For a constant k, the inequality 2(3 - x) - k(x + 5) > 4 has the solution set x < -3. What… — SAT

For a constant $k$ , the inequality $2(3 - x) - k(x + 5) > 4$ has the solution set $x < -3$ . What is the value of $k$ ?

The correct answer is

4

. Expanding the inequality

2(3 - x) - k(x + 5) > 4

gives

6 - 2x - kx - 5k > 4

. Grouping the

x

terms yields

(-2 - k)x > 5k - 2

. Because the given solution set is

x < -3

, the inequality sign must flip, indicating that the coefficient

-2 - k

is negative. Dividing by this negative coefficient yields

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

. Setting the boundary expression equal to the boundary of the solution set gives

\frac{5k - 2}{-2 - k} = -3

. Multiplying both sides by

-2 - k

results in

5k - 2 = 6 + 3k

. Subtracting

3k

and adding

2

to both sides gives

2k = 8

, which simplifies to

k = 4

Distribute the constants in the inequality

6 - 2x - kx - 5k > 4

To clear parentheses and prepare to group terms.

Group the

x

terms and constant terms

(-2 - k)x > 5k - 2

To isolate the variable

x

on one side of the inequality.

Divide by the coefficient of

x

and flip the inequality direction

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

Since the solution set is

x < -3

, the direction of the inequality must flip from

>

<

, meaning the coefficient

-2 - k

must be negative.

Set the boundary value equal to

-3

and solve for

k

k = 4

The boundary of the solution set must be equal to

-3

. Solving

\frac{5k - 2}{-2 - k} = -3

yields

5k - 2 = 6 + 3k

, which simplifies to

2k = 8

, so

k = 4

Solving linear inequalities in one variable with symbolic coefficients, accounting for direction flips when dividing by negative quantities.

Question