# Solving Algebraic Equations

This section covers:

# Solving Linear Equations

Let’s practice solving algebra (linear) problems.  It seems really difficult at first, but I promise you, after some practice, it will get easier.  The thing I love about math is that you have a bunch of rules and all you have to do is apply them (memorize less than say History!).

Let’s solve for x in the following problems.  At this point, don’t worry about where we got the variables and numbers – let’s just go through the mechanics.  We’ll do real algebra word problems later.

Again, you want to get the x by itself – when it’s by itself, we get the answer of what x is!  To do this, you have to move things around by doing the opposite of what you have.  For example, if the problem has addition in it, you have to subtract from both sides; if the problem has division, you have to multiply on both sides.  Remember also that the sign of a variable or constant is what is in front of it; sometimes it’s an “invisible +” if it’s at the beginning.

• Step 1: First combine the numbers (called “constants”) and the same letters (variables) on each side if you need to.  This is called combining like terms, meaning you can add the same letters together – in this case the x‘s.    As we saw earlier, this is because of the Distributive Property, and later we’ll see we can do the same for x2, or actually any term with identical variables. Remember that when we just have an x, it’s the same as 1x.  We can change the order of our terms using the Commutative Property of Addition.  So let’s say we are trying to solve   for x:

• Step 2:  Add or subtract the “numbers” (“constants”) from each side to get rid of them on the side where the variable (the x) is.  You want to do the opposite or (additive inverse) of what we have to get rid of.   If you are subtracting a number, you want to add on both sides, and if you are adding a number, you want to subtract on each side.  Line up the letters and numbers vertically.  In our case, we are subtracting on each side, and using the Subtraction Property of Equality:

• Step 3:  If there’s a number before the x (which means it’s multiplied by the x), divide both sides by that number (using the Multiplicative Inverse).  If there’s a number belowthe x (which means it’s divided by the x), multiply both sides by that number.  If there’s a fraction before the x, we need to multiply both sides of the reciprocal of the fraction.  (The numbers immediately before the variables are called coefficients.)  In our case, we have a number before the x, and we are using the Division Property of Equality.

Now we have solved to get what x is:  x = 3!

• Step 4:  We should always try to check our answer, by plugging the answer we got back in for wherever x is:

Yeah — we did it!

Here are more examples, since solving these is an important foundation of algebra.  Just remember to keep doing the above steps until you get x by itself.

It doesn’t really matter what order you do the operations, as long as you do the addition and/or subtraction first, and then do the multiplication and/or division.  Do you see how this is sort of the opposite of PEMDAS – it’s SADMEP!   So first do the addition and subtraction, then the multiplication and division, and then worry about things in parentheses.  (We’ll talk about solving with exponents later).

Make sure you can follow each example before you go on to the next section.  Do you recognize the algebraic properties we saw earlier?

These will look complicated, but really, you are just doing the same things over and over again — like a puzzle!

Here is one more problem:

Here’s another example with fractions.  When we have fractions on both sides, we can always multiply both sides by the least common denominator to get rid of the fractions:

# Solving Literal Equations (Transforming Formulas)

In this next example, we’ll actually solve for the variable xin terms of” some other variables (meaning that we still solve for x, but may not get a number); notice that we treat variables just like numbers!

This is called solving Literal Equations, since sometimes variables are called literals.  These are also called Transforming Formulas, or solving Multivariable Equations.

Let’s use the formula that converts Celsius temperatures (the type of temperatures they use in Paris!) to Fahrenheit (the type of temperatures they use here in the United States) to get the formula that converts Fahrenheit to Celsius.  Celsius is the “metric system” way to figure temperature, and we use Fahrenheit.  See, this stuff can really be useful!

Let’s say we are given F in terms of C (meaning F is by itself).  Let’s solve for C in terms of F (meaning get C by itself):

See – algebra is pretty cool (no pun intended)!

You’ll see later that these two equations are called inverses, since one is sort of the opposite of the other.

Here’s one more example that’s a little bit more difficult, since it has a lot of fractions:

Let’s do the literal equation problem from above again, this time using the multiply-through fraction approach:

# Algebraic Equations with Absolute Value

As we saw earlier, an absolute value (designated by |  |) means take the positive value of whatever is between the two bars.  The absolute value is always positive, so you can think of it as the distance from 0.  So  and .  It’s as simple as that!

(Note that we also address absolute values here in the Linear Inequalities section, here in the Solving Absolute Value Equations and Inequalities sectionhere in the Piecewise Functions section, and  here in the Graphing Rational Functions,  including Asymptotes section.)

The absolute value does get a little more complicated when we’re dealing with variables, since we don’t know the sign of the variable.  For example, if we have , we don’t know if x is 3 or –3 .  So we’d have to say that the solution to the equation  is  3 and –3.

Once we get answers, we must always check them, since they may be extraneous (not work).  This is just the nature of the beast!

Here are examples of solving absolute value equations:

Here’s one more that’s a bit tricky, since we have two expressions with absolute value in it.  In this case, we have to separate in four cases, just to be sure we cover all the possibilities.  We then must check for extraneous solutions, possible solutions that don’t work.

Learn these rules and practice, practice, practice!

Click on Submit (the arrow to the right of the problem) to solve this problem. You can also type in more problems, or click on the 3 dots in the upper right hand corner to drill down for example problems.

If you click on “Tap to view steps”, you will go to the Mathway site, where you can register for the full version (steps included) of the software.  You can even get math worksheets.

You can also go to the Mathway site here, where you can register, or just use the software for free without the detailed solutions.  There is even a Mathway App for your mobile device.  Enjoy!

On to Linear Inequalities – you are ready!

## 6 thoughts on “Solving Algebraic Equations”

1. 1/|2−7x|=1
How would you do this problem when the absolute value signs are in the denominator?

• Good question, Laura, and sorry it took so long to reply (and I see you got it – great!). I’ll add it to my site – you would treat the absolute value as a variable and cross multiply or whatever you need to do to get it on one side (so it’s not in a denominator), and then divide it into two equations. I got x = 1/7 and x = 3/7.

2. For the problem, |3x-2| + |x+2|=6
when the first expression is positive and the second is negative, resulting in 5, you state above that 5 doesn’t work (extraneous expression). In fact, 5 does work when plugged back into the equation.

• Thanks for writing! I still get that 5 doesn’t work: |3(5) – 2| + |5 + 2| = 13 + 7 = 20. Does that make sense? Lisa