Solving and Manipulating Equations | SAT Algebra | SAT Math

Introduction

This lesson combines two different kinds of SAT questions because they share a similar foundational skill: isolating a variable. On the one hand, solving for a single variable is the very heart of algebra; we will briefly review that skill but will focus primarily on a specific application of the skill: isolating one variable among many. The question is, how do you isolate a variable in an equation with three or more variables? The good news is that the process doesn’t significantly differ from that involving just one variable. We’ll talk about that process in further detail below.

Approach Question

The equation for converting degrees Kelvin, K, to degrees Fahrenheit, F, is as follows:

$K = (\frac{5}{9}) (F - 32) + 273.15$

Which of the following correctly expresses F in terms of K?

A. F=(9/5)(K-273.15)+32
B. F=(9/5)(K-273.15+32)
C. F=(5/9)(K+273.15)+32
D. F=(5/9)(K-273.15)+32

Explanation

Both of the key skills in this lesson–solving one-variable equations and isolating a variable in multi-variable equations–hinge on the same score skill inverse operations. As you learned in algebra class, moving a term (a variable, a constant, or a combination of the two like 2x) from one side of an equation to the other requires reversing the operation. If a term is added in an equation, it must be subtracted, and vice versa. If part of an equation (usually a number) is multiplied by something else, then it must be divided from both sides, and vice versa. Finally (and this aspect of inverse operations is often overlooked), this applies to exponents and roots as well: if a variable in an equation is squared and that term (say, x^2) is isolated on one side, we may take the square root of both sides to solve for x. Conversely, if an entire side of an equation is under a square root, we can square both sides to enable solving for the variable.

The equation in this problem has two variables; one is currently isolated, but we are called upon to isolate the other. One helpful principle here might be called “reverse PEMDAS.” If we follow the acronym PEMDAS (Parentheses, then Exponents, then Multiplication and Division, then Addition and Subtraction) for order of operations in working with numbers, we typically reverse that order when solving equations (there is one notable exception, which we will cover below).

“Reverse PEMDAS” means that we should begin with addition or subtraction if presented with the opportunity. This means we should subtract $273.15$ from both sides, resulting in:

$K - 273.15 = \frac{5}{9} (F - 32)$

What next? Moving right to left through PEMDAS, we come to multiplication and division. If a number is multiplied by the rest of the equation, now is the time to divide both sides of the equation by that number. In this case, that number is a fraction, so we follow the “keep-change-flip” process of dividing by that fraction. This means that the reciprocal of that fraction appears on the other side:

$\frac{9}{5} (K - 273.15) = F - 32$

To finish the process, note that we no longer need parentheses on the right side of the equation because there is no longer multiplication by the binomial $F - 32$ . By removing the parentheses, we have “reset” the right side, so we now do whatever is necessary to fully isolate the variable F. Inverse operations tells us to add $32$ to both sides, giving us:

$\frac{9}{5} (K - 273.15) + 32 = F$

or choice A.

Note that choice B is very close, but it brings the $+ 32$ inside the parentheses; this doesn’t make sense because we add $32$ at the last step, so it must simply be adding on the end of the expression already present. Two other wrong answers fail to flip the fraction $\frac{5}{9}$ when dividing both sides by that fraction.

Definitions

In Terms Of: As seen in the approach question, the SAT may present you with an equation with two or more variables and ask you to solve for one variable “in terms of” the other. The phrase here means “in the language of” or “expressed using” the variable in question. A good way to translate this phrase is to remember that you are always asked to solve for the variable before the phrase “in terms of.” For instance, “solve for $y$ in terms of $x$ ” means to focus on isolating the y variable; you can almost ignore the phrase with “in terms of” if you understand this concept.
Variable: A symbol used to represent a number in equations and expressions.
Constant: A value that does not change. Put another way, a number that occurs without a variable.
Coefficient: A number that is multiplied by a variable so that the two elements together make one term. Examples: the $2$ in $2 x$ , the $\frac{1}{3}$ in $\frac{1}{3} y$

Topics for Cross-Reference

Variations

Rather than simply asking you to solve for $x$ , the SAT likes to set up equivalent expressions. For instance, if $5 x + 35 = 15$ , what should $x + 7$ equal? You could certainly solve for $x$ and plug that result in for $x + 7$ , but once you notice that $5 x + 35$ is a multiple of $x + 7$ , it’s much quicker to divide both sides of the whole equation by $5$ , thereby arriving at the value of $x + 7$ in just one step.

This is also a good place to mention one exception to the “reverse PEMDAS” principle. When one side of the equation is simply a fraction (even if the fraction has many terms in either the numerator or denominator, or both), your first step is always to multiply both sides by the denominator. Consider this example:

$\frac{x + 45}{7} = 5 x - 13$

Though we might want to begin adding or subtracting in order to get the $x$ terms on one side and the constant terms on the other, we must multiply both sides by $7$ first. Once you’ve done that, you have a “parallel” equation and can proceed with the usual “reverse PEMDAS” principles.

Strategy Insights

Since “Reverse PEMDAS” is so helpful in solving equations, make sure to review PEMDAS if you don’t remember it! It is also known as the order of operations; don’t forget that the MD and the AS essentially go together as blocks of 2. In other words, multiplication does not go before division, nor vice versa; addition does not precede subtraction, nor vice versa.
Look for hidden similarities! SAT algebra is about this principle as much as just about anything else. You might be wondering why the SAT pairs the equation $2 x - 3 = 5$ with the exponential fraction $\frac{4 ^{3 x}}{4 ^{x + 3}}$ , but when you apply the quotient rule for exponents, you realize that the result is $4^{2 x - 3}$ . You already know that $2 x - 3 = 5$ , so now you plug in $5$ for the exponent and calculate $4^{5}$ . The hidden similarity unlocked the relationship!
As noted under “Definitions,” ignore the phrase “in terms of …” All that matters is the variable that comes before that phrase. If it asks you to solve for $y$ in terms of $z$ , all you should care about is isolating $y$ .

Flashcard Fodder

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.

Also, make a flashcard for PEMDAS if you are hazy on the concept!

Sample Questions

Difficulty 1

Which of the following equations has the same solution as the equation $4 a - 7 = 21$ ?

A. $4 a = 3$
B. $4 a = 14$
C. $4 a = 28$
D. $4 a = 147$

(spoiler)

The answer is $4 a = 28$ . This is the kind of question, as explained in the lesson, where there is a faster approach than solving all the way through. The UnCLES method reminds us to always notice the multiple-choice possibilities. What do they have in common? $4 a$ isolated on one side! So why solve completely for a when you can stop at $4 a$ ? Simply add $7$ to both sides, and you’ll see that $4 a = 28$ . The wrong answers all result from doing the wrong (inverse) operation: division, subtraction, or multiplication.

Difficulty 2

$2 n + 11 = 4 p - 10 q$

The equation above relates the numbers $n$ , $p$ , and $q$ . Which of the following correctly expresses $n$ in terms of $p$ and $q$ ?

A. $n = 8 p - 20 q - 22$
B. $n = 4 p - 10 q - 11$
C. $n = 2 p - 10 q + \frac{11}{2}$
D. $n = 2 p - 5 q - \frac{11}{2}$

(spoiler)

The answer is $n = 2 p - 5 q - \frac{11}{2}$ . As we did in the Approach question, we focus on the first variable mentioned (after the word “expresses”) and essentially ignore the “in terms of” phrase. Everything rides on isolating n. Following the “reverse PEMDAS” principle, we don’t yet address the $2$ multiplied by the $n$ ; rather, we take care of the $+ 11$ by subtracting 11 from both sides. (Notice that we are addressing the left side of the equation, because that’s where the variable we want to isolate resides.)

After subtraction, we have $2 n = 4 p - 10 q - 11$ . All that remains is to divide both sides by $2$ in order to fully isolate $n$ . That might lead us to expect an answer that is all one fraction, with a denominator of $2$ . That is certainly one possible result, but remember that it is also permissible to divide every term on the right side individually by $2$ . $4$ divided by $2$ is $2$ and $10$ divided by $2$ is $5$ ; as for the $11$ , typically the SAT will render the result as an improper fraction ( $11/2$ in this case) rather than a decimal ( $5.5$ ), though a decimal result is certainly possible. Only one answer has all the correct coefficients and constants.

Difficulty 3

If the equation $\frac{3 y}{7} = \frac{4 w}{z}$ , which of the following correctly expresses $w$ in terms of $y$ and $z$ ?

A. $w = \frac{3 yz}{28}$
B. $w = \frac{28 y}{3 z}$
C. $w = 3 yz - 28$
D. $w = \frac{3 y}{28 z}$

(spoiler)

The answer is $w = \frac{3 yz}{28}$ . When confronted by fractions on both sides of an equation, with no other terms present, the best recourse is cross-multiplication. (You will see this again in the Rates and Proportions lesson.) We multiply the first numerator by the second denominator to create one side of the equation and the first denominator by the second numerator to create the other side, like so:

$\frac{3 y}{7} (3 y) (z) 3 yz = \frac{4 w}{z} = (7) (4 w) = 28 w$

Now we’re very close to isolating $w$ ; all that remains is to divide both sides by $28$ . That $28$ simply “lives” below the $3 yz$ , since no simplifying by greatest common factor is possible.

Experienced solvers of this type of problem may notice a quicker way: if we’re looking to isolate $w$ and we understand the $w$ is multiplied by $\frac{4}{z}$ , then we can isolate $z$ in one step by multiplying both sides by the reciprocal, $\frac{z}{4}$ . This yields the answer in one step.

Difficulty 4

If $27 y - 15 x = 8$ , what is the value of $45 y - 25 x$ ?

A. 8/3
B. 12
C. 40/3
D. 16

(spoiler)

The answer is 40/3. This problem is an excellent example of how the UnCLES method and a determination to notice similarities can help us arrive at the answer much faster than the “textbook” approach. While you could solve by substitution, isolating either $x$ or $y$ and then substituting the result into the expression $45 y - 25 x$ , that approach would create some unhelpful fractions and be altogether messy. There’s a better way!

The key is to notice that there is a common ratio at work here. The ratio between the two $y$ coefficients ( $27$ and $45$ ) is the same as the ratio between the $x$ coefficients ( $15$ and $25$ ). A calculator can help you realize that, in both cases, the ratio is $3 : 5$ , which can also be written as the fraction $3/5$ . That means that if we multiply the equation by $5/3$ all the way (since $5/3$ is the reciprocal of $3/5$ ), we should arrive at exactly $45 y - 25 x$ . Indeed we do, which means that the correct answer is the original right side ( $8$ ) multiplied by $5/3$ . That comes to $40/3$ .

Difficulty 5

$40 c - 27 d = \frac{3 ( 7 c - 9 )}{b}$

The equation above relates the positive numbers $b$ , $c$ , and $d$ . Which of the following correctly expresses $c$ in terms of $b$ and $d$ ?

A. $c = \frac{b ( 27 b d + 1 )}{19}$
B. $c = \frac{27 ( b d + 1 )}{40 b - 21}$
C. $c = \frac{27 ( b d - 27 )}{40 b + 21}$
D. $c = \frac{27 b d - 27}{40 b + 21}$

(spoiler)

The answer is $c = \frac{27 ( b d + 1 )}{40 b - 21}$ . The added difficulty in this question lies largely in the fact that there are two terms that contain the key variable $c$ . That would not be a significant issue if the two terms end up as like terms, such that they can be simplified by addition or subtraction. But in this case, the first step in solving must be to multiply both sides by $b$ , which results in one of the $c$ terms also containing $b$ . This makes it not a like term with the terms that end up being $21 c$ .

The answer to this difficulty is factoring by the greatest common factor. We’ll see why that is, but first we need to put in motion the process described under “Variations” in this lesson. Since the entire side of our equation is a fraction, we should begin by multiplying by the denominator $b$ . That cancels the $b$ on the right side and multiplies by $b$ on the left; the equation is now $40 b c - 27 b d = 3 (7 c - 9)$ . Let’s distribute the $3$ on the right and then do the work necessary to get all the terms with $c$ on one side, since $c$ is the variable we’re trying to solve for. Those two steps look like this:

$40 b c - 27 b d 40 b c - 21 c - 27 b d 40 b c - 21 c = 21 c - 27 = - 27 = 27 b d + 27$

Now we’re ready to use GCF factoring to unlock the next steps. Once all the terms on one side share the same variable, the next step is always to factor out that variable. Doing so will give us a form in which the variable is multiplied by the rest of the expression on that side of the equation, and then we can divide by that expression to fully isolate our variable. Those two steps look like this:

$c (40 b - 21) c = 27 b d + 27 = \frac{27 b d + 27}{40 b - 21}$

But believe it or not, there is one more step! Did you notice the common coefficient in the expression $27 b d + 27$ ? To recognize the final answer, we need to factor out that $27$ , yielding $27 (b d + 1)$ for that portion of the answer. And finally, at long last, we’re there!

For Reflection

How will you approach questions that involve manipulating equations on test day? Write down at least three takeaways from this module.
Rate the difficulty of these questions for you from 1 (no problem) to 5 (problem!). This will help you decide when to answer them and when to skip them on test day.
Do you feel confident in your ability to manipulate equations? If not, you should practice; search for worksheets on multi-step equations on the internet and you’ll find plenty of material.

Introduction

Approach Question

The equation for converting degrees Kelvin, K, to degrees Fahrenheit, F, is as follows:

$K = (\frac{5}{9}) (F - 32) + 273.15$

Which of the following correctly expresses F in terms of K?

A. F=(9/5)(K-273.15)+32
B. F=(9/5)(K-273.15+32)
C. F=(5/9)(K+273.15)+32
D. F=(5/9)(K-273.15)+32

Explanation

“Reverse PEMDAS” means that we should begin with addition or subtraction if presented with the opportunity. This means we should subtract $273.15$ from both sides, resulting in:

$K - 273.15 = \frac{5}{9} (F - 32)$

$\frac{9}{5} (K - 273.15) = F - 32$

$\frac{9}{5} (K - 273.15) + 32 = F$

or choice A.

Definitions

In Terms Of: As seen in the approach question, the SAT may present you with an equation with two or more variables and ask you to solve for one variable “in terms of” the other. The phrase here means “in the language of” or “expressed using” the variable in question. A good way to translate this phrase is to remember that you are always asked to solve for the variable before the phrase “in terms of.” For instance, “solve for $y$ in terms of $x$ ” means to focus on isolating the y variable; you can almost ignore the phrase with “in terms of” if you understand this concept.
Variable: A symbol used to represent a number in equations and expressions.
Constant: A value that does not change. Put another way, a number that occurs without a variable.
Coefficient: A number that is multiplied by a variable so that the two elements together make one term. Examples: the $2$ in $2 x$ , the $\frac{1}{3}$ in $\frac{1}{3} y$

Topics for Cross-Reference

Variations

$\frac{x + 45}{7} = 5 x - 13$

Strategy Insights

Since “Reverse PEMDAS” is so helpful in solving equations, make sure to review PEMDAS if you don’t remember it! It is also known as the order of operations; don’t forget that the MD and the AS essentially go together as blocks of 2. In other words, multiplication does not go before division, nor vice versa; addition does not precede subtraction, nor vice versa.
Look for hidden similarities! SAT algebra is about this principle as much as just about anything else. You might be wondering why the SAT pairs the equation $2 x - 3 = 5$ with the exponential fraction $\frac{4 ^{3 x}}{4 ^{x + 3}}$ , but when you apply the quotient rule for exponents, you realize that the result is $4^{2 x - 3}$ . You already know that $2 x - 3 = 5$ , so now you plug in $5$ for the exponent and calculate $4^{5}$ . The hidden similarity unlocked the relationship!
As noted under “Definitions,” ignore the phrase “in terms of …” All that matters is the variable that comes before that phrase. If it asks you to solve for $y$ in terms of $z$ , all you should care about is isolating $y$ .

Flashcard Fodder

Also, make a flashcard for PEMDAS if you are hazy on the concept!

Sample Questions

Difficulty 1

Which of the following equations has the same solution as the equation $4 a - 7 = 21$ ?

A. $4 a = 3$
B. $4 a = 14$
C. $4 a = 28$
D. $4 a = 147$

(spoiler)

Difficulty 2

$2 n + 11 = 4 p - 10 q$

The equation above relates the numbers $n$ , $p$ , and $q$ . Which of the following correctly expresses $n$ in terms of $p$ and $q$ ?

A. $n = 8 p - 20 q - 22$
B. $n = 4 p - 10 q - 11$
C. $n = 2 p - 10 q + \frac{11}{2}$
D. $n = 2 p - 5 q - \frac{11}{2}$

(spoiler)

Difficulty 3

If the equation $\frac{3 y}{7} = \frac{4 w}{z}$ , which of the following correctly expresses $w$ in terms of $y$ and $z$ ?

A. $w = \frac{3 yz}{28}$
B. $w = \frac{28 y}{3 z}$
C. $w = 3 yz - 28$
D. $w = \frac{3 y}{28 z}$

(spoiler)

$\frac{3 y}{7} (3 y) (z) 3 yz = \frac{4 w}{z} = (7) (4 w) = 28 w$

Difficulty 4

If $27 y - 15 x = 8$ , what is the value of $45 y - 25 x$ ?

A. 8/3
B. 12
C. 40/3
D. 16

(spoiler)

Difficulty 5

$40 c - 27 d = \frac{3 ( 7 c - 9 )}{b}$

The equation above relates the positive numbers $b$ , $c$ , and $d$ . Which of the following correctly expresses $c$ in terms of $b$ and $d$ ?

A. $c = \frac{b ( 27 b d + 1 )}{19}$
B. $c = \frac{27 ( b d + 1 )}{40 b - 21}$
C. $c = \frac{27 ( b d - 27 )}{40 b + 21}$
D. $c = \frac{27 b d - 27}{40 b + 21}$

(spoiler)

$40 b c - 27 b d 40 b c - 21 c - 27 b d 40 b c - 21 c = 21 c - 27 = - 27 = 27 b d + 27$

$c (40 b - 21) c = 27 b d + 27 = \frac{27 b d + 27}{40 b - 21}$

For Reflection

How will you approach questions that involve manipulating equations on test day? Write down at least three takeaways from this module.
Rate the difficulty of these questions for you from 1 (no problem) to 5 (problem!). This will help you decide when to answer them and when to skip them on test day.
Do you feel confident in your ability to manipulate equations? If not, you should practice; search for worksheets on multi-step equations on the internet and you’ll find plenty of material.