Question

solve the following inequality for x. y ≤ -ax + 5 a x ≥ -\frac{1}{a}y - 5 b x ≤ -\frac{1}{a}y - 5 c x ≤ \frac{5 - y}{a} d x ≥ \frac{5 - y}{a}

Explanation:

Step1: Start with the inequality

We have the inequality \( y \leq -ax + 5 \). Our goal is to solve for \( x \). First, we'll isolate the term with \( x \) by subtracting 5 from both sides.
\( y - 5 \leq -ax \)

Step2: Divide both sides by \(-a\)

When we divide or multiply an inequality by a negative number, the direction of the inequality sign flips. So, dividing both sides by \(-a\) (assuming \( a
eq 0 \)):
\( \frac{y - 5}{-a} \geq x \)
Which can be rewritten as:
\( x \leq \frac{5 - y}{a} \) (by multiplying numerator and denominator by -1, which also flips the inequality sign back? Wait, no, let's re - examine.

Wait, starting over:

Given \( y \leq -ax + 5 \)

Subtract 5 from both sides: \( y - 5 \leq -ax \)

Now, divide both sides by \(-a\). If \( a>0 \), then \(-a < 0\), so the inequality sign flips:

\( \frac{y - 5}{-a}\geq x \)

Simplify \( \frac{y - 5}{-a}=\frac{-(5 - y)}{-a}=\frac{5 - y}{a} \) when \( a>0 \)? Wait, no:

\( \frac{y - 5}{-a}=\frac{-(5 - y)}{-a}=\frac{5 - y}{a} \) only if we cancel the negatives. But the inequality after dividing by \(-a\) (negative) is \( \frac{y - 5}{-a}\geq x \), which is equivalent to \( x\leq\frac{5 - y}{a} \) (because \( \frac{y - 5}{-a}=\frac{5 - y}{a} \) when we multiply numerator and denominator by - 1: \( \frac{y - 5}{-a}=\frac{-(5 - y)}{-a}=\frac{5 - y}{a} \))

Wait, let's do it step by step:

\( y\leq - ax + 5 \)

Subtract 5: \( y - 5\leq - ax \)

Multiply both sides by - 1 (and flip the inequality sign): \( 5 - y\geq ax \)

Now divide both sides by \( a \) (assuming \( a>0 \), if \( a < 0 \), the inequality sign would flip again, but since the answer choices have \( \frac{5 - y}{a} \) or similar, we'll proceed with the general case where we can get to the form in the options.

From \( 5 - y\geq ax \), divide both sides by \( a \) (assuming \( a>0 \), the inequality sign remains as is):

\( \frac{5 - y}{a}\geq x \)

Which is the same as \( x\leq\frac{5 - y}{a} \)

Looking at the options, option C is \( x\leq\frac{5 - y}{a} \) (wait, the options:

Option A: \( x\geq-\frac{1}{a}y - 5 \)

Option B: \( x\leq-\frac{1}{a}y - 5 \)

Option C: \( x\leq\frac{5 - y}{a} \)

Option D: \( x\geq\frac{5 - y}{a} \)

Let's re - express our result. From \( y\leq - ax + 5 \)

\( - ax\geq y - 5 \)

Multiply both sides by \( -\frac{1}{a} \). If \( a>0 \), then \( -\frac{1}{a}<0 \), so the inequality sign flips:

\( x\leq\frac{5 - y}{a} \) (because \( - ax\geq y - 5\Rightarrow x\leq\frac{y - 5}{-a}=\frac{5 - y}{a} \))

Yes, so the correct inequality is \( x\leq\frac{5 - y}{a} \), which is option C.

Answer:

C. \( x\leq\frac{5 - y}{a} \)