If x is an integer that satisfies both |2x + 1| \le 7 and \frac{5 - x}{-2} < -1, how many … — GRE General Test

If $x$ is an integer that satisfies both $|2x + 1| \le 7$ and $\frac{5 - x}{-2} < -1$ , how many possible values of $x$ are there?

There are 7 possible integer values of

x

Solving

|2x + 1| \le 7

yields

-4 \le x \le 3

. Solving

\frac{5 - x}{-2} < -1

requires flipping the inequality symbol when multiplying by

-2

, resulting in

5 - x > 2

, which simplifies to

x < 3

. Taking the intersection of

-4 \le x \le 3

and

x < 3

gives the interval

-4 \le x < 3

. The integers in this range are

-4, -3, -2, -1, 0, 1, 2

, totaling 7 integers.

Solve the absolute value inequality

|2x + 1| \le 7

-7 \le 2x + 1 \le 7 \implies -8 \le 2x \le 6 \implies -4 \le x \le 3

An absolute value inequality of the form

|u| \le k

(for

k \ge 0

) expands to

-k \le u \le k

Solve the rational inequality

\frac{5 - x}{-2} < -1

5 - x > (-1)(-2) \implies 5 - x > 2 \implies -x > -3 \implies x < 3

Multiplying or dividing an inequality by a negative number requires reversing the direction of the inequality sign.

Find the intersection of the two solution sets and count integer solutions.

-4 \le x < 3

. The integer values satisfying this inequality are

-4, -3, -2, -1, 0, 1, 2

, which gives a total of 7 integer values.

The solution must satisfy both conditions simultaneously.

Solving combined linear and absolute value inequalities while correctly reversing inequality signs when multiplying or dividing by negative values.

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