## SAT Test Prep

**CHAPTER 6**

WHAT THE SAT MATH IS REALLY TESTING

WHAT THE SAT MATH IS REALLY TESTING

**Lesson 4: Simplifying Problems**

**Beeline, Substitute, Combine, and Cancel**

When a problem seems overwhelming, try one of these four simplification strategies: *beelining, substituting, combining*, and *canceling*.

**Look for the Beeline—The Direct Route**

Many SAT problems have “beelines”—direct paths from the given information to the answer. We sometimes miss the “beeline” because we get trapped in a knee-jerk response—for instance, automatically solving every equation or using the Pythagorean theorem on every right triangle. Avoid the knee-jerk response. Instead, step back and look for the “beeline.”

If and , what is the value of ?

This problem looks tough because of all the unknowns. You might do the knee-jerk thing and try to solve for *a, b*, and *c*. Whoa, there! Step back. The question doesn’t ask for *a, b*, and *c*. It asks for a fraction that you can get much more directly. Notice that just multiplying the two given fractions gets you almost there: . This is close to what you want—all you have to do is divide by 3 to get Substituting the given values of the fractions gives you which is the value of

**Simplify by Substituting**

The simplest rule in algebra is also the most powerful: *Anything can be substituted for its equal*. When you notice a complicated expression on the SAT, just notice if it equals something simpler, and substitute!

If and , then what is the value of

Again, take a deep breath. Both the equation and the fraction look complicated, but you can simplify by 16 2 3 5 just remembering that *anything can be substituted for its equal*. Notice that appears in both the equation and the fraction. What does it equal? Subtract *y* from both sides of the equation to get . If you substitute for in the fraction, you get . Nice!

**Simplify by Combining or Canceling**

Many algebraic expressions can be simplified by combining or canceling terms. Always keep your eye out for *like terms* that can be combined or canceled and for *common factors in fractions* that can be canceled.

If *m* and *n* are positive integers such that and what is the value of ?

To simplify this one, it helps to know a basic factoring formula from __Chapter 8__, Lesson 5: . If you factor the numerator and denominator of the fraction, a common factor reveals itself, and it can be canceled: Since must equal 9.

If and , then for how many values of *x* does ?

(A) 0

(B) 1

(C) 2

(D) 3

(E) 4

Remember the simple rule that *anything can be substituted for its equal*, and then cancel to simplify. Since , you can say that .

So the answer is (A) 0, right? Wrong! Remember that the question asks *for how many values of x* are the function values equal. Since we only got one solution for *x*, the answer is (B)1.

**Concept Review 4: Simplifying Problems**

Simplify the following expressions.

Solve the following problems with substitution.

__6.__ If and , then what is the value of *x*?

__7.__ If , what is the value of

__8.__ For all real numbers . What is the value of ?

**SAT Practice 4: Simplifying Problems**

**1**__.__ If and , then

(A) 12

(B) 6

(C) 4

(D) 0

(E) –4

**2**__.__ If , , and , then

(A) 4

(B) 2

(C) 1

(D) 0

(E) –1

**3**__.__ If and , what is the value of ?

(D) 2

(E) 8

**Questions 4 and 5** pertain to the following definition:

For all non-zero real numbers *k*, let

**4**__.__ Which of the following is equivalent to ?

(C) ^1

(E) ^2

**5**__.__ Other than 0, what is the only value of *k* for which ^^*k* is undefined?

**Answer Key 4: Simplifying Problems**

**Concept Review 4**

__6.__ Substitute into to get

**SAT Practice 4**

__3.__ **D** Start by simplifying the expressions:

Substituting into the original equations gives

__4.__ **D** Begin by simplifying by substitution: But be careful not to pick (A) ^1/6 because

Notice that the choices must be evaluated first before we can see which one equals 1/6. Notice that choice (D)

__5.__ **1** Begin by simplifying ^^*k* by substitution:

Yikes! That doesn’t look simple! But think about it: why is it that *k* can’t be 0? Because division by 0 is undefined, and *k* is in a denominator. But notice that is also in a denominator, so can’t be 0, either!

Then check by noticing that ^^1 is undefined:

, and 1/0 is undefined.