How many integer values of y satisfy the inequality |2y - 7| \le |y + 2|? — GMAT

How many integer values of $y$ satisfy the inequality $|2y - 7| \le |y + 2|$ ?

Cevap: 8

The total number of integer values of

y

satisfying the inequality is 8.

Squaring both sides of

|2y - 7| \le |y + 2|

gives

3y^2 - 32y + 45 \le 0

, which factors into

(3y - 5)(y - 9) \le 0

. The solution range for

y

\frac{5}{3} \le y \le 9

. Since

\frac{5}{3} \approx 1.67

, the integer values of

y

satisfying this range are

2, 3, 4, 5, 6, 7, 8, 9

, giving a total of 8 integers.

Square both sides of the inequality

|2y - 7| \le |y + 2|

(2y - 7)^2 \le (y + 2)^2

Since both sides of an absolute value expression are non-negative, squaring both sides maintains the inequality direction.

Expand terms and move all terms to the left side

3y^2 - 32y + 45 \le 0

Expanding gives

4y^2 - 28y + 49 \le y^2 + 4y + 4

. Subtracting

(y^2 + 4y + 4)

from both sides produces the standard quadratic inequality.

Factor the quadratic expression to find critical points

(3y - 5)(y - 9) \le 0

, yielding

\frac{5}{3} \le y \le 9

The roots are

y = \frac{5}{3}

and

y = 9

. A quadratic with a positive leading coefficient is non-positive between its roots.

Determine all integers within the range

\left[\frac{5}{3}, 9\right]

2, 3, 4, 5, 6, 7, 8, 9

(8 integers total)

Because

\frac{5}{3} \approx 1.67

, the smallest integer within the range is 2 and the largest is 9.

Solving absolute value inequalities of the form

|A| \le |B|

by squaring both sides and determining integer solutions.

Soru