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When an equation has more than one variable, we can't just say "solve the equation." That's like telling you to "name the One Tenor." You can't do that, because there are three of them. Not that many people could name the Three Tenors, either, but you get the point. #CarrerasIsTheRingoOfTheTenors
We need to specify which variable we want to get all by itself at the end. Rearranging an equation so that x (or w, or y, etc.) is all by itself is referred to as solving the equation for x (or w, or y, etc.) You may think this step will give you a headache, but just try following the plot of an opera sung entirely in Italian.
To solve an equation for a particular variable, we can perform the same actions we did when solving equations that only had one variable. We can add the same number to both sides, multiply both sides by the same number, etc.
The end goal is still to get a variable all by itself on one side of the = sign, but now the variable needs to be specified since there are several to choose from. You're like a mathematician in a candy store, except all the candy is made of variables.
Solve the equation xy = 3z + 2 for z.
First we subtract 2 from both sides of the equation.
xy – 2 = 3z
Then divide both sides by 3.
Since a formula is an equation that expresses one variable in terms of other variables, and life has lots of formulas, we often want to solve a formula for a particular variable. You may not have come across a ton of these real-life formulas yet, but you will. There are a lot of inequalities out there.
Let A be the area of a rectangle, l its length, and w its width. The formula A = lw expresses the area of the rectangle in terms of its length and width. Hey, if you're trying to figure out the area of a rectangular desk to see if it'll fit in your new room, this could be one of those very real-life formulas we mentioned. Solve this formula for w.
This is a one-step problem. We divide both sides of the formula by l, write down and we're ready to roll.
Equations with multiple variables lend themselves to problems that ask you to find the value of some variable given the values of some other variable(s). Sometimes these are straightforward and require using a formula. See how we made a formula seem like a good thing?
Suppose a rectangle has area 30 cm2 and length 6 cm. What's the width of the rectangle?
We use the same formula as before, now substituting 30 for A and 6 for l, to find that
30 = 6w
This is an equation with only one variable: w. We know how to do this problem. We eat single variables for breakfast. (Okay, let's be honest: we eat Sugar-O's.) We divide both sides by 6 to find that w = 5 cm.
When given area and length, finding width isn't so bad. We write down the formula for area, substitute the given numbers for area and length to get an equation with one variable, and solve the equation. However, if we were asked to find the widths of two hundred different rectangles, we'd get tired of solving similar equations over and over and over again. Hopefully you're not thinking of setting up two hundred desks in your new apartment.
But, even if you are, there's an easier way. Earlier, we solved the formula A = lw for w to find the new formula
We can use this formula from now on to find widths of rectangles.
We can also solve equations for expressions that aren't single variables. We can solve an equation for x2, or 2x, for example. We can sense your eager anticipation. We won't make you wait any longer.
Solve the equation A = πr2 for r2.
Divide both sides by π to find that .
Because we have r2 all by itself on one side of the equation, and no copies of r on the other side, that's it. Gosh, it seems like so little work for such an ugly-looking expression. Good thing we're all about inner beauty here.
Solve the equation y = 2x – 9 for 2x.
Add 9 to both sides to find that y + 9 = 2x. We're already done. Whoa, we barely started! If we were solving for x, we would divide both sides by 2, but since we're solving for 2x, we don't even need to bother. We can use all that extra time we saved to volunteer at a homeless shelter. Or...to play another level of Call of Duty.
When we're solving for an expression involving x—say, x2—we want to end up with a formula for x2 that doesn't have any x terms in it. Having a formula like isn't helpful, because if we knew x then we would know x2 without needing to use some fancy formula. Even though we'd probably try to figure out the fancy formula before realizing that. Doh.
Be Careful. Remember to answer the question that's actually being asked. If you're told to solve for 2x, don't do extra work to solve for x if you don't need to. Your time is precious. Time is money. That's money in the bank. You can take that to the bank. We think you see where we're going with this train.
Algebra may not be naturally pretty, but if we have the right foundation, blush, and lipstick, we can enhance its natural features. Of course, we won't be using literal makeup, since that makes the monitor all smeary.
When solving for a variable, we might wind up with a formula that involves some kind of fraction. When that happens, make sure to give the fraction in reduced form.
Solve the equation 4z + 2y = 8 for z.
By rearranging, we find that This is a correct answer, but it's not as nice as it could be. Time to apply the mascara. Multiply the fraction by (a "clever form of 1") to find that
Ooh la la! Hey equation—are you a model?
Another way to find the same answer is to divide both sides of the original equation by 2 to get:
2z + y = 4
And then solve for z. Thankfully, we find the same answer either way. No going back to the drawing board for us.
Factoring is another useful thing to keep in our bag of tricks—right between the levitating card and the disappearing coin—when solving equations for particular variables. While we've been solving a lot of equations by simplifying with the distributive property, sometimes we need to do things the other way around and factor instead. Sit back and take a load off, distributive property. You're on vacation.
What's the solution to the equation 2x + xy – y = 5?
This one is a little tricky. How do we get the x all by itself? We can't exactly declare a quarantine. Instead, look at the left-hand side of the equation, and factor out x:
2x + xy
x(2 + y)
Multiply this out, and you'll see that it works. We can rewrite the original equation as:
x(2 + y) – y = 5
From there, add y to both sides and divide by (2 + y) to find that:
Solve the equation xy + yz = xz for y.
Hey, where did all the numbers go? No worries: we can do this step with only variables just as easily. Again, we need to factor. If we factor y out of the first two terms, we get y(x + z) = xz, and so
Solve the equation 3x + xy = 4x – 2 for x.
We need to factor out x, but first, we get all the x terms on one side by subtracting 4x from each side:
-x + xy = -2
Now we factor out x:
x(-1 + y) = -2
Almost there. Now divide both sides by (-1 + y):
Be careful: When we factor out x from itself, we leave 1 behind. This is unfortunate, as we swore to 1 that we would never leave him behind, but it had to be done. We only hope he can forgive us.
If you don't remember them, it's a good idea to review your geometry formulas. They'll come back to haunt you all the time, and it's better if they're more like Casper the Friendly Ghost than Jasper the Grumpy Ghost Who Will Scare the Bejeezus out of You in the Middle of the Night. So strange that that Saturday morning cartoon never got off the ground.
If a triangle has a height of 4 cm and an area of 20 cm2, how long is the base of the triangle?
This is a fairly straightforward question, as long as you've memorized your geometry formulas. Ahem. The area A of a triangle is given by , so just plug in the numbers given in the problem: h = 4 and A = 20.
We're multiplying the fraction, the variable, and the 4 together. The variable we can't do anything about for the time being, but we can multiply the fraction and the 4 together to get 2.
20 = 2b
We can then divide both sides by 2 to find that b = 10 cm.
In the picture below, find a formula for a2 in terms of the shaded area. Let A denote the shaded area.
The shaded area A is equal to the area of the square minus the area of the circle. There's no special formula for that part—you can figure out that part by eyeballing it. Or by smelling it, or whatever your most reliable sense is. The side length of the square is a, so the area of the square is a2.
The radius of the circle is (that's half the square's side length), which means the area of the circle is:
Putting together all the puzzle pieces, we subtract the circle from the square.
We're almost done. The only thing left is to actually do what the question asks, which is to provide a formula for a2 in terms of A. With whipped cream and a cherry on top, if possible. We need to rearrange our current formula a bit. First, simplify the right-hand side.
There are two ways we can go from here. Technically, there are infinitely many ways we can go from here, but only two correct ones. Since time is limited, we'll only go over the correct ones.
Way 1: Factor out the a2 from each term in the right-hand side.
Divide each side by the quantity in parentheses, and here we are:
Way 2: Instead of factoring out the a2 right away, multiply both sides of the simplified formula for A by 4 to get rid of fractions.
4A = 4a2 – πa2
Which is much prettier. In fact, we'd kiss it on the mouth if it had one.
Now factor out a2 to get:
4A = a2(4 – π)
And divide by the stuff in parentheses to finish up.
This answer might look a little different from the one we got using Way 1, but they're totally identical.