Question

linear inequalities with decimals
question
solve the following inequality:
-8x ≥ 1 + 10(1.2x + 2.9) + 10x
answer
attempt 1 out of 2

Explanation:

Step1: Expand the right side

First, we expand \(10(1.2x + 2.9)\) using the distributive property \(a(b + c)=ab+ac\). So, \(10\times1.2x=12x\) and \(10\times2.9 = 29\). The inequality becomes \(-8x\geq1 + 12x+29 + 10x\).

Step2: Combine like terms on the right

Combine the \(x\) terms and the constant terms on the right. The \(x\) terms: \(12x+10x = 22x\). The constant terms: \(1 + 29=30\). So now the inequality is \(-8x\geq22x + 30\).

Step3: Move all x terms to the left

Subtract \(22x\) from both sides to get all \(x\) terms on the left. \(-8x-22x\geq22x + 30-22x\), which simplifies to \(-30x\geq30\).

Step4: Solve for x

Divide both sides by \(-30\). Remember that when we divide or multiply an inequality by a negative number, we reverse the inequality sign. So, \(\frac{-30x}{-30}\leq\frac{30}{-30}\), which simplifies to \(x\leq - 1\).

Answer:

\(x\leq - 1\)