If x and y are real numbers satisfying the inequalities |2x - 7| \le 5 and |3y + 2| < 8, w… — GRE General Test

If $x$ and $y$ are real numbers satisfying the inequalities $|2x - 7| \le 5$ and $|3y + 2| < 8$ , which of the following inequalities must be true? Select all that apply.

$x - y > -1$ Answer
$xy < 12$ Answer
C
$x + y > 0$
D
$|y| < 3$
E
$y - x > -3$

Answer

The inequalities that must be true are

x - y > -1

and

xy < 12

Solving

|2x - 7| \le 5

yields

1 \le x \le 6

, and solving

|3y + 2| < 8

yields

-\frac{10}{3} < y < 2

. Combining

x \ge 1

with

-y > -2

gives

x - y > -1

, which is always true. Furthermore, since

x

is positive and bounded above by

6

while

y

is bounded above by

2

, the product

xy

must be strictly less than

12

Step-by-Step Solution

Solve the absolute value inequality for

x

-5 \le 2x - 7 \le 5 \implies 2 \le 2x \le 12 \implies 1 \le x \le 6

Unfolding the absolute value

|2x - 7| \le 5

gives a compound linear inequality.

Solve the absolute value inequality for

y

-8 < 3y + 2 < 8 \implies -10 < 3y < 6 \implies -\frac{10}{3} < y < 2

Unfolding the absolute value

|3y + 2| < 8

gives a strict compound linear inequality.

Evaluate the inequality

x - y > -1

Since

x \ge 1

and

y < 2 \implies -y > -2

, adding the inequalities gives

x + (-y) > 1 + (-2) = -1

This establishes that

x - y > -1

is always true.

Evaluate the inequality

xy < 12

Since

1 \le x \le 6

(all positive) and

y < 2

, if

y > 0

xy < 6 \cdot 2 = 12

. If

y \le 0

xy \le 0 < 12

In all cases within the domain,

xy < 12

holds strictly.

Test counterexamples for the remaining statements.

For

x + y > 0

x=1, y=-3 \implies x+y=-2 \ngtr 0

. For

|y| < 3

y=-3.2 \implies |-3.2|=3.2 \nless 3

. For

y - x > -3

x=6, y=0 \implies y-x=-6 \ngtr -3

Counterexamples disprove that these remaining statements must be true.

Key Concept

Linear Inequalities and Absolute Value Bounds

Question

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

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