Question

1. solve the inequality:

$5 + 2(-3x + 2) \geq -x - 11$
what is the solution set for $x$?
$\square \\{x \mid x \leq 4\\}$
$\square \\{x \mid x \geq 1.25\\}$
$\square \\{x \mid x \geq 4\\}$
$\square \\{x \mid x \leq 1.25\\}$

Explanation:

Step1: Expand the left - hand side

First, we use the distributive property \(a(b + c)=ab+ac\) to expand \(2(-3x + 2)\).
\(2(-3x+2)=2\times(-3x)+2\times2=-6x + 4\)
So the original inequality \(5 + 2(-3x + 2)\geq -x-11\) becomes \(5-6x + 4\geq -x-11\).

Step2: Combine like terms on the left - hand side

Combine the constant terms \(5\) and \(4\) on the left - hand side: \(5 + 4-6x\geq -x-11\), which simplifies to \(9-6x\geq -x-11\).

Step3: Move the terms with \(x\) to one side and constants to the other side

Add \(6x\) to both sides of the inequality to get rid of the \(-6x\) on the left - hand side:
\(9-6x + 6x\geq -x-11 + 6x\)
Simplifying both sides, we have \(9\geq5x-11\).
Then add \(11\) to both sides: \(9 + 11\geq5x-11 + 11\)
Which gives \(20\geq5x\).

Step4: Solve for \(x\)

Divide both sides of the inequality \(20\geq5x\) by \(5\): \(\frac{20}{5}\geq\frac{5x}{5}\)
Simplifying, we get \(4\geq x\), or \(x\leq4\).

Answer:

\(\{x\mid x\leq4\}\)