A 4-Step Guide to Solving Equations (Part 2)
Do you think that solving equations is hard? Would you believe that it doesn’t have to be? Want to learn an easier way to do it? Keep on reading to find out!
Jason Marshall, PhD
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A 4-Step Guide to Solving Equations (Part 2)
We learned quite a few important things in the last episode. First, we learned that the infamous Knot Dude was once upon a time challenged to a sort of mathematical duel by a group of seafaring pyramid builders. The group’s skipper said, “If I wanted you to build me a small rectangular pyramid with one side that’s 15 feet long and a diagonal that’s 17 feet long, could you figure out how long the shorter side would have to be?”
We next learned that Knot Dude figured out that he could easily squash this challenge and humiliate his father’s pyramid building rivals if he could simply figure out how to solve the Pythagorean Theorem equation, a^2 + b^2 = c^2, for the variable b. And we learned that while doing so, Knot Dude developed an easy 4-step method—the first two steps of which we learned last time—for solving equations that’s still in use!
Today we’re going to learn the final two steps to Knot Dude’s method and we’re going to put it all together and figure out precisely how Knot Dude solved his problem and sent those nautically inclined pyramid builders sailing away in defeat..
Step 1: Simplify Each Side of the Equation
As we learned last time, the first step in solving an equation is to make the equation as simple as possible. This means that you need to start by using the golden rule of equation solving and the order of operations, PEMDAS, to make the expression on each side of the equals sign as simple as possible. In the example that we talked about last time, we added, subtracted, multiplied, and divided until we turned the equation
2 + x – 2 • 5 = 4 / 2 – x
into the very much simplified version of itself
x – 8 = 2 – x
These two equations may not look the same, but as we verified last time, they really are just different ways of writing the same underlying equation!
Step 2: Move Variable to One Side
The next step in solving an equation for a particular variable is to use addition and/or subtraction to move every part of the equation that contains the variable you’re solving for to one side of the equals sign. In Knot Dude’s problem of solving for b in the Pythagorean Theorem, a^2 + b^2 = c^2, it’s most convenient for us to find a way to isolate b on the left side and move everything else to the other side.
As we learned last time, we can do that by subtracting a^2 from both sides (to keep everything balanced according to the golden rule). When we do that, we get a slightly different form of the Pythagorean Theorem equation that looks like
b^2 = c^2 – a^2
Which is great, because we’re now very close to solving for b.
Step 3: Isolate the Variable
The name of the game in the third step is to do whatever you have to do to get a single power of the variable you’re solving for all by itself. In other words, we have b^2 on the left side of our equation right now, but we really want to solve for b. The exact process used will be different in different problems. Sometimes you’ll need to use multiplication or division to get the variable by itself, sometimes you’ll need to raise your variable to a power, and sometimes—as in our problem—you’ll need to take a square root.
If we take the square root of both sides of our equation, we get
√(b^2) = √(c^2 – a^2)
Of course, √(b^2) is just equal to b, which gives us
b = ±√(c^2 – a^2)
Great, we’ve solved for b! But whoa—what’s up with that “±” sign? Well, equations like x^2 = “stuff” always have two solutions—one is equal to +“stuff” and the other is equal to –“stuff”. For example, x^2 = 4 has the two solutions x = √4 = 2 and x = –√4 = –2 (plug them in and see). But instead of writing both solutions out, we just say x = ±2.
You can check that b = ±√(c^2 – a^2) are both valid solutions by plugging each into the original Pythagorean equation. Try it and see what happens! For Knot Dude’s problem, we really only care about one of these solutions…the positive one. Why? Well, since we’re building a pyramid, we absolutely know that the solution must be a positive number. What would a negative length for a pyramid even mean?
So for this problem, the solution that we’re after is
b = √(c^2 – a^2)
As you’ll recall, this whole problem started with a challenge to Knot Dude to see if he could figure out the value of b when a = 15 and c = 17. And now we know how he can do it—he just needs to plug the values of a and c into this solution to get:
b = √(17^2 – 15^2) = √(289 – 225) = √64 = 8
Step 4: Check Your Solution!
Success, we have a solution! But no, we’re not done yet! And that’s because you should always check that your solution works. In other words, you should always take your solution and put it back in the original equation. If you did everything correctly, the original equation must balance with your solution…meaning its left and right hand sides must end up having the same value.
For Knot Dude’s problem, the left hand side of the original Pythagorean equation gives: a^2 + b^2 = 15^2 + 8^2 = 289. And the right hand side gives: c^2 = 17^2 = 289. So, both sides have the same value—which means that our answer is correct!
Practice Problems
To sum things up, just remember these 4 steps when solving equations:
And also remember that every equation is different, so the only way to get good at solving them is to practice. With that in mind, here are a few equations for you to practice with. Each requires a slightly different approach for solving for x:
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2 + x – 2 • 5 = 4 / 2 – x
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x / 3 + 3 = x / 2
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√x – 4 = –√x
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2x^2 – 3 + x^2 = 9 + x^2
If you get stuck, take a look at the solutions at the very end of the article.
Wrap Up
If you want to learn more about algebra, please check out my book The Math Dude’s Quick and Dirty Guide to Algebra.
Remember to become a fan of the Math Dude on Facebook where you’ll find lots of great math posted throughout the week. If you’re on Twitter, please follow me there, too. Finally, please send your math questions my way via Facebook, Twitter, or email at mathdude@quickanddirtytips.comcreate new email.
Until next time, this is Jason Marshall with The Math Dude’s Quick and Dirty Tips to Make Math Easier. Thanks for reading, math fans!
Practice Problem Solutions
1. 2 + x – 2 • 5 = 4 / 2 – x
Step 1:
Simplify 2 • 5 and 4 / 2: 2 + x – 10 = 2 – x
Simplify 2 – 10: –8 + x = 2 – x
Step 2:
Add x to both sides: –8 + x + x = 2 – x + x
Simplify: –8 + 2x = 2
Add 8 to both sides: –8 + 2x + 8 = 2 + 8
Simplify: 2x = 10
Step 3:
Divide both sides by 2: 2x / 2 = 10 / 2
Simplify: x = 5
Step 4:
Plug-in and check LHS: 2 + 5 – 2 • 5 = 2 + 5 – 10 = 7 – 10 = –3
Plug-in and check RHS: 4 / 2 – 5 = 2 – 5 = –3
LHS → –3 and RHS → –3 ?
2. x / 3 + 3 = x / 2
Step 1:
Multiply both sides by 3: (3x / 3) + 3 • 3 = 3x / 2
Simplify 3x / 3 and 3 • 3: x + 9 = 3x / 2
Multiply both sides by 2: 2x + 18 = 3x
Step 2:
Subtract 2x from both sides: 2x + 18 – 2x = 3x – 2x
Simplify: 18 = x
Step 3: Nothing to do!
Step 4:
Plug-in and check LHS: 18 / 3 + 3 = 6 + 3 = 9
Plug-in and check RHS: 18 / 2 = 9
LHS → 9 and RHS → 9 ?
3. √x – 4 = –√x
Step 1: Nothing to do!
Step 2:
Add √x to both sides: √x – 4 + √x = –√x + √x
Simplify: 2√x – 4 = 0
Add 4 to both sides: 2√x – 4 + 4 = 0 + 4
Simplify: 2√x = 4
Step 3:
Divide both sides by 2: 2√x / 2 = 4 / 2
Simplify: √x = 2
Square both sides: (√x)^2 = 2^2
Simplify: x = 4
Step 4:
Plug-in and check LHS: √4 – 4 = 2 – 4 = –2
Plug-in and check RHS: –√4 = –2
LHS → –2 and RHS → –2 ?
4. 2x^2 – 3 + x^2 = 9 + x^2
Step 1:
Combine 2x^2 and x^2: 3x^2 – 3 = 9 + x^2
Step 2:
Subtract 3x^2 from both sides: 3x^2 – 3 – x^2 = 9 + x^2 – x^2
Simplify: 2x^2 – 3 = 9
Add 3 to both sides: 2x^2 – 3 + 3 = 9 + 3
Simplify: 2x^2 = 12
Step 3:
Divide both sides by 2: 2x^2 / 2 = 12 / 2
Simplify: x^2 = 6
Take the square root of both sides: √(x^2) = ±√6 (remember both solutions!)
Simplify: x = ±√6
Step 4:
Check solution x = √6
Plug-in and check LHS: 2(√6)2 – 3 + (√6)^2 = 2 • 6 – 3 + 6 = 15
Plug-in and check RHS: 9 + (√6)^2 = 9 + 6 = 15
LHS → 15 and RHS → 15 ?
Check solution x = –√6
Plug-in and check LHS: 2(–√6)^2 – 3 + (–√6)^2 = 2 • 6 – 3 + 6 = 15
Plug-in and check RHS: 9 + (–√6)^2 = 9 + 6 = 15
LHS → 15 and RHS → 15 ?
About the Author
Jason Marshall is the author of The Math Dude’s Quick and Dirty Guide to Algebra. He provides clear explanations of math terms and principles, and his simple tricks for solving basic algebra problems will have even the most math-phobic person looking forward to working out whatever math problem comes their way.