Question

question
select the correct answer.
what is the solution of the given inequality?
$-4(2x + 1) \leq 3(x - 5)$
options (partially visible):
x ≥ 1
x ≥ 3
x ≥ 7
x ≥ -5

Explanation:

Step1: Expand both sides

First, we expand the left - hand side and the right - hand side of the inequality \(- 4(2x + 1)\leq3(x - 5)\).
Using the distributive property \(a(b + c)=ab+ac\), the left - hand side: \(-4(2x + 1)=-4\times2x+(-4)\times1=-8x - 4\)
The right - hand side: \(3(x - 5)=3x-15\)
So the inequality becomes \(-8x - 4\leq3x-15\)

Step2: Move the terms with x to one side

We add \(8x\) to both sides of the inequality to get all the \(x\) terms on the right - hand side.
\(-8x - 4+8x\leq3x-15 + 8x\)
Simplifying, we have \(-4\leq11x-15\)

Step3: Move the constant terms to the other side

We add 15 to both sides of the inequality:
\(-4 + 15\leq11x-15+15\)
Simplifying, \(11\leq11x\)

Step4: Solve for x

We divide both sides of the inequality by 11:
\(\frac{11}{11}\leq\frac{11x}{11}\)
Which simplifies to \(1\leq x\) or \(x\geq1\)

Answer:

\(x\geq1\) (assuming the first option is \(x\geq1\))