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solve the inequality for x. $3(x + 6) + 6 \\leq 6x - 3$ $\\bigcirc\\ x …

Question

solve the inequality for x.
$3(x + 6) + 6 \leq 6x - 3$
$\bigcirc\\ x \leq 5$
$\bigcirc\\ x \geq 9$
$\bigcirc\\ x \leq 9$
$\bigcirc\\ x \geq 5$

Explanation:

Step1: Expand the left side

First, we expand \(3(x + 6)\) using the distributive property \(a(b + c)=ab+ac\). So \(3(x + 6)=3x+18\). Then the left side of the inequality \(3(x + 6)+6\) becomes \(3x + 18+6\). Combining like terms, we get \(3x+24\). So the inequality is now \(3x + 24\leq6x-3\).

Step2: Move all x terms to one side and constants to the other

Subtract \(3x\) from both sides: \(3x - 3x+24\leq6x-3x - 3\), which simplifies to \(24\leq3x-3\). Then add 3 to both sides: \(24 + 3\leq3x-3 + 3\), so \(27\leq3x\).

Step3: Solve for x

Divide both sides by 3: \(\frac{27}{3}\leq\frac{3x}{3}\), which gives \(9\leq x\) or \(x\geq9\). Wait, no, wait, let's check again. Wait, when we had \(3x + 24\leq6x-3\), subtract \(3x\) from both sides: \(24\leq3x - 3\), then add 3: \(27\leq3x\), divide by 3: \(9\leq x\), so \(x\geq9\)? Wait, no, wait, maybe I made a mistake. Wait, original inequality: \(3(x + 6)+6\leq6x - 3\). Let's re - do it.

First, expand left side: \(3x+18 + 6=3x + 24\). So \(3x+24\leq6x-3\). Subtract \(3x\) from both sides: \(24\leq3x-3\). Add 3 to both sides: \(27\leq3x\). Divide by 3: \(9\leq x\), so \(x\geq9\). Wait, but let's test with \(x = 9\): left side \(3(9 + 6)+6=3\times15+6 = 45 + 6=51\), right side \(6\times9-3=54 - 3 = 51\), so \(51\leq51\) which is true. Test \(x = 10\): left side \(3(10 + 6)+6=3\times16+6 = 48+6 = 54\), right side \(6\times10-3=60 - 3 = 57\), \(54\leq57\) which is true. Test \(x = 8\): left side \(3(8 + 6)+6=3\times14+6 = 42+6 = 48\), right side \(6\times8-3=48 - 3 = 45\), \(48\leq45\) is false. So \(x\geq9\) is correct. Wait, but let's check the steps again.

Wait, maybe I messed up the direction. Let's start over.

Original inequality: \(3(x + 6)+6\leq6x-3\)

Step 1: Expand left - hand side: \(3x+18 + 6=3x + 24\). So inequality is \(3x + 24\leq6x-3\)

Step 2: Subtract \(3x\) from both sides: \(24\leq3x-3\)

Step 3: Add 3 to both sides: \(27\leq3x\)

Step 4: Divide both sides by 3: \(9\leq x\) or \(x\geq9\)

Answer:

B. \(x\geq9\)