*Multivariable equations* is the language we will use in this course to describe equations with two or more variables in which you are called on to solve for one variable in terms of the other(s). In this problem type, we will also see word problems that are modeled by multiple variables (like the Approach Question below) as well as systems of equations with at least three equations, since those present a unique challenge not fully addressed in our lesson on systems.

Mary has a collection of 30 fruits consisting of apples, bananas, and cherries. If she has twice as many apples as bananas and five fewer cherries than apples, how many bananas does she have?

A. 7

B. 9

C. 11

D. 14

The textbook approach to this problem begins with assigning variables to each unknown element. Notice how this problem makes that assignment as easy as ABC: $a$ = apples, $b$ = bananas, $c$ = cherries. We then create equations using the relationships we are given. The first relationship is between apples and bananas. Does that relationship mean our equation is $a=2b$ or $b=2a$? This is an area of common confusion. One way to think about this clearly and consistently is to think about which variable needs “help” to reach the other. In other words, the smaller number needs “help” with addition or multiplication in order to reach the larger number. In this case, bananas are the smaller group, so the $b$ gets multiplied by two.

In addition to $a=2b$, we also have $a−5=c$, because this is the “help” principle in reverse: there are more apples so they need to “help” the cherries through subtraction. (Alternatively, we could “help” the cherries by saying $c+5=a$; either equation will work.)

Because we have three variables, we need one more equation in order to solve. Using the UnCLES method and recovering every important fact, we come upon the quantity of 30, the total number of all the fruits. This naturally leads to the equation $a+b+c=30$.

The best approach from here will be to consider $a+b+c=30$ the “parent” equation and use substitution to reduce the left side to just one variable used repeatedly. Since both of our equations contain a, it’s best to substitute expressions in terms of $a$ for $b$ and $c$. That’s easy enough in the case of $c$: $c=a−5$. For $b$, however, we need to first isolate $b$ by dividing both sides of that equation by $2$: $b=(1/2)a$. We’re now ready to substitute: $a+(½)a+a−5=30$. Simplifying, we get $2.5a−5=30$, and our solving steps are $2.5a=35$ and therefore $a=14$.

Don’t forget the $C$ step of UnCLES! We’re not asked for apples, but for bananas. It’s *always* a good idea to double-check this in the case of a word problem. Since there are half as many bananas as apples, our answer is **A**.

An alternative, “backdoor” approach to this problem involves plugging in the answers. If you do this, plug in one of the middle choices first; this helps because CLT answer choices are typically ordered from least to greatest or vice versa, so if you start in the middle, you can often “triangulate” your way there pretty quickly.

Let’s try this by plugging in Choice $B$. If we begin with the assumption there are $9$ bananas, we can infer that there must then be $18$ apples. There are five fewer cherries than applies, so that’s $13$. But $9+13+18$ is well above $30$. Here’s the good news: since $B$ is too large, *A must be the answer!* (We can, of course, plug it in to make sure: $7$ bananas means $14$ apples and $9$ cherries, so the numbers add up.)

Did you notice how much quicker the “backsolving” approach was than the textbook approach in this case? This doesn’t mean you should always plug in the answers in a case like this, but it does strongly suggest that, in the case of word problems with numerical answers, you should seriously consider this option.

Many multivariable equations problems involve substitution, like the Approach Question above or like questions that present you with a system of three equations. But look out for questions that are more about *manipulation*: moving variable terms around an equation by means of algebra’s “inverse operations” process: that is, adding a variable to both sides when the variable is originally subtracted, dividing both sides by a term when that term is multiplied, etc.

This concept does not require much in the way of memorization, but you want to make a flashcard from the definition of the phrase “in terms of” above. Otherwise, one equation often used in multivariable equation word problems is distance = speed x time. Make sure you are familiar with that equation and can manipulate it to solve for speed or time.

If $a=3b$, what is $b$ in terms of $a$?

A. $b=−3a$

B. $b=(⅓)a$

C. $b=a$

D. $b=3a$

(spoiler)

The answer is **B**. Our objective is to isolate the variable $b$, which here involves a single step: dividing both sides by 3. Keep in mind that choice B could also have been written as $b=a/3$; it’s important to recognize both forms of a fractional variable term.

A basketball player creates a statistic for herself she calls her “hustle index.” She decides that the formula for the hustle index will be:

$minutes playedcharges taken + loose balls recovered $

If $m$ denotes her minutes played for the season and her hustle index is $0.35$, what is the total number of charges she took plus the number of loose balls she recovered, in terms of $m$?

A. $0.35$

B. $m0.35 $

C. $0.35m $

D. $0.35m$

(spoiler)

The answer is **D**. We need to remember that (charges taken + loose balls recovered) is the numerator of the fraction, so to isolate that number we need to multiply both sides of the equation by the denominator. If $m$ is the denominator and the hustle index of $0.35$ occupies the other side of the equation, then $0.35$ and $m$ get multiplied together. One way to simplify your thinking is to use a variable like $n$ to represent all the charges taken plus loose balls recovered. Then the equation can be represented more compactly as $0.35=n/m$, which is the same thing as $0.35m=n$.

If $3y=z4 $, which of the following is equal to $z_{2}$?

A. $144y_{2}$

B. $12y$

C. $3y4 $

D. $9y_{2}16 $

(spoiler)

The answer is **D**. Since z is the focus of the question, we need to isolate the variable $z$. This takes two stops: multiplying both sides by $z$ (in algebra, it’s wise to get rid of the denominator as soon as possible) and then dividing both sides by $3y$. That yields $z=3y4 $. But be careful: the answer is not C, because the question asks for the equivalent of $z$ *squared* (keep using that UnCLES method!). Answers A and B make the mistake of *multiplying* both sides by $3y$ instead of dividing.

The relationship between pressure, $P$, volume, $V$, and temperature, $T$, of a gas is given by $PV=nrT$ according to the ideal gas law. In this equation, $n$ is the number of moles and $R$ is a constant. If the volume of a certain gas is doubled while the temperature is cut in half, what happens to the pressure of the gas? (Assume the number of moles of the gas is held constant.)

A. It is divided by 4.

B. It is divided by 2.

C. It remains the same.

D. It is doubled.

(spoiler)

The answer is **A**. It would be easy to fall for answer $C$ here, thinking the doubling and the dividing in half will cancel each other out. But remember that volume and temperature are on *opposite* sides of the equation. Doubling a variable on side while halving a variable on the other will unbalance the equation, and we must fix that by adjusting the pressure accordingly. It is not enough simply to halve the pressure, as that would cancel out the doubling of the volume but the equation would remain unbalanced because of the change in temperature. We must divide the pressure by $4$, because doubling a quantity and then dividing it by $4$ results in the quality being divided in half. This results in both sides of the equation being divided in half. Balance is restored!

We can also approach this problem by plugging in numbers. There are a lot of variables, but we can choose “friendly” numbers like $10$ for $P$ and $6$ for $V$. We choose the $6$ in order to get a product of $60$ for the left side, a product which we can also create with three factors on the right side. Let’s say that $n=3$, $r=5$, and $T=4$. These numbers multiply to $60$ and we have also chosen an even number for $T$, which will be helpful when need to divide $T$ by $2$.

Now let’s double the volume to $12$ and cut the temperature in half, to $2$. Leaving $P$ as unknown, our equation is now:

$12P12PP =(3)(5)(2)=30=2.5 $

Whereas $P$ started out as $10$ with the numbers we chose, it ended up at $2.5$. It has been divided by $4$.

Note that plugging in good numbers on hard problems takes practice! If you like the plugging in strategy on this problem, make sure you understand why we plugged in the numbers we did.

A car travels at a constant speed $s$ and covers a distance $d$ in time $t$. If the speed increases by $20%$ and the distance remains the same, what happens to the time?

A. It decreases by $20%$.

B. It decreases by $16⅔%$.

C. It increases by $16⅔%$.

D. It increases by $20%$.

(spoiler)

The answer is **B**. We first must identify the correct equation: $distance=speed×time$ ($d=st$). We are holding the distance constant, so the other side of the equation needs to remain the same as well. That means that if speed is increased, time must decrease, so we can eliminate answers $C$ and $D$. (Note how it makes sense that if you go faster, you take less time. It never hurts to confirm with common sense!)

To distinguish between choices $A$ and $B$, we should replace $s$ in the equation with $1.2s$, since the speed has increased by $0.2$ of its original value, $s$. If we now solve for time, we will divide both sides by $1.2s$ and come up with $t=d/1.2s$. That result is not very helpful, but let’s follow the rules of simplifying and get rid of the decimal inside the fraction. We can multiply the top and bottom of the fraction by $10$, then reduce $10/12$ to $5/6$. This tells us that the new time is $65 $ of the old time since the old time can be represented as $d/s$. But what is $65 $ as a percent? You could try to calculate without a calculator, but it should be sufficient to note that the answer is *not* a 20% decrease because that would be shown by $54 $, not $65 $.

If that process seems confusing to you, try plugging in numbers. Let’s say the original speed was $50$ (units don’t matter here) and the time was $2$. That means the distance was 100. If we increase the speed to 60 but leave the distance constant, the means the new time is $60100 $, or $35 $. The question now is what change has taken place from a time of $2$ to a time of $35 $. Subtracting, we get $31 $, and dividing that $31 $ by the original two, we have $61 $. The time has decreased by $61 $. Again, that’s not easy to convert to a percent without a calculator, but since $20%$ is equal to $51 $, not $61 $, the answer can’t be $A$ but must therefore be $B$.

All rights reserved ©2016 - 2024 Achievable, Inc.