## ACT Math For Dummies (2011)

### Part II. Building Your Pre-Algebra and Elementary Algebra Skills

### Chapter 5. It’s Elementary Algebra, My Dear Watson

*In This Chapter*

Knowing two main ideas and six key words of algebra

Evaluating, simplifying, and factoring expressions

Solving a wide variety of equations

Turning words into an equation you know how to solve

Algebra is a key focus of ACT math. Luckily, in this chapter and in Chapter 7, you gain the algebra skills you need most to ace the ACT. I start at the very beginning, reviewing the basic concepts and vocabulary of algebra so you gain a good foundation for what follows. Then I show you three key skills that are essential for working with algebraic expressions: evaluating, simplifying, and factoring. At that point, you’re ready to solve a variety of equations that you’ll see more of on the ACT: rational equations, square root (radical) equations, absolute value equations, and exponential equations. I also show you how to handle equations that have more than one variable. To wrap up, I cover some basics of word problems, showing you how to do direct translation of words into equations.

**Knowing the Two Big Ideas of Algebra**

At its core, solving an algebra equation really comes down to two big ideas — the *how *and the *what *of algebra:

**How algebra works:** Every step you take must keep the equation in balance.

**What you need to do:** Isolate the variable on one side of the equal sign.

These two main points may seem a little scary, but don’t worry. In this section, I discuss everything you need to know to easily solve any algebra equation that the ACT throws at you.

**Every step you take: Keeping equations balanced**

You can change an equation in all sorts of ways, provided every step you take keeps it in balance. When I say you have to keep an equation balanced, what I mean is that you need to make the same changes to both sides of the equation.

For example, here’s an equation you know is true:

2 + 2 = 4

You can add the same number to or subtract the same number from both sides of this equation. For example:

2 + 2 + 3 = 4 + 3

The result here is still true because both sides equal 7. Similarly, you can multiply or divide both sides by the same number. For example:

5(2 + 2) = 5(4)

Again, the equation stays in balance because both sides equal 20. You can even raise both sides of the equation to a power and still keep it in balance. For example:

(2 + 2)^{2} = 4^{2}

Both sides of the equation equal 16, so the equation is still in balance.

**Walkin’ a lonely road: Isolating the variable**

Your main objective in algebra is to *isolate the variable* — that is, you want to get the variable alone on one side of the equation and everything else on the other side. The *variable* is simply a letter that stands in for a specific number. The most common variable is *x,* but you can use any letter you want.

What is the value of *x* if 3*x* + 4 = 12 – *x*?

(A) 2

(B) 4

(C) 6

(D) 8

(E) 10

To answer this question, you want to get all the *x*’s on one side of the equation and all the numbers on the other side. To do so, begin by adding *x* to both sides:

This step brings all the *x*’s to the left side of the equation. Next, subtract 4 from both sides:

To finish, divide both sides by 4:

Therefore, *x* = 2, so the correct answer is Choice (A).

Although most of the algebra questions you’ll see on the ACT will be more difficult than this one, every algebra equation comes back to this simple strategy of isolating the variable.

**Discovering Six Choice Words about Algebra**

Algebra is chock-full of words that are useful but often misunderstood. Here are a few favorites:

**Variable:** A *variable *is any letter that stands for a number. The most commonly used letters are *x* and *y,* but you can use any letter.

**Constant:** A *constant* is a number without a variable. For example: The equation 6*m* + 7 = –*m* has one constant, the number 7.

**Equation:** An *equation* is any string of numbers and symbols that makes sense and includes an equal sign. For example, here are three equations:

2 + 2 = 4 3*x* – 7 = 46

**Expression:** An *expression* is any string of numbers and symbols that makes sense when placed on one side of an equation. For example, here are four expressions:

2 + 2 3*x* – 7 –*x*^{2} – 9*x* + 11

**Term:** A *term *is any part of an expression that’s separated from the other parts by either a plus sign (+) or a minus sign (–). ** Important: **A term always includes the sign that immediately precedes it. For example:

The expression 3*x* – 7 has two terms: 3*x* and –7.

The expression –*x*^{2} – 9*x* + 11 has three terms: –*x*^{2}, –9*x*, and 11.

The expression has one term: .

**Coefficient:** A *coefficient *is the numerical part of a term, including the sign that precedes it (+ or –). ** Important: **Every term has a coefficient.

**When a term appears to have no coefficient, its coefficient is either 1 or –1, depending on the sign. For example:**

The term 3*x* has a coefficient of 3.

The term –9*x* has a coefficient of –9.

The term –*x*^{2} has a coefficient of –1.

Being clear about the meanings of these six words can help you with the rest of this chapter and with any other math you study.

**Express Yourself: Working with Algebraic Expressions**

An *expression* is any string of numbers and symbols that makes sense when placed on one side of an equation, as I discuss in the earlier section “Discovering Six Choice Words about Algebra.” In this section, I show you three important skills for working with expressions:

**Evaluation:** Plugging in numbers for variables to find the value of the expression

**Simplifying:** Reducing the complexity of an expression

**Factoring:** Pulling a common factor out of each term of an expression

Working with expressions is important for solving equations. Additionally, some ACT questions may test these skills directly.

**Evaluating expressions**

When you *evaluate *an expression, you find out what number it’s equal to. To evaluate an algebraic expression — that is, an expression with at least one variable — you need to know the values of all the variables in that expression. Then you can *substitute,* or plug in, the value of each variable and use the normal order of operations (PEMDAS) to find the value of the expression. For more on the order of operations, check out Chapter 3.

** Remember:** Make sure you know how to evaluate expressions correctly. If you need more review on this topic, I cover it in much greater detail in

*Basic Math and Pre-Algebra For Dummies*(Wiley).

What is the value of 3*a*^{2} – *ab* – 2*b*^{2} if *a* = 4 and *b* = –5?

(A) 10

(B) 16

(C) 18

(D) 20

(E) 25

To begin, substitute 4 for *a* and –5 for *b* throughout the expression:

3*a*^{2} – *ab* – 2*b*^{2} = 3(4)^{2} – 4(–5) – 2(–5)^{2}

Now evaluate the expression using the standard order of operations. First, evaluate the powers:

= 3(16) – 4(–5) – 2(25)

Next, multiply:

= 48 + 20 – 50

Finally, add and subtract:

= 18

Therefore, the correct answer is Choice (C).

**Simplifying expressions**

When you *simplify* an expression, you make it more compact and easier to work with without changing its value. In this section, I show you a few basic techniques for simplifying expressions.

**Combining like terms**

Two terms in an expression are *like terms* when they have exactly the same variables, complete with the exactly the same exponents on each variable. Here are three examples of like terms:

3*x* and 4*x* 10*k*^{2} and –9*k*^{2} 2*a*^{3}*b*^{2}*c *and –2*a*^{3}*b*^{2}*c*

When you *combine like terms,* you put the terms together by adding their coefficients and keeping the same variable. For example:

3*x* + 4*x* = 7*x* 10*k*^{2} + (–9*k*^{2}) = *k*^{2} 2*a*^{3}*b*^{2}*c *+ (–2*a*^{3}*b*^{2}*c*) = 0

As you can see in the last case, when the coefficients of a pair of like terms add up to 0, the two terms cancel each other out.

Which of the following expressions is equivalent to 2*x* – 3*y* + 7*xy* – *y* – 2*x*?

(A) –2*xy*

(B) –3*y* + 7*xy*

(C) –4*y* + 7*xy*

(D) 4*x* + 4*y* + 7*xy*

(E) 4*x* – 4*y* + 7*xy*

The expression contains two *x* terms that cancel each out, because 2*x *+ (–2*x*) = 0:

2*x* – 3*y* + 7*xy* – *y* – 2*x* = –3*y* + 7*xy* – *y*

It also contains two *y* terms, which combine as –3*y *+ (– *y*) = –4*y,* leaving you with

= –4*y *+ 7*xy*

Thus, the correct answer is Choice (C).

**Removing parentheses**

Some expressions contain parentheses that prevent you from combining like terms (as I describe in the preceding section). However, after you remove the parentheses from an expression, you may find that you can simplify the expression further. Generally speaking, three different cases require you to remove a single set of parentheses. Here are the rules for all three cases:

**Parentheses preceded by a plus sign (+):** Drop the parentheses. For example:

2*x* + (3 – 4*y*) = 2*x* + 3 – 4*y*

**Parentheses preceded by a minus sign (–): **Negate — that is, change the sign of — every term inside the parentheses, and then drop the parentheses. For example:

2*x* – (3 – 4*y*) = 2*x* – 3 + 4*y*

**Parentheses preceded by a term outside the parentheses: **Distribute the term outside the parentheses — that is, multiply it by every term inside the parentheses — and then drop the parentheses. For example:

2*x*(3 – 4*y*) = 6*x* – 8*xy*

** Remember:** When simplifying an expression with

*nested parentheses*— that is, one set of parentheses inside another — always begin with the inside set and move outward.

Which of the following is equivalent to 4*x*^{2} – (3 – 2*x*(5 – *x*))?

(A) *x*^{2} + 10*x *+ 3

(B) *x*^{2} – 13*x*

(C) 2*x*^{2} + 10*x *– 3

(D) 2*x*^{2} – 7*x*

(E) 6*x*^{2} – 13*x *– 3

To simplify this expression, begin with the inside set of parentheses. The term –2*x* precedes the parentheses, so distribute this term (multiply it by 5 and then by –*x*), and then drop this set of parentheses:

4*x*^{2} – (3 – 2*x*(5 – *x*)) = 4*x*^{2} – (3 – 10*x *+ 2*x*^{2})

Now a minus sign precedes the remaining set of parentheses, so negate each term (change its sign) and drop the parentheses:

= 4*x*^{2} – 3 + 10*x *– 2*x*^{2}

To finish, combine the two like terms (4*x*^{2} and –2*x*^{2}) and rearrange the terms in descending order by their exponents:

= 2*x*^{2} – 3 + 10*x*

= 2*x*^{2} + 10*x *– 3

So the correct answer is Choice (C).

**Getting FOILed**

When simplifying two sets of adjacent parentheses, multiply every term in one set of parentheses by every term in the other. This process often is called *FOILing,* an acronym that stands for *First,* *Outside,* *Inside,* *Last.* The following example shows you how to FOIL an expression.

Which of the following is equivalent to the expression (2*x* + 3)(4*x* – 1)?

(A) 6*x* – 3

(B) 6*x*^{2} – 10*x* – 3

(C) 8*x*^{2} + 10*x* + 3

(D) 8*x*^{2} + 10*x* – 3

(E) 8*x*^{2} + 14*x* – 3

The acronym *FOIL* helps you keep track of the four multiplication calculations you need to perform in order to drop the parentheses. Multiply the two *first* terms, the two *outside* terms, the two *inside* terms, and finally the two*last* terms:

**First:** (2*x*)(4*x*) = 8*x*^{2}

**Outside: **(2*x*)( –1) = –2*x*

**Inside: **(3)(4*x*) = 12*x*

**Last:** (3)(–1) = –3

Place these four terms in one expression:

8*x*^{2} – 2*x* + 12*x* – 3

Combine like terms:

8*x*^{2} + 10*x* – 3

So the correct answer is Choice (D).

**Simplifying expressions with exponents**

When an exponent appears outside a set of parentheses, multiply everything in the parentheses by itself the number of times indicated. Often, doing so requires you to FOIL the contents of the parentheses as I show you in the previous section.

When simplified, how many terms are in the expression (*x* + 1)^{3}?

(A) 1

(B) 2

(C) 3

(D) 4

(E) More than 4

To begin, change the power to multiplication:

(*x* + 1)^{3} = (*x* + 1)(*x* + 1)(*x* + 1)

Now FOIL the contents of the first two parentheses. I do this in two steps:

= (*x*^{2} + *x* + *x *+ 1)(*x* + 1)

= (*x*^{2} + 2*x *+ 1)(*x* + 1)

Next, distribute the contents of the parentheses. To do this, multiply each of the three terms inside the first set of parentheses by both terms inside the second set — six multiplications in all:

= *x*^{3} + *x*^{2} + 2*x*^{2} + 2*x* + *x* + 1

To finish, combine both sets of like terms:

= *x*^{3} + 3*x*^{2} + 3*x* + 1

The simplified form has four terms, so the correct answer is Choice (D).

**Factoring expressions**

*Factoring* an expression reverses the simplification process and introduces parentheses into the expression. Some equations are much easier to solve after you do some factoring. In this section, I show you two of the most useful ways to factor to solve elementary algebra problems. (By the way, I save factoring quadratic expressions for Chapter 7, where you need this skill for solving quadratic equations.)

**Finding common factors**

The most straightforward way to factor an expression is to look for the *greatest common factor* (*GCF*) among all the terms. You probably already have experience at finding the GCF of two or more integers (if not, check out Chapter 4). When finding the GCF among a set of terms, you use this skill and add in one more step: You identify the lowest exponent for each individual variable.

What is the greatest common factor among 6*a*^{2}*b*^{5}*c*^{4}, 12*a*^{3}*b*^{3}*c*^{3}, 15*a*^{4}*b*^{8}*c*?

(A) 2*a*^{2}*b*^{3}*c*^{4}

(B) 3*a*^{2}*b*^{3}*c*

(C) 3*a*^{2}*b*^{5}*c*

(D) 3*a*^{4}*b*^{8}*c*^{4}

(E) 6*a*^{2}*b*^{3}*c*^{3}

To begin, find the GCF of the three coefficients — 6, 12, and 15. In other words, find the highest integer that divides all three of them. The GCF of these three numbers is 3, so 3 must be the coefficient of the GCF; therefore, you can rule out Choices (A) and (E).

Next, focus on the variable *a* and find the *lowest* exponent among all the *a*’s in the three terms. The exponents of *a* are 2, 3, and 4, so the lowest exponent is 2; therefore, the GCF you’re looking for must have the variable *a*^{2}, so you can rule out Choice (D). Now focus on the variable *b* and find the lowest exponent among all the *b*’s. The exponents of *b* are 5, 3, and 8, so the lowest is 3; therefore, the GCF has the variable* b*^{3}, so you can rule out Choice (C). Therefore, the correct answer is Choice (B).

**Unearthing three useful ways to factor squares and cubes**

Three handy formulas exist for factoring different combinations of squared and cubed variables. Most important is the formula for the *difference of two squares,* which looks like this:

*x*^{2} – *y*^{2} = (*x* + *y*)(*x* – *y*)

This formula comes in handy for solving some algebra problems that may otherwise force you to solve a system of equations (as I discuss in Chapter 7). I illustrate this in the following example.

If *x*^{2} – *y*^{2} = 3 and *x* – *y* = 6, what is the value of *x* + *y*?

(A) 1

(B) 2

(C) 3

(D)

(E)

You could solve this problem as a system of equations, first solving for *x* and then for *y* and then adding the results. But this would be time-consuming and tedious. Instead, noticing that the first equation contains the difference of two squares gives you a much simpler way to go.

Begin with the formula for the difference of two squares:

*x*^{2} – *y*^{2} = (*x* + *y*)(*x* – *y*)

Plug in 3 for *x*^{2} – *y*^{2} and 6 for (*x* – *y*):

3 = (*x* + *y*)(6)

You want to isolate the expression *x *+ *y*, so divide both sides by 6 and simplify:

Therefore, the correct answer is Choice (D).

Also useful are the formulas for the *sum of two cubes *and the *difference of two cubes:*

*x*^{3}* + y*^{3} = (*x* + *y*)(*x*^{2} – *xy *+ *y*^{2}) *x*^{3} – *y*^{3} = (*x* – *y*)(*x*^{2} + *xy *+ *y*^{2})

As with the formula for the difference of squares, when these formulas are useful, the problems almost scream for you to use them!

Suppose that *x* + *y* = *a*, *x*^{2} + *y*^{2} = *b*, and *x*^{3}* + y*^{3} = *c*. What is the value of *xy* in terms of *a,* *b,* and *c*?

(F)

(G)

(H)

(J)

(K)

This problem threatens to be a math nightmare. But the right approach turns it into, well, merely a slightly unpleasant dream. Begin with the formula for the sum of two cubes:

*x*^{3}* + y*^{3} = (*x* + *y*)(*x*^{2} – *xy *+ *y*^{2})

Rearrange the terms inside the second set of parentheses:

*x*^{3}* + y*^{3} = (*x* + *y*)(*x*^{2} + *y*^{2 }– *xy*)

Now plug in *a* for *x* + *y,* *b* for *x*^{2} + *y*^{2}, and *c* for *x*^{3}* + y*^{3}:

*c* = *a*(*b* – *xy*)

Now the trick is to isolate* xy:*

This solution doesn’t quite match any of the answers provided, so multiply the numerator and the denominator of the fraction by –1 and rearrange the variabes in the top of the fraction alphabetically:

Thus, the correct answer is Choice (G). This problem is by no means easy, but knowing the right formula makes it solvable using the algebra you already know.

**Solving Everything but the Kitchen Sink**

Although the main ideas behind algebra are simple (as I discuss in the earlier section “Knowing the Two Big Ideas of Algebra”), executing them isn’t always a walk in the park. Each type of ACT question requires a different strategy to find the correct answer quickly. So to help you get a firm grasp on them, in this section I show you how to answer a bunch of different elementary algebra questions.

**Approaching rational equations rationally**

At some point on the ACT, you’ll likely come across a rational equation question. A *rational equation *includes one or more fractions with variables. A good first step with rational equations is to eliminate the fractions, turning the equation into something more friendly. When both sides of an equation are entirely fractions, the quickest way to eliminate the fractions is to *cross-multiply* — multiply the numerator of each fraction by the denominator of the other fraction and set the two products equal.

If , then 7*n* =

(A) 1

(B) 2

(C) 3

(D) 4

(E) 5

To cross-multiply, multiply 21*n* – 6 by 3 and 4*n* + 1 by 7, and then set the two values equal to each other:

3(21*n* – 6) = 7(4*n* + 1)

Now distribute both sides of the equation and solve for *n:*

63*n* – 18 = 28*n* + 7

35*n* – 18 = 7

35*n* = 25

Divide both sides by 5:

7*n* = 5

This equation answers the question that’s being asked, so you don’t have to solve further; the correct answer is Choice (E).

For a rational equation with more than one fraction, you may need stronger medicine. One approach is to find a common denominator for all the terms in the equation and multiply every term by this number, like I show you in the following example.

If , then *x* =

(F)

(G)

(H)

(J)

(K)

To eliminate the three fractions, first find a common denominator — that is, the least common multiple of 2, 4, and 6, which is 12. Multiply each term of the equation by this number:

Now simplify by dividing 12 by the denominator and multiplying the result by the numerator. Do this in two steps so you don’t get confused and make a mistake:

3(3*x*) + 2(7) = 6(5*x*)

9*x* + 14 = 30*x*

At this point, you can solve the equation easily:

Therefore, the correct answer is Choice (H).

**Rooting out ways to solve square root equations**

To solve an equation with a square root (or radical), isolate the root on one side of the equation. Then you can square both sides of the equation to remove the root.

You must square the *entire* expression on each side of the equation. If you square each term individually, you’ll get the wrong answer.

Which of the following values satisfies the equation ?

(A)

(B)

(C)

(D)

(E)

To isolate the square root expression, add 3 to both sides of the equation:

To undo the square root, square both sides, using parentheses to make sure you square the *entire* expression on each side:

Now simplify. On the left side of the equation, the square undoes the square root, leaving you with the expression inside. On the right side, the square means you have to multiply the expression by itself and then FOIL it:

6*x* + 17 = (*x* + 3)(*x* + 3)

6*x* + 17 = *x*^{2} + 6*x* + 9

Simplify further, like so:

17 = *x*^{2 }+ 9

8 = *x*^{2}

Take the square root of both sides and simplify again. Disregard the negative value , because all five answers are positive:

Therefore, the correct answer is Choice (C).

**Gaining absolute confidence with absolute value**

The *absolute value* of a number is its positive value, as I discuss in Chapter 4. Solving algebraic equations with absolute value presents a unique challenge. To unpack the value inside the absolute value bars, split the equation into two separate equations:

The same equation with the bars removed

The same equation with the bars removed *and* the opposite side of the equation negated

Splitting the equation usually results in two solutions rather than just one. In some cases, the question asks you for just one solution. In others, it may ask you to use both solutions to calculate the correct answer, as in the next example.

If |3*x* + 7| = 5*x* – 4, what is the resulting product if you multiply all values of *x* that satisfy the equation?

(A)

(B)

(C)

(D)

(E)

To begin, split the equation |3*x* + 7| = 5*x* – 4 into two separate equations:

3*x* + 7 = 5*x* – 4 3*x* + 7 = –(5*x* – 4)

Solve the first equation:

Now solve the second equation:

Multiply these two solutions:

So the correct answer is Choice (D).

Before splitting an equation, sometimes you may need to do some preliminary work to isolate the absolute value expression on one side of the equation. The following example shows you how.

Which of the following is a solution for the equation ?

(F) 1

(G)

(H)

(J)

(K)

Your first step is to isolate |8*k* + 2| on one side of the equal sign. To do this, multiply both sides of the equation by *k* – 3 and then simplify:

Now you can split the equation:

8*k* + 2 = 3*k* – 9 8*k* + 2 = –(3*k* – 9)

Solve the first equation for *k:*

This isn’t one of the solutions given, so solve the second equation:

Thus, the correct answer is Choice (H).

**Exposing variables in the exponent**

A tricky type of ACT question asks you to solve an equation with one or more variables in the exponent. For example, a question may ask you to solve for *n* in the following equation:

At first glance, this problem appears unsolvable. But with a few tricks, you’ll find that it’s not so bad.

The main thing to know when solving this type of problem is how to express a variety of values as exponents with a common base. You can express whole numbers, fractions, and even square roots as exponents. For example, using a base of 5, Table 5-1 shows the numbers that you can express as exponents:

And here’s one more expression that you may need:

If , then *n* =

(A)

(B)

(C)

(D)

(E)

Your first goal is to express both sides of the equation in terms of a common base — in this case, 5. Replace both and by their equivalents:

On the right side of the equation, multiply the two exponents together to make a single exponent:

This problem still doesn’t look very easy to solve, but in fact you’ve made a great improvement: Because the bases on both sides of the equation are equal, the exponents also are equal. Therefore, you can throw away the bases and keep the exponents:

This equation is now pretty easy to solve for *n.* Begin by multiplying both sides by 2 to eliminate the fraction, and then distribute and solve:

Therefore, the correct answer is (C).

What is the value of *k* if

(F)

(G)

(H)

(J)

(K)

In this case, the numbers and 32 can both be expressed as powers of 2:

Substitute these two values into the equation:

Multiply the exponents on the right side of the equation:

2^{–3k} = 2^{5}

The two bases are equal, so the exponents also are equal; therefore, drop the bases and keep the exponents:

–3*k* = 5

Now this equation is easy to solve:

The correct answer is Choice (K).

**Taking extra care with extra variables**

Generally speaking, when an equation has more than one variable, you can’t solve it. However, some equations on the ACT may include one or more extra variables. In most cases, when you have an equation with extra variables, you must solve the equation *in terms of* the other variables — that is, isolate one variable on one side of the equation.

In the equation 2*pq* + 5*qr* = 3*pr,* what is the value of *p* in terms of *q* and *r*?

(A)

(B)

(C)

(D)

(E)

To answer this question, isolate *p* on one side of the equation. Begin by moving all the *p* terms to one side:

2*pq* + 5*qr* = 3*pr*

5*qr* = 3*pr* – 2*pq*

Now factor out *p* on the right side of the equation:

5*qr* = *p*(3*r* – 2*q*)

Divide both sides by 3*r* – 2*q*:

So the correct answer is Choice (E).

Sometimes when an equation has more than one variable, one variable drops out of the equation (surprise!), allowing you to solve for the remaining variable. Check out the following example to see how.

What is the value of *x* in the equation 3(*x* + 4*y*) = 6(2*y* + 5)?

(F) 10

(G) 14

(H) 17

(J) 23

(K) Cannot be determined from the information given

At first glance, the equation doesn’t look solvable. But don’t be too quick to jump to this conclusion. You can, in fact, solve this problem. Begin by distributing to remove the parentheses:

3(*x* + 4*y*) = 6(2*y* + 5)

3*x* + 12*y* = 12*y* + 30

Now subtract 12*y* from both sides of the equation:

3*x* = 30

Magically, the *y* term has dropped out, leaving you with an equation that you can easily solve:

*x* = 10

The correct answer is Choice (F).

Finally, you can sometimes solve an equation with extra variables for an expression that includes both variables.

If , what is the value of *x* – *y*?

(A) –2

(B) –1

(C) 1

(D) 2

(E) 4

This equation can’t be solved for either *x* or *y.* However, you can isolate the expression *x* – *y* on one side of the equation to answer the question. Begin by multiplying both sides by *y* – 1:

Now distribute on the right side of the equation:

2*x* – 4 = 2*y* – 2

Isolate the *x* and *y* terms on one side of the equation:

2*x* – 2*y* – 4 = –2

2*x* – 2*y* = 2

Factor out a 2 from each term on the left side:

2(*x* – *y*) = 2

Divide both sides by 2:

*x* – *y* = 1

So the correct answer is Choice (C).

**Lost in Translation: Translating Words into Equations**

Some ACT questions give you information about what happens when you perform various operations on an unknown number. To answer these questions, let *x* (or any variable you choose) equal the number you’re looking for. Then build an equation and solve it using any of the methods you already know.

When you divide a certain number by 2 and then add 2, the number you end up with is the same as when you add 24 and then divide this result by 3. What is the number?

(A) 30

(B) 36

(C) 42

(D) 48

(E) 56

To begin, let *x* be the number you’re looking for. The first part tells you to divide *x* by 2 and then add 2:

The next part tells you to add 24 to *x* and then divide the result by 3:

The number you end up with is the same in both cases, so you can link these two expressions with an equal sign:

Now you have an equation that you can solve using any of the tools and tricks you know. Begin by multiplying every term here by the common denominator of 6 and simplify as I show you in the earlier section “Approaching rational equations rationally”:

Now simplify and solve for *x*:

3*x* + 12 = 2*x* + 48

*x* + 12 = 48

*x* = 36

So the correct answer is Choice (B).