Question

svlc algebra 1a - standard (15260) solving one-variable inequalities what value of x is in the solution set of 3(x - 4) ≥ 5x + 2? -5 10 -10 5

Explanation:

Step1: Expand the left side

First, we expand \(3(x - 4)\) using the distributive property \(a(b - c)=ab - ac\). So, \(3(x - 4)=3x-12\). The inequality becomes \(3x - 12\geq5x + 2\).

Step2: Move variable terms to one side

Subtract \(3x\) from both sides of the inequality to get the \(x\) terms on one side. We have \(3x-3x - 12\geq5x-3x + 2\), which simplifies to \(-12\geq2x + 2\).

Step3: Move constant terms to the other side

Subtract 2 from both sides: \(-12 - 2\geq2x+2 - 2\), so \(-14\geq2x\).

Step4: Solve for \(x\)

Divide both sides by 2: \(\frac{-14}{2}\geq\frac{2x}{2}\), which gives \(-7\geq x\) or \(x\leq - 7\). Now we check the given options:

• For \(x=-5\): \(-5>-7\), so it does not satisfy \(x\leq - 7\).
• For \(x = 10\): \(10>-7\), does not satisfy.
• For \(x=-10\): \(-10\leq - 7\), satisfies the inequality.
• For \(x = 5\): \(5>-7\), does not satisfy.

Answer:

\(-10\) (the option with \(-10\))