Q. \(3x + 2y \geq 24\).

Q: How do I rewrite 3x + 2y \geq 24 in slope–intercept form?

Subtract 3x and divide by 2: y \geq -\tfrac{3}{2}x + 12 . This shows slope -\tfrac{3}{2} and y-intercept 12..

Q: How do I graph 3x+2y \geq 24?

Graph the boundary line y = -\tfrac{3}{2}x + 12 as a solid line. Test (0,0): 0 \geq 12 is false, so shade the opposite side of the line (the half-plane satisfying the inequality).

Q: What are the x- and y-intercepts?

Set y=0: 3x = 24 so x = 8, intercept (8,0). Set x=0: 2y = 24 so y = 12, intercept (0,12).

Q: Is the boundary line solid or dashed?

Solid, because the inequality is \geq and points on the line (where 3x+2y = 24 ) are included..

Q: Which side of the line is the solution region: above or below?

Since y \geq -\tfrac{3}{2}x + 12 , the solution is the region on or above the line (the half-plane where y -values are greater than or equal to the line).

Q: How do I find integer solutions x,y quickly?

For integer x, require y \geq \tfrac{-3x+24}{2}. Choose integer x and take integer y \geq \left\lceil \tfrac{-3x+24}{2}\right\rceil. Examples: (8,0), (6,3), (0,12), (9,0)..

Q: What is the minimal y for a given real x?.

For real y the minimum satisfying the inequality is y_{\min} = -\tfrac{3}{2}x + 12. For integer y take y \geq \left\lceil -\tfrac{3}{2}x + 12 \right\rceil.

Answer

Subtract \( 3x \): \( 2y \geq 24 – 3x \).

Divide by 2 (positive): \( y \geq \dfrac{24 – 3x}{2} = 12 – \dfrac{3}{2}x \).

Final result: \( y \geq 12 – \dfrac{3}{2}x \)

Detailed Explanation

Write the inequality to solve:\[3x + 2y \ge 24\]
Isolate y by first subtracting 3x from both sides:\[2y \ge 24 – 3x\]Then divide both sides by 2. Because 2 is positive, the direction of the inequality does not change:\[y \ge \frac{24 – 3x}{2}\]
Simplify the right-hand side:

\[y \ge 12 – \frac{3}{2}x\]

It is customary to write the linear expression in slope-intercept form (slope first):

\[y \ge -\tfrac{3}{2}x + 12\]
Identify the boundary line and its properties. The boundary is the equation obtained when the inequality sign is replaced by equality:\[y = -\tfrac{3}{2}x + 12\]Slope and intercepts:
- Slope \(m = -\tfrac{3}{2}\). This means that for every 2 units you move right, the line falls 3 units.
- y-intercept: set \(x=0\). Then \(y = 12\), so the y-intercept is \((0,12)\).
- x-intercept: set \(y=0\). Solve \(0 = -\tfrac{3}{2}x + 12\). Multiply both sides by 2: \(0 = -3x + 24\). Then \(-3x = -24\), so \(x = 8\). Thus the x-intercept is \((8,0)\).
Determine which side of the boundary is part of the solution. Since the inequality is \(\ge\), the boundary line itself is included in the solution set (so draw it as a solid line when graphing).To decide which half-plane to shade, use a test point not on the line. The origin \((0,0)\) is usually convenient. Substitute into the original inequality:\[3(0) + 2(0) \ge 24 \quad\text{gives}\quad 0 \ge 24,\]which is false. Therefore the half-plane containing the origin is not part of the solution. The solution set is the opposite half-plane (the set of points on or above the line).
As an additional check, substitute a point known to be above the line, for example \((0,13)\):

\[3(0) + 2(13) \ge 24 \quad\text{gives}\quad 26 \ge 24,\]

which is true, confirming that points above the line satisfy the inequality.
Final description of the solution set:\[\{(x,y)\in\mathbb{R}^2 \mid y \ge -\tfrac{3}{2}x + 12\}\]On a graph: draw the solid line \(y = -\tfrac{3}{2}x + 12\) through \((0,12)\) and \((8,0)\), then shade the region above that line (the side that does not contain the origin).

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Algebra FAQs

How do I rewrite \(3x + 2y \geq 24\) in slope–intercept form?

Subtract 3x and divide by 2: \( y \geq -\tfrac{3}{2}x + 12 \). This shows slope \(-\tfrac{3}{2}\) and \(y\)-intercept \(12\)..

How do I graph \(3x+2y \geq 24\)?

Graph the boundary line \(y = -\tfrac{3}{2}x + 12\) as a solid line. Test \((0,0)\): \(0 \geq 12\) is false, so shade the opposite side of the line (the half-plane satisfying the inequality).

What are the \(x\)- and \(y\)-intercepts?

Set \(y=0\): \(3x = 24\) so \(x = 8\), intercept \((8,0)\). Set \(x=0\): \(2y = 24\) so \(y = 12\), intercept \((0,12)\).

Is the boundary line solid or dashed?

Solid, because the inequality is \( \geq \) and points on the line (where \( 3x+2y = 24 \)) are included..

Which side of the line is the solution region: above or below?

Since \( y \geq -\tfrac{3}{2}x + 12 \), the solution is the region on or above the line (the half-plane where \( y \)-values are greater than or equal to the line).

Are solutions convex and bounded?

How do I find integer solutions \(x,y\) quickly?

For integer \(x\), require \(y \geq \tfrac{-3x+24}{2}\). Choose integer \(x\) and take integer \(y \geq \left\lceil \tfrac{-3x+24}{2}\right\rceil\). Examples: \((8,0)\), \((6,3)\), \((0,12)\), \((9,0)\)..

What is the minimal y for a given real x?.

For real \(y\) the minimum satisfying the inequality is \(y_{\min} = -\tfrac{3}{2}x + 12\). For integer \(y\) take \(y \geq \left\lceil -\tfrac{3}{2}x + 12 \right\rceil\).

If additionally \(x,y \geq 0\), what is the feasible region?

It is the portion of the half-plane inside the first quadrant: the triangle edge between \(8,0\) and \(0,12\) inclusive, plus the region above that segment inside \(x\geq0,y\geq0\) (unbounded upward/right as allowed)..

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