High School

Solve for [tex]$x$[/tex].

[tex]-6x + 11 \leq -9x + 2[/tex]

A. [tex]x \leq -3[/tex]
B. [tex]x \geq -\frac{13}{15}[/tex]
C. [tex]x \geq 3[/tex]
D. [tex]x \leq \frac{13}{15}[/tex]

Answer :

Certainly! Let's solve the inequality step by step:

We want to solve the inequality:
[tex]\[ -6x + 11 \leq -9x + 2 \][/tex]

Step 1: Move the terms involving [tex]\( x \)[/tex] to one side of the inequality.

To do this, we'll add [tex]\( 9x \)[/tex] to both sides:
[tex]\[ -6x + 9x + 11 \leq 2 \][/tex]

Step 2: Combine the like terms.

On the left side, combine [tex]\( -6x + 9x \)[/tex]:
[tex]\[ 3x + 11 \leq 2 \][/tex]

Step 3: Isolate the term with [tex]\( x \)[/tex] by moving the constant on the left side to the right side.

Subtract 11 from both sides:
[tex]\[ 3x \leq 2 - 11 \][/tex]

Step 4: Simplify the right side.

Calculate [tex]\( 2 - 11 \)[/tex]:
[tex]\[ 3x \leq -9 \][/tex]

Step 5: Solve for [tex]\( x \)[/tex] by dividing both sides by 3.

Divide both sides of the inequality by 3:
[tex]\[ x \leq \frac{-9}{3} \][/tex]

Simplifying the division gives:
[tex]\[ x \leq -3 \][/tex]

Therefore, the solution to the inequality is [tex]\( x \leq -3 \)[/tex].

The correct answer is:
[tex]\[ \text{A. } x \leq -3 \][/tex]