=6$

Solution

Before we do anything else, we need to make sure that all the terms are on one side of the equation. As you can see above, the formula is based on the idea that we have 0 on one side.

Subtract 6 from both sides to get:

$x^2-x-6=0$

Now that we have it in this form, we can see that:

$a=1$ , $b= -1$ , $c=-6$

Why are $b$ and $c$ negative? The formula is based off the form $ax^2+bx+c=0$ where all the numerical values are being added and we can rewrite $x^2-x-6=0$ as $x^2 + (-x) + (-6) = 0$. To keep it simple, just remember to carry the sign into the formula.

Once you have the values of $a$, $b$, and $c$, the final step is to substitute them into the formula and simplify.

$\begin{align}x &= \dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\\ &=\dfrac{-(-1)\pm\sqrt{(-1)^2-4(1)(-6)}}{2(1)} \\ &=\dfrac{1\pm\sqrt{1+24}}{2} \\ &=\dfrac{1\pm\sqrt{25}}{2}\end{align}$

At this stage, the plus or minus symbol ($\pm$) tells you that there are actually two different solutions:

$\begin{align} x &= \dfrac{1+\sqrt{25}}{2}\\&=\dfrac{1+5}{2}\\&=\dfrac{6}{2}\\&=3\end{align}$

and

$\begin{align} x &= \dfrac{1- \sqrt{25}}{2}\\ &= \dfrac{1-5}{2}\\ &=\dfrac{-4}{2}\\ &=-2\end{align}$

Therefore the final answer would be:

$x= \bbox[border: 1px solid black; padding: 2px]{3}$ , $x= \bbox[border: 1px solid black; padding: 2px]{-2}$

This particular quadratic equation could have been solved using factoring instead, and so it ended up simplifying really nicely. Often, there will be a bit more work – as you can see in the next example.

Example

Solve: $2x^2+2x-7=0$

Solution

This time we already have all the terms on the same side. So, we just need to determine the values of $a$, $b$, and $c$.

$a=2$, $b=2$ , $c=-7$

Applying the formula:

$\begin{align}x&=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\\ &=\dfrac{-2\pm\sqrt{(2)^2-4(2)(-7)}}{2(2)}\\ &=\dfrac{-2\pm\sqrt{4+56}}{4} \\ &=\dfrac{-2\pm\sqrt{60}}{4}\\ &=\dfrac{-2\pm 2\sqrt{15}}{4}\end{align}$

Notice that 2 is a FACTOR of both the numerator and denominator, so it can be cancelled.

$x =\dfrac{-1\pm\sqrt{15}}{2}$

This answer can not be simplified anymore, though you could approximate the answer with decimals. Therefore the final answer is:

$x=\bbox[border: 1px solid black; padding: 2px]{\dfrac{-1+\sqrt{15}}{2}}$ ,

$x=\bbox[border: 1px solid black; padding: 2px]{\dfrac{-1-\sqrt{15}}{2}}$

Complex Solutions

When using the quadratic formula, it is possible to find complex solutions – that is, solutions that are not real numbers but instead are based on the imaginary unit, $i$. Recall the following definition:

$i = \sqrt{-1}$

If a negative square root comes up in your work, then your equation has complex solutions which can be written in terms of $i$.

Example

Solve: $x^2-2x+5=0$

Solution

Just as in the previous example, we already have all the terms on one side. So, we will just determine the values of $a$, $b$, and $c$ and then apply the formula.

$a=1$, $b=-2$ , $c=5$

Applying the formula:

$\begin{align}x &=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\\ &=\dfrac{-(-2)\pm\sqrt{(-2)^2-4(1)(5)}}{2(1)}\\ &=\dfrac{2\pm\sqrt{4-20}}{2} \\ &=\dfrac{2\pm\sqrt{-16}}{2}\end{align}$

A negative value under the square root means that there are no real solutions to this equation. However, there are complex solutions. Using the definition of $i$, we can write:

$\sqrt{-16} = 4i$

This allows us to keep simplifying.

$\begin{align} x &=\dfrac{2\pm 4i}{2}\\ &=1 \pm 2i\end{align}$

The solutions to this quadratic equation are:

$x= \bbox[border: 1px solid black; padding: 2px]{1+2i}$ , $x = \bbox[border: 1px solid black; padding: 2px]{1 – 2i}$

There are three cases with any quadratic equation: one real solution, two real solutions, or no real solutions (complex solutions). Each case tells us not only about the equation, but also about its graph as each of these represents a zero of the polynomial.

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Summary

Understanding the quadratic formula really comes down to memorization. As long as you can check that your equation is in the right form and remember the formula correctly, the rest is just arithmetic (even if it is a little complicated).

A Quadratic Equation Solver from Wolfram Alpha

To use this solver, your quadratic equation should be written in the form $ax^2+bx+c=0$ . Simply input the coefficients (a,b,c) into the space below! While it will show you the steps, be aware that wolfram alpha (the program behind this widget) likes to complete the square for just about everything which may not be the most straightforward way to get your answer.

(I will probably add cool wolfram alpha widgets here over time but if you are looking for one for your class of website head on over this way: http://www.wolframalpha.com/widgets/

The laws of exponents

The laws of exponents (sometimes called the rules of exponents) allow us to simplify expressions by using our understanding of what an exponent means in much the same way that combining like terms uses the idea of what multiplication means. In both cases, the idea is to write algebraic expressions in the most simple way possible.

As a brief reminder, an exponent tells you how many times to multiply a number, variable or expression times itself. For example:

$x^4=x \cdot x\cdot x\cdot x$

$(x+5)^3=(x+5)(x+5)(x+5)$

All of the rules require that the terms have the SAME BASE. The base is whatever is being taken to a power. Above, $x$ is the base in the first example and $x+5$ is the base in the second example.

Multiplying terms with exponents

exponents-rules-multiplying

Examples:

$x^{9}x^{3}=x^{9+3}=x^{12}$
$y^{4}y^{10}=y^{4+10}=y^{14}$

Why does this work?

If we have $3^43^2$ , this means to multiply $3^4$ by $3^2$ . But, $3^4=3\cdot3\cdot3\cdot3$ while $3^2=3\cdot3$ . Looking at the original example, $3^43^2 = (3\cdot3\cdot3\cdot3)(3\cdot3)$ and if you count the number of 3’s being multiplied, there are 6. So really, $3^43^2$ is a complicated way to write $3^6$ .

Dividing terms with exponents

exponents-rules-dividing

Examples:

$\dfrac{y^7}{y^4}=y^{7-4}=y^3$
$\dfrac{x^2}{x^5}=x^{2-5}=x^{-3}=\dfrac{1}{x^3}$

Why does this work?

As an example, think about $\dfrac{5^4}{5^2}$ . This can be rewritten by writing out the multiplication $\dfrac{5\cdot5\cdot5\cdot5}{5\cdot5}$ . At this point, we can cancel any factors that are shared by the numerator and the denominator and are left with $\dfrac{5\cdot5}{1}=5^2$ . The “rule” is essentially doing this work for us by subtracting.

Taking a “power to a power”

exponents-rules-power-to-power

Examples:

$(x^4)^4=x^{16}$
$(-5x)^2=(-5)^2x^2=25x^2$

Why does this work?

Let’s use another example: $(8^2)^3$ . If an exponent tells us how many times to multiply something by itself, then this means to multiply $8^2$ by itself 3 times: $8^28^28^2$ . Now, we have three terms with exponents multiplied that all have the same base. This is the first rule of exponents so we add them: $8^28^28^2 = 8^6$ . This rule shortcuts this process.

Other properties of exponents

When you begin simplifying expressions with exponents, there will more other properties you may find helpful.

Anything to the 1 power is just the same term back again. $23^1=23$
Anything other than zero the 0 power is 1. $(x+y+z^{12})^0=1$ (this is not true for $0^0$ which is an indeterminate form)
Exponents distribute over multiplication and division only! $(xy)^2=x^{2}y^{2}$ but $(x+y)^2 \neq x^2 + y^2$ . To find $(x+y)^2$ you must use FOIL.
Negative exponents and fractional exponents have a specific meaning, which you can review here: http://www.mathbootcamps.com/understanding-exponents/.

Multiplying two binomials: FOIL method

A binomial is a polynomial with two terms that are either added or subtracted from each other like $3x+2$ or even something more complicated like $5xy-3x^2$. When multiplying any two polynomials, the ultimate goal is to make sure every term in the first polynomial has been multiplied by every term in the second polynomial. When multiplying binomials, this process can be simplified into an easy to remember process that you have probably heard of: FOIL. In this lesson, we will review this process and look at some examples of applying it.

Table of Contents

What does FOIL stand for?
Examples of applying FOIL
Summary
Additional reading

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What does FOIL stand for?

FOIL is shorthand for remembering the process of multiplying binomials. The letters stand for:

First: multiply the first terms of each binomial

Outer: multiply the “outer” terms in each binomial

Inner: multiply the “inner” terms in each binomial

Last: multiply the “last” terms in each binomial

If you think about it, this makes sure every term in each binomial is multiplied by every other term. Take the first term for instance. Once you have FOILed, it has been multiplied by the first term of the second binomial and the last (by the “outer” step). This is true for every term!

Once this process is complete, it is very likely you have some cleaning up to do in the sense of combining like terms, so you should make sure to simplify your answer once you are done.

Examples of multiplying binomials using FOIL

In the following examples, we will apply this method to multiply some binomials. Pay close attention to how we apply each part of FOIL and also how we simplify to get the final answer.

Example

Multiply: $(3x+1)(2x-3)$

Solution

For this first example, we will look at every step individually.

First: The first two terms are the $3x$ and the $2x$. Multiplying these will give: $(3x)(2x) = 6x^2$

Outer: The “outer terms” are the $3x$ and the –3. Multiplying these terms will give: $(3x)(-3) = -9x$

Inner: The inner terms are the 1 and the $2x$. Multiplying these terms will give: $(1)(2x) = 2x$

Last: The last terms are the 1 and the -3. Multiplying these terms will give: $(1)(-3) = -3$

The final answer would be the terms we got in each step above added together. The steps all together would look like this:

Apply FOIL:

$(3x+1)(2x-3) = \underbrace{(3x)(2x)}_\text{First}+\underbrace{(3x)(-3)}_\text{Outer}+\underbrace{(1)(2x)}_\text{Inner}+\underbrace{(1)(-3)}_\text{Last}$

Simplify:

$\begin{align} (3x)(2x)+(3x)(-3)+(1)(2x)+(1)(-3) &= 6x^2-9x+2x-3\\ &= \boxed{6x^2-7x-3}\end{align}$

Let’s try another example, but this time one that looks a little bit different.

Example

Multiply: $(x+y)^2$

Solution

If you think about what an exponent means, you can rewrite this as $(x+y)(x+y)$. Now we can apply FOIL and simplify to get the final answer:

$\begin{align} (x+y)^2 &=(x+y)(x+y)\\ &=\underbrace{x^2}_\text{First} + \underbrace{xy}_\text{Outer} + \underbrace{yx}_\text{Inner} + \underbrace{y^2}_\text{Last}\\ &= \boxed{x^2 + 2xy + y^2}\end{align}$

When simplifying here, remember that $xy$ is the same as $yx$. That’s why we combined them to get $2xy$.

Careful! A common mistake is to write $(x + y)^2$ as $x^2 + y^2$. This is incorrect. You cannot distribute a power. You must use FOIL.

This shows a special rule that can be used for squaring a binomial. For example, if we have $(x + 3)^2$, we could use the result in the last example and let $y = 3$ to show:

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

But, it is not necessary to memorize this since you can always just write out the binomials and apply FOIL.

Summary

FOIL is a method for multiplying binomials, but only when you have two binomials. In cases where you have three or more, you will need to combine FOIL along with other methods. When multiplying two binomials however, you can simply remember to multiply the First terms, the Outer terms, the Inner terms, and then the Last terms. In other words, FOIL!

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Additional reading

MathBootCamps has many other articles and lessons for algebra topics. You can find those here: http://www.mathbootcamps.com/algebra-guides-and-articles/

What are Polynomials?

In math, there are some words that are practically foundations to everything else. Just like any topic, math has its own vocabulary and one of the first words you should get comfortable with in algebra is the word “polynomial”. Polynomials are expressions involving terms which may have exponents that are multiplied, added, or subtracted from each other. In the following lesson, we will look at different examples of polynomials and learn about some special types of polynomials.

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Examples of polynomials

As mentioned above, in a polynomial, terms consist of variables or constants (numbers) that are either multiplied, added, or subtracted from each other. Additionally, any exponents must be positive whole numbers. For example, each of the expressions below would qualify as a polynomial:

$5x^2$
$2x^{4}-5x+11$
$\dfrac{2}{3}z^5+2xyz-xy$

In each case above, notice that the exponents were positive integers (positive whole numbers). If an exponent is negative, then this implies you are dividing by that term (based on the definition of a negative exponent), and polynomials cannot have any division involving the variables. A polynomial can have fractions involving just the numbers in front of the variables (the coefficients), but not involving the variables.

Examples of expressions which are not polynomials

Keeping the explanation above in mind, the following are not polynomials:

$x^{-2}+x-1$
There is a negative exponent.
$xy^{\frac{1}{2}}+2$
The exponent is a fraction.
$\log(x) + 3x^4$
This involves another function: $\log(x)$.

Special types of polynomials

There are a few special types of polynomials which get their own names. Remember, we are thinking of a term as something like $3xy$ or 5. In other words, it is one of the pieces of the polynomial that is being added or subtracted from the other pieces/terms. When counting terms, you are assuming that none of the terms are like terms.

Monomials

Monomials are polynomials with a single term. Some examples are:

$-3x^4$
$10xy$
$2x$

Binomials

Binomials have two terms. One of the most common types of binomials you will see in algebra are those like $3x + 5$ or $x + 4$. These involve one variable term and one constant term and are often the factors found when working with quadratic equations.

Some other examples of binomials are:

$\dfrac{1}{2}x^5 + 2x$
$4x – 9$
$12x^2y – 4xy^2$

When it comes to multiplying this type of polynomial, there is a special rule: FOIL.

Trinomials

Trinomials have three terms. One place you will see these a lot is the classic “quadratic equation” like $2x^2 + 5x + 1 = 0$. The expression on the left-hand side of this equation is a trinomial.

Some other example of trinomials are:

$-1 + 4x + 3y$
$2x^2y + 3x + 7$
$x^2 + 3x + 1$

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Additional reading

As you continue studying algebra, you may find the following articles and guides useful:

What is Algebra?

Isn’t it funny that so many people talk about algebra, do algebra, or even fear algebra but have never thought about what it actually is? Well, let’s take care of that right now. (and for the record, I’m talking about elementary algebra not abstract algebra which is a completely awesome, but a different thing).

Back to arithmetic

Think back to when you were learning adding and subtracting, multiplying and dividing – all that stuff. When you do arithmetic, you deal with specific numbers. In algebra, we shift that perspective into something more general and more abstract. Instead of talking about specific numbers, we will start including variables, which represent either a specific unknown number or a general number that we could allow to be anything.

Why would variables be useful?

Let’s take a look at a simple example to think about this a bit.

Kaitlyn’s mom always gives $5 more to Kaitlyn than to her brother Alex.

This means that if Alex gets $10, then Kaitlyn gets $15 right? In fact, we could keep doing this forever! If the mom wins the lotto and gives Alex $50,000, then Kaitlyn should be getting $50,005. (Kind of weird but go with it.)

Variables can describe what is happening in general. If we let $x$ be the amount that Alex gets, then we know that $x+5$ is how much Kaitlyn will get.

Not only does this show the relationship between the two in a compact way, but it also allows a deeper analysis. Imagine if the situation was more complicated. You can’t just explain it in words to a computer so it can figure it out. Variables would have to be used!

It’s just a different way of looking at things

In the end, think of algebra as a shift in perspective. You now have the freedom to let a symbol stand in when you don’t know the value of a number or want to know what happens when it is any number. This is a powerful ability! Algebra really is the study of expressions (and equations) with unknown values. What general rules apply? Are there certain situations where some rule is not true? These are the questions you can answer with algebra.

Three Common Algebra Mistakes You MUST Avoid

Whether it is a calculus class or a first algebra class, I have seen students at every level and every natural ability make these mistakes. Why would so many fall into these traps? Well, it really comes down to working too quickly and applying rules without stopping to think about whether or not they apply. If you can avoid these mistakes, you will find yourself well ahead of the curve. In fact, these are the kinds of things people writing multiple choice tests think about.

Distributing a Power
It seems as though everyone wants $(a+b)^2=a^2+b^2$ , and who wouldn’t? Every math exercise in the world becomes much easier if you use this! Unfortunately, in general, $(a+b)^2\neq a^2+b^2$ . Its true in a trivial case – when a or b equals zero (plug zero in for either one to check). Otherwise, this is a no go and if you aren’t sure, try it out with some nonzero numbers: What if a=1 and b=1?

$(1+1)^2=2^2=4$ while $1^2+1^2=1+1=2$ . Clearly we get two different things. Now this isn’t a mathematical proof since I used two specific numbers but the fact that it does not work here should give you pause. In fact, you would have to FOIL the left hand side to find its equivalent expression. In other words, $(a+b)^2=(a+b)(a+b)^2=a^2+2ab+b^2$ .
Forgetting to Completely Distribute
What if you needed to simplify $-(x+2y-5)$ . Many students would write $-x+2y-5$ . Can you see what is wrong?

The negative in front of the parentheses is basically a negative 1 multiplying all of the terms within those parentheses. That means it must be distributed to EVERYTHING, not just the first term! $-(x+2y-5)=-x-2y+5$ . This works regardless of what is in front of the parentheses. For instance, $2(5x+4x^2-1)=10x+8x^2-2$ .

Most often, it seems that people understand this but make this mistake when they are going too fast. This is why I advocate writing down every single step when you do any math problem. (or just double check yourself anytime you have had to distribute)
Applying the Zero-Product Rule to Everything
The zero product rule is what allows you to solve an equation like $(x+2)(x-3)=0$ . By realizing that the only way two things can multiply to give us zero is if one or both are zero, we can say $x+2=0$ and $x-3=0$ . This is all based on the fact that there is a zero on the right hand side!

If instead, we had $(x+2)(x-3)=2$ we couldn’t immediately take this approach. This does not imply that $x+2=2$ and $x-3=2$ . Again, you can only apply this if you manage to get a product of terms equal to zero. (to solve the new equation, you could foil the left hand side and then bring the two over – going from there)

There are plenty of other mistakes that I have seen but these are by far the most common. In the end, it comes down to applying properties in situations where they do not apply. Learn to ask yourself if the rule you are applying makes sense in a given situation and you will be well on your way to avoiding these!

PEMDAS: The Order of Operations

PEMDAS is a way of remembering the order of operations, which are used to simplify numerical expressions and are even represented in how we approach solving equations. In this lesson, we will review what PEMDAS is and look at several examples of how it works.

Table of Contents

Why do we need PEMDAS?
What does PEMDAS stand for?
Examples of using PEMDAS and the order of operations

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Why do we need PEMDAS?

Consider the calculation $1+3\times 5$. Which of these is correct?

$1+3 \times 5 = 4 \times 5 = 20$

$1 + 3\times 5 = 1 + 15 = 16$

If you think about it, both seem like reasonable ways to approach the calculation. The problem is, the two approaches give different answers! To avoid confusion, there is a standard agreed upon order in which to perform the operations (where an operation is something like addition or multiplication) known as the order of operations.

By the way, the correct calculation is: $1 + 3\times 5 = 1 + 15 = 16$

What is PEMDAS?

The easiest way to remember the order of operations is with the shorthand “PEMDAS”. You can even add in a way to remember this shorthand: “Please Excuse My Dear Aunt Sally”. This stands for:

Parentheses: Perform all operations that are inside parentheses. If there are many operations, you must follow the order of operations inside the parentheses as well.

Exponents: Calculate any exponents you see.

Multiplication and Division: Perform any multiplication or division you see, generally from left to right.

Addition and Subtraction Perform any addition and subtraction you see, from left to right.

Examples applying PEMDAS / the order of operations

Let’s try a couple of examples to see how to apply these rules.

For the following examples, you must remember that an exponent represents multiplication. So, $3^2 = 9 \times 9$ and $2^4 = 2 \times 2 \times 2 \times 2$. If you need to review exponents, you can do that here: http://www.mathbootcamps.com/understanding-exponents/

Example

Simplify: $1 + 3 \times 4^2$

Solution

Thinking about PEMDAS, we should first start with parentheses. But, since there are no parentheses, we will move on to “E”, which tells us to calculate any exponents. Since $4^2 = 16$, we can write:

$1 + 3 \times 4^2 = 1 + 3 \times 16$

Next, we are to perform any multiplication or division (MD). You can verify with your calculator that $ 3 \times 16 = 48$. Now we have:

$1 + 3 \times 16 = 1 + 48$

Finally, we will add/subtract (AS) to get our final answer:

$1 + 48 = \boxed{49}$

As you can see, we just need to walk though each part of PEMDAS and do each calculation carefully. Let’s try another example applying the same rules, but just take a look at the math itself.

Example

Simplify: $3(4-3 \times 10)+2^2$

Solution

We will follow the same steps as used above. Remember, anything inside of parentheses must be simplified first AND you must follow order of operations within the parentheses.

$\begin{align} 3(4-3 \times 10)+2^2 &=3(4-30)+2^2\\ &=3(-26)+2^2\\ &=3(-26)+4\\ &= -78+47\\ &= \boxed{-74}\end{align}$

As you can see, we worked inside of the parentheses first and then followed the order of operations outside of the parentheses once we got down to one number.

Example

Simplify: $2+3[-2^2+4(2+1)^3]$

Solution

In this example, there are multiple sets of parentheses. Even though it looks complicated, we still apply the same set of rules. Just start with the innermost parentheses are work your way out.

$\begin{align} 2+3[-2^2+4(2+1)^3] &= 2+3[-2^2+4(3)^3]\\ &=2+3[-4+4(27)]\\ &=2+3[-4+108]\\ &=2+3[104]\\ &=2+312\\ & = \boxed{314}\end{align}$

These examples have hopefully shown you how by just applying the order of operations carefully and methodically, you can simplify even the most complicated expressions. Now, we will look at one last example which seems like it should be simple, but shows the importance of parentheses when working with numbers.

Example

Simplify: $-3^2$

Solution

This may seem confusing since there is only one operation (the exponent), however there are really two operations here. You can actually write $-3^2$ as $-1\times3^2$. Now by the order of operations:

$-1\times3^2 = -1\times9 = -9$

In other words $-3^2 = -9$.

You may be thinking “BUT WAIT I know that anytime you square a negative number you get a positive number!!” . YES! That is true, but here we were not squaring a negative number, we were squaring 3. If we wanted to square –3, we would write $(-3)^2 = 9$. See the difference? If there are parentheses, it means you are squaring the entire term, otherwise the negative “rides along” as part of a multiplication.

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Conclusion

The order of operations applies not just to numerical examples, but also to any mathematical problem including those seen later in algebra and calculus. Make sure you take the time to work with these rules so that they are second nature and don’t slow you down in advanced courses! You should also take special care when using graphing calculators and apply parentheses correctly since they will always use the order of operations when making a calculation.

Solving linear equations with one variable

Linear equations in one variable are equations where the variable has an exponent of 1, which is typically not shown (it is understood). An example would be something like $12x = x – 5$. To solve linear equations, there is one main goal: isolate the variable. In this lesson, we will look at how this is done through several examples.

Table of Contents

Examples of solving one-step equations
Examples of solving two-step equations
Examples of equations where you must simplify first
Infinitely many or no solutions
Summary

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Examples of solving one-step linear equations

After all your hard work solving the equation, you know that you want a final answer like $x=5$ or $y=1$. In both of these cases the variable is isolated, or by itself.

So we need to figure out how to isolate the variable. How we do this depends on the equation itself! If it was multiplied by something, we will divide. If something was added to it, we will subtract. By doing this, we will slowly be getting the variable by itself.

Let’s use an example to see how this works.

Example

Solve the equation: $4x = 8$

Solution

In this example, the 4 is multiplying the $x$. Therefore, to isolate $x$, you must divide that side by 4. When doing this, you must remember one important rule: whatever you do to one side of the equation, you must do to the other side. So we will divide both sides by 4.

$\begin{align}4x &= 8 \\ \dfrac{4x}{\color{red}{4}} &= \dfrac{8}{\color{red}{4}}\end{align}$

Simplifying:

$x = \boxed{2}$

That’s it, one step and we are done. (That’s why equations like these are often called “one-step” equations)

Check

Anytime you are solving linear equations, you can always check your answer by substituting it back into the equation. If you get a true statement, then the answer is correct. This isn’t 100% necessary for every problem, but it is a good habit so we will do it for our equations.

In this example, our original equation was $4x = 8$. To check this, verify the following is true:

$\begin{align}4x &= 8\\ 4(2) &= 8 \\ 8 &= 8\end{align}$

This is a true statement, so our answer is correct.

For any equation, whatever operation you do to one side must also be done to the other side

Let’s try a couple more examples before moving on to more complex equations.

Example

Solve: $3x=12$

Solution

Since $x$ is being multiplied by 3, the plan is to divide by 3 on both sides:

$\begin{align}3x &=12\\ \dfrac{3x}{\color{red}{3}} &=\dfrac{12}{\color{red}{3}}\\ x&= \boxed{4}\end{align}$

Check

To check our answer, we will let $x = 4$ and substitute it back into the equation:

$\begin{align}3x &= 12\\3(4) &= 12 \\ 12 &= 12\end{align}$

Just as before, since this is a true statement, we know our answer is correct.

In the next example, instead of the variable being multiplied by a value, a value is being subtracted from the variable. To “undo” this, we will add that value to both sides.

Example

Solve: $y-9=21$

Solution

This time, 9 is being subtracted from y. So, we will undo that by adding 9 to both sides.

$\begin{align}y-9&=21\\ y-9 \color{red}{+9}&=21\color{red}{+9}\\y&=30\end{align}$

Next we will look at what are commonly called “two-step” equations. In these equations, we will need to undo two operations in order to isolate the variable.

Examples of two step equations

In each of the examples above, there was a single step to perform before we had our answer. In these next examples, you will see how to work with equations that have two steps instead. If there is more than one operation, it is important to remember the order of operations, PEMDAS. Since you are undoing the operations to $x$, you will work from the “outside in”. This is easier to understand when you see it in an example.

Example

Solve: $2x-7=13$

Solution

Notice the two operations happening to $x$: it is being multiplied by 2 and then having 7 subtracted. We will need to undo these. But, only the $x$ is being multiplied by 2, so the first step will be to add 7 to both sides. Then we can divide both sides by 2.

Adding 7 to both sides:

$\begin{align} 2x-7 &= 13\\ 2x-7 \color{red}{+7} & =13 \color{red}{+7}\\ 2x&=20\end{align}$

Now divide both sides by 2:

$\begin{align} 2x &=20 \\ \dfrac{2x}{\color{red}{2}}&=\dfrac{20}{\color{red}{2}}\\ x&= \boxed{10}\end{align}$

Check

Just like with simpler problems, you can check your answer by substituting your value of $x$ back into the original equation.

$\begin{align}2x-7&=13\\ 2(10) – 7 &= 13\\ 13 &= 13\end{align}$

This is true, so we have the correct answer.

Let’s look at one more two-step example before we jump up in difficulty again. Make sure that you understand each step shown and work through the problem as well.

Example

Solve: $5w + 2 = 9$

Solution

As above, there are two operations: $w$ is being multiplied by 5 and then has 2 added to it. We will undo these by first subtracting 2 from both sides and then dividing by 5.

$\begin{align}5w + 2 &= 9\\ 5w + 2 \color{red}{-2} &= 9 \color{red}{-2}\\ 5w &= 7\\ \dfrac{5w}{\color{red}{5}} &=\dfrac{7}{\color{red}{5}}\\w=\boxed{\dfrac{7}{5}}\end{align}$

The fraction on the right can’t be simplified, so that is our final answer.

Check

Let $w = \dfrac{7}{5}$. Then:

$\begin{align}5w + 2 &= 9\\ 5\left(\dfrac{7}{5}\right) + 2 &= 9\\ 7 + 2 &= 9\\ 9 &= 9 \end{align}$

So, we have the correct answer once again!

Simplifying before solving

In the following examples, there are more variable terms and possibly some simplification that needs to take place. In each case, the steps will be to first simplify both sides, then use what we have been doing to isolate the variable. We will first take an in depth look at an example to see how this all works.

To understand this section, you should be comfortable with combining like terms.

Example

Solve: $3x+2=4x-1$

Solution

Since both sides are simplified (there are no parentheses we need to figure out and no like terms to combine), the next step is to get all of the x’s on one side of the equation and all the numbers on the other side. The same rule applies – whatever you do to one side of the equation, you must do to the other side as well!

It is possible to either move the $3x$ or the $4x$. Suppose you moved the $4x$. Since it is positive, you would do this by subtracting it from both sides:

$\begin{align}3x+2 &=4x-1\\ 3x+2\color{red}{-4x} &=4x-1\color{red}{-4x}\\ -x+2 & =-1\end{align}$

Now the equation looks like those that were worked before. The next step is to subtract 2 from both sides:

$\begin{align}-x+2\color{red}{-2} &= -1\color{red}{-2}\\-x=-3\end{align}$

Finally, since $-x= -1x$ (this is always true), divide both sides by $-1$:

$\begin{align}\dfrac{-x}{\color{red}{-1}} &=\dfrac{-3}{\color{red}{-1}}\\ x&=3\end{align}$

Check

You should take a moment and verify that the following is a true statement:

$3(3)+ 2 = 4(3) – 1$

In the next example, we will need to use the distributive property before solving. It is easy to make a mistake here, so make sure that you distribute the number in front of the parentheses to all the terms inside.

Example

Solve: $3(x+2)-1=x-3(x+1)$

Solution

First, distribute the 3 and –3, and collect like terms.

$\begin{align} 3(x+2)-1 &=x-3(x+1)\\ 3x+6-1&=x-3x-3 \\ 3x+5&=-2x-3\end{align}$

Now we can add 2x to both sides. (Remember you will get the same answer if you instead subtracted 3x from both sides)

$\begin{align} 3x+5\color{red}{+2x} &=-2x-3\color{red}{+2x}\\ 5x+5& =-3\end{align}$

From here, we can solve as we did with other two-step equations.

$\begin{align}5x+5\color{red}{-5} &=-3\color{red}{-5}\\ 5x &=-8\\ \dfrac{5x}{\color{red}{5}}&=\dfrac{-8}{\color{red}{5}}\\ x &= \dfrac{-8}{5} \\ &=\boxed{-\dfrac{8}{5}}\end{align}$

Check

This was a tough one, so remember to check your answer and make sure no mistake was made. To do that, you will be making sure that the following is a true statement:

$3\left(-\dfrac{8}{5}+2\right)-1=\left(-\dfrac{8}{5}\right)-3\left(-\dfrac{8}{5}+1\right)$

(Note: it does work – but you have to be really careful about parentheses!)

Infinitely many solutions and no solutions

There are times when you follow all of these steps and a really strange solution comes up. For example, when solving the equation $x+2=x+2$ using the steps above, end up with $0=0$. This is certainly true but what good does it do?

If you get a statement such as this, it means that the equation has infinitely many solutions. Any $x$ you could think of would satisfy the equation $x+2=x+2$. The appropriate answer in this case is “infinitely many solutions”.

The other situation comes up when you simplify an equation down into a statement that is never true such as $3=4$ or $0=1$. This happens with the equation $x+5=x-7$ which will lead to $5= -7$, something that is certainly never true. This means that no $x$ would satisfy this equation. In other words “no solution”. In summary:

If you get a statement that is always true like $5 = 5$ or $0 = 0$, then there are infinitely many solutions.
If you get a statement that is always false like $10 = 11$ or $1 = 5$, then there are no solutions.

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Summary

Solving linear equations is all about isolating the variable. Depending on the equation, this may take as little as one step or many more steps. Always check if you need to simplify one or both sides of the equation first, and always check your answer.

Using Wolfram Alpha to Solve Equations

If you aren’t familiar with it, Wolfram Alpha is a powerful online tool for calculations and other fun (for example, type in your birthday and see what comes up!). For a student who is trying to understand a new math concept, this tool is great for really taking apart a problem and seeing how it works.

The only downside with wolfram alpha is the input. You have to make sure you state what you want it to do very specifically or it may misinterpret things and give you output that isn’t useful. Today, I will show you how to have it solve equations since this is one of the easier things as far as input goes.

Suppose that we wanted to solve the equation $x^4-4x^3-6x^2+4x+5=0$ . In fact, suppose we wanted to find EXACT values (no decimals!). As it is, we would have to use some pretty advanced algebra techniques to get the answer. Heading to wolfram alpha I type in “solve x^4-4x^3-6x^2+4x+5=0”. (I said the input for solving something was easy! The symbol ^ is commonly used for typing exponents)

Once you type this in, it is as simple as clicking the equals button or pressing enter on your keyboard! You can see the exact output here. The initial output simply gives the answers and a graph:

However, if you click “show steps” that I have circled above, it will show you HOW it got those results making this tool into a powerful solutions manual to help with your studying! Two things to note:

Sometimes the steps are unavailable. You will see this in the output link I have given! I have not yet figured out a pattern to this, but sometimes it won’t show the steps. Even when this happens, it still is a great way to at least check answers. (if you are curious how this problem works look at this link for the factorization and set it equal to zero)
The steps can be unnatural. You may be thinking of just copying down the steps and handing in the work to your professor or teacher, but if you do this you will be found out pretty easily. A lot of the time the steps are in an order no human would ever use! So, it is a good guide to get started but realize there is likely a nicer way to do things. This is especially true for calculus problems.

So, go play! You can try telling it to “factor”,”solve”, “find” etc. I always try the most direct input and see how it interprets it. Like any tool, you have to understand its limitations and how to work with it well to get any help!