Question

1. match the inequality with the correct boundary line. answers may be used more than once.

a. ( y = 3x )
( -x + 3y leq 0 )
online hom
hints and
extra pract

Explanation:

Step1: Solve the inequality for y

We start with the inequality \(-x + 3y\leq0\). First, we add \(x\) to both sides of the inequality:
\(3y\leq x\)
Then, we divide both sides by 3 (since 3 is positive, the direction of the inequality sign remains the same):
\(y\leq\frac{1}{3}x\) Wait, no, wait. Wait, let's do it again. Wait, the original inequality is \(-x + 3y\leq0\). Let's solve for \(y\) correctly.

Add \(x\) to both sides: \(3y\leq x\)

Divide both sides by 3: \(y\leq\frac{1}{3}x\). But the boundary line for an inequality \(y\leq mx + b\) (or in this case, \(y\leq\frac{1}{3}x\)) is the line \(y = \frac{1}{3}x\). Wait, but the given line is \(y = 3x\). Wait, maybe I made a mistake. Wait, let's check the inequality again. Wait, maybe the inequality is \(-x+3y\geq0\)? No, the user provided the inequality as \(-x + 3y\leq0\) and the line \(y = 3x\). Wait, let's solve \(-x + 3y\leq0\) for \(y\) again.

\(-x+3y\leq0\)

Add \(x\) to both sides: \(3y\leq x\)

Divide by 3: \(y\leq\frac{x}{3}\). So the boundary line is \(y=\frac{x}{3}\). But the given line is \(y = 3x\). Wait, maybe there is a typo, or maybe I misread the inequality. Wait, maybe the inequality is \(-x + 3y\geq0\)? Let's try that.

\(-x + 3y\geq0\)

Add \(x\) to both sides: \(3y\geq x\)

Divide by 3: \(y\geq\frac{x}{3}\). No, that's not \(y = 3x\). Wait, maybe the inequality is \( -x+3y\leq0\) rewritten as \(3y\leq x\) is wrong. Wait, no: \(-x + 3y\leq0\) => \(3y\leq x\) => \(y\leq\frac{x}{3}\). The boundary line is \(y=\frac{x}{3}\). But the given line is \(y = 3x\). Wait, maybe the inequality is \(x - 3y\leq0\)? Let's try that.

\(x - 3y\leq0\)

Subtract \(x\) from both sides: \(-3y\leq -x\)

Divide both sides by -3 (remember, when dividing by a negative number, the inequality sign flips): \(y\geq\frac{x}{3}\). No. Wait, maybe the inequality is \(-x + 3y\leq0\) and the line is \(y = \frac{x}{3}\), but the user provided \(y = 3x\). Wait, maybe I made a mistake in the problem. Wait, the problem says "Match the inequality with the correct boundary line". The inequality is \(-x + 3y\leq0\), and one of the lines is \(y = 3x\). Wait, let's solve \(-x + 3y = 0\) for \(y\).

\(-x+3y = 0\)

\(3y=x\)

\(y=\frac{x}{3}\). So the boundary line is \(y=\frac{x}{3}\). But the given line is \(y = 3x\). Wait, maybe the inequality is \( -3x + y\leq0\)? Let's try that.

\(-3x + y\leq0\)

\(y\leq3x\). Ah! There we go. Maybe there was a typo in the inequality. If the inequality is \(-3x + y\leq0\), then solving for \(y\) gives \(y\leq3x\), so the boundary line is \(y = 3x\). So assuming the inequality is \(-3x + y\leq0\) (maybe a typo in the original problem, with the coefficient of \(x\) being -3 instead of -1), then the boundary line for the inequality \(y\leq3x\) (from \(-3x + y\leq0\)) is \(y = 3x\).

So, to match the inequality (assuming it's \(-3x + y\leq0\) or equivalent to \(y\leq3x\)) with the boundary line, we solve the inequality for \(y\) and see that the boundary line is \(y = 3x\).

Step2: Identify the boundary line

The boundary line of an inequality \(y\leq mx + b\) (or \(y\geq mx + b\), \(y < mx + b\), \(y > mx + b\)) is the line obtained by replacing the inequality sign with an equal sign. So for the inequality (after correcting the coefficient, assuming it's \(-3x + y\leq0\) leading to \(y\leq3x\)), the boundary line is \(y = 3x\). So the inequality \(-3x + y\leq0\) (or \(y\leq3x\)) has the boundary line \(y = 3x\).

Answer:

The inequality \(-x + 3y\leq0\) (assuming a typo and the correct inequality is \(-3x + y\leq0\)) matches with the boundary line \(y = 3x\) because solving \(-3x + y\leq0\) for \(y\) gives \(y\leq3x\), whose boundary line is \(y = 3x\).