Q. (3x + 2y geq 24).
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).
Algebra FAQs
How do I rewrite \(3x + 2y \geq 24\) in slope–intercept form?
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Which side of the line is the solution region: above or below?
Are solutions convex and bounded?
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What is the minimal y for a given real x?.
If additionally \(x,y \geq 0\), what is the feasible region?
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