Question

1. solve the compound inequality: $2a + 10 \leq 6$ or $3 + 3a > 3$

a $a > 1$
b $a < -9$
c $a \leq -2$ or $a > 0$
d all real numbers

Explanation:

Step1: Solve \(2a + 10 \leq 6\)

Subtract 10 from both sides: \(2a + 10 - 10 \leq 6 - 10\)
Simplify: \(2a \leq -4\)
Divide by 2: \(\frac{2a}{2} \leq \frac{-4}{2}\)
Result: \(a \leq -2\)

Step2: Solve \(3 + 3a > 3\)

Subtract 3 from both sides: \(3 + 3a - 3 > 3 - 3\)
Simplify: \(3a > 0\)
Divide by 3: \(\frac{3a}{3} > \frac{0}{3}\)
Result: \(a > 0\)

Step3: Combine solutions (OR)

The compound inequality is "or", so we combine \(a \leq -2\) and \(a > 0\). This means all real numbers because any real number is either \(\leq -2\), \(> 0\), or between them (but "or" includes all cases where either is true, and together they cover all real numbers? Wait, no—wait, \(a \leq -2\) or \(a > 0\) does not cover between -2 and 0. Wait, no, I made a mistake. Wait, no—wait, let's re - check.

Wait, step 1: \(2a+10\leq6\) gives \(a\leq - 2\)

Step 2: \(3 + 3a>3\) gives \(a > 0\)

But the "or" means the solution is all \(a\) such that \(a\leq - 2\) or \(a > 0\)? Wait, no, that can't be. Wait, no, wait the options: option D is all real numbers. Wait, maybe I made a mistake in solving.

Wait, re - solve \(3 + 3a>3\):

\(3+3a>3\)

Subtract 3: \(3a>0\)

Divide by 3: \(a > 0\)

And \(2a + 10\leq6\):

\(2a\leq6 - 10=-4\)

\(a\leq - 2\)

Now, the compound inequality is "or", so the solution is the union of \(a\leq - 2\) and \(a > 0\). But that's not all real numbers. Wait, but the options have D as all real numbers. Wait, maybe I messed up the solving.

Wait, wait, let's check the options again. The options are:

A. \(a>1\)

B. \(a < - 9\)

C. \(a\leq - 2\) or \(a > 0\)

D. all real numbers

Wait, when we solve \(2a + 10\leq6\), we get \(a\leq - 2\)

When we solve \(3+3a>3\), we get \(a>0\)

But the "or" in the compound inequality means that a number is a solution if it satisfies either inequality. Now, let's check if there is any real number that doesn't satisfy either. Take a number between - 2 and 0, say \(a=-1\). Does it satisfy \(2a + 10\leq6\)? \(2*(-1)+10 = 8\), which is not \(\leq6\). Does it satisfy \(3 + 3a>3\)? \(3+3*(-1)=0\), which is not \(>3\). Wait, so \(a = - 1\) does not satisfy either. So my previous conclusion was wrong. But the options have D as all real numbers. There must be a mistake in my solving.

Wait, wait, re - solve \(3 + 3a>3\):

\(3+3a>3\)

Subtract 3: \(3a>0\)

Divide by 3: \(a>0\)

And \(2a + 10\leq6\):

\(2a\leq - 4\)

\(a\leq - 2\)

But the options: option C is \(a\leq - 2\) or \(a > 0\), option D is all real numbers. Wait, maybe the original problem was written incorrectly? Or maybe I misread the problem.

Wait, the problem is \(2a + 10\leq6\) or \(3 + 3a>3\)

Wait, let's check the options again. Option C is \(a\leq - 2\) or \(a > 0\), option D is all real numbers.

Wait, maybe I made a mistake in the first inequality. Let's re - solve \(2a+10\leq6\):

\(2a\leq6 - 10=-4\)

\(a\leq - 2\) (correct)

Second inequality: \(3 + 3a>3\)

\(3a>0\)

\(a > 0\) (correct)

So the solution is \(a\leq - 2\) or \(a > 0\), which is option C? But that's not all real numbers. Wait, the options given:

A. \(a>1\)

B. \(a < - 9\)

C. \(a\leq - 2\) or \(a > 0\)

D. all real numbers

Wait, maybe there is a typo in the problem. Or maybe I misread the coefficients. Wait, the first inequality: \(2a + 10\leq6\)

If it was \(2a+1\leq6\), then \(2a\leq5\), \(a\leq2.5\), but no. Wait, the original problem is \(2a + 10\leq6\) or \(3 + 3a>3\)

Wait, let's check with \(a=-1\): does it satisfy either?

For \(2a + 10\): \(2*(-1)+10 = 8\), \(8\leq6\)? No.

For \(3 + 3a\): \(3+3*(-1)=0\), \(0>3\)? No. So \(a = - 1\) is not a solution. So the solution is \(a\leq - 2\) or \(a > 0\), which is option C. But the user's options have D as all real numbers. Maybe there is a mistake in the problem or in my solving.

Wait, wait, maybe the second inequality is \(3+3a\geq3\)? If it was \(\geq\), then \(3a\geq0\), \(a\geq0\), and then \(a\leq - 2\) or \(a\geq0\), which still doesn't cover between - 2 and 0.

Alternatively, maybe the first inequality is \(2a + 1\leq6\), then \(2a\leq5\), \(a\leq2.5\), and the second is \(3 + 3a>3\) ( \(a > 0\) ), then the solution would be \(a\leq2.5\) or \(a > 0\), which is all real numbers (since \(a > 0\) is part of \(a\leq2.5\) or \(a > 0\), and together they cover all real numbers). But with the given problem, the solution is \(a\leq - 2\) or \(a > 0\), which is option C. But the options have D as all real numbers. There must be a mistake.

Wait, maybe the first inequality is \(2a+10\leq6a\)? No, the problem says \(2a + 10\leq6\) or \(3 + 3a>3\)

Wait, let's check the options again. The options are:

A. \(a>1\)

B. \(a < - 9\)

C. \(a\leq - 2\) or \(a > 0\)

D. all real numbers

Given that when we solve the two inequalities, we get \(a\leq - 2\) or \(a > 0\), which is option C. But maybe the problem was supposed to be \(2a + 1\leq6\) or \(3 + 3a>3\), but as per the given problem, the solution is option C. But wait, let's check with \(a = 1\):

For \(2a + 10\): \(2*1+10 = 12\), \(12\leq6\)? No.

For \(3 + 3a\): \(3+3*1 = 6\), \(6>3\)? Yes. So \(a = 1\) is a solution (since it satisfies the second inequality).

For \(a=-3\): \(2*(-3)+10 = 4\), \(4\leq6\)? Yes. So \(a=-3\) is a solution.

For \(a = 0.5\): \(2*0.5+10 = 11\), \(11\leq6\)? No. \(3+3*0.5 = 4.5\), \(4.5>3\)? Yes. So \(a = 0.5\) is a solution.

For \(a=-1\): \(2*(-1)+10 = 8\), \(8\leq6\)? No. \(3+3*(-1)=0\), \(0>3\)? No. So \(a=-1\) is not a solution. So the solution is \(a\leq - 2\) or \(a > 0\), which is option C.

But the user's options have D as all real numbers. There is a contradiction. Wait, maybe I made a mistake in solving \(3 + 3a>3\). Let's re - solve:

\(3+3a>3\)

Subtract 3: \(3a>0\)

Divide by 3: \(a > 0\) (correct)

First inequality: \(2a+10\leq6\)

\(2a\leq - 4\)

\(a\leq - 2\) (correct)

So the solution is \(a\leq - 2\) or \(a > 0\), which is option C.

Answer:

C. \(a \leq -2\) or \(a > 0\)