Solving Multi-Step Linear Equations

The variable is on the left-hand side (LHS) of the equation. It is currently multiplied by seven, and then it has a two added to it. I need to undo the "times seven" and the "plus two".

There is no rule about which "undo" operation I should do first. However, if I first divide through by 7, I'm definitely going to create fractions. Personally, I prefer to avoid fractions if possible, so I almost always do any plus / minus before any times / divide. I might end up having to deal with fractions anyway, but at least I can put them off until closer to the end of my work.

Starting with the "plus two" first, I'll subtract two from either side of the equation. Only then will I divide through by the seven. My work looks like this:

By doing the plus / minus first, I avoided fractions. As you can see, the answer doesn't involve fractions, so I did myself a favor by doing the dividing-through last. My solution is:

In this equation, the variable (on the left-hand side) is multiplied by a minus five, and then a seven is subtracted from it. In hopes (as always!) of avoiding fractions, I'll add seven to either side of the equation first. Only then will I divide through by the minus five. My work looks like this:

I've shown my work neatly. Now I'll clearly rewrite my solution at the end of my work:

The variable (on the left-hand side of the equation) is multiplied by a three, and then a nine is subtracted from it. I'll take care of the nine first, and then the three:

In this case, again, my solution has no fractions:

In this equation, I have two terms on the left-hand side that contain variables. So my first step is to combine these "like terms" on the left. Then I can solve:

So now my equation is:

Even though it might initially have looked more complicated, this is actually a one-step equation. I'll solve by dividing through by twelve:

My answer is:

In this equation, I've got terms with variables on either side of the equation. To solve, I need to get those variable terms all on one side of the equation.

There is no rule saying which of the two terms I should move, the 4x or the 6x. However, I've learned from experience that, to avoid negative coefficients on my variables, I should move the x term with the smaller coefficient. That means, in this case, that I'll subtract the 4x from the left-hand side over to the right-hand side:

And now I have a one-step equation, which I'll solve by dividing through by two:

My solution is:

In this equation, I've got variables on either side of the equation, and also loose numbers on either side. I need to get the variable terms on one side, and the loose numbers on the other side. Because I'd like to avoid negative coefficients on my variables, I'll be moving the smaller of the two terms; namely, the −4x that's currently on the right-hand side. To get the loose numbers on the side opposite the variable terms, I'll be moving the −1 that's currently on the left-hand side. There is no particular "right" order for doing these steps; since they're both a matter of adding, people usually do them together in one step. I'll do the variable terms first, and then the loose numbers:

At this point, I've got a one-step equation which requires one division to solve:

Then my answer is:

8x - 1 = 23 - 4x
+4x +1 +1 +4x
-----------------
12x = 24
--- --
12 12

x = 2

This equation is all kinds of messy! Before I can solve, I'll need to combine the like terms on either side of the equation:

Now that I've simplified each side of the equation, I can do the solving.

I added the (usually unstated) 1 to the variable term on the right-hand side of the original equation in order to help me keep track of what I was doing; it isn't "necessary". And it isn't expected on the final answer, which is properly stated as:

This equation solves just like all the other linear equations I've done. It just looks worse because of the decimals. But that's easy to fix!

Whatever is the largest number of decimal places in any of the coefficients, I can multiply through on both sides by "1" followed by that number of zeroes. In this case, all of the decimals have one decimal place, so I'll multiply through by 10:

Now I can solve as usual:

Just because the original equation had decimal places, doesn't mean that I'm stuck working with them. File this trick away for later; it'll come in handy.

Ick! Fractions! But, just as with the decimals in the previous exercise, I don't have to be stuck with fractions. In this case, I'll be multiplying through to "clear" the denominators, giving me a nicer equation to solve.

To simplify my computations for equations with fractions, I will first multiply through on both sides by the common denominator of the various fractions. For this equation, the common denominator is 12, so I'll multiply everything by 12 (or, when multiplying against a fraction, I'll multiply by ):

This is now a much nicer equation to work with. I'll continue my solution by subtracting the smaller 2x from either side:

I'll remove the 1 from the variable when I write my final answer: