Question

which shaded region in the diagram is defined by the linear inequality (2x - 5y < 10)?

Explanation:

Step1: Rewrite inequality to slope-intercept

Rearrange \(2x-5y<10\) to solve for \(y\):
\[ \begin{align*} -5y&< -2x + 10\\ y&> \frac{2}{5}x - 2 \end{align*} \]
(Note: Inequality flips when dividing by negative -5)

Step2: Identify line style

The inequality uses \(<\) (strict inequality), so the boundary line \(2x-5y=10\) must be dashed. This eliminates the first and third regions (solid lines).

Step3: Test a point to find region

Use the origin \((0,0)\) to test the inequality \(2(0)-5(0)<10\), which simplifies to \(0<10\) (true). Check which dashed-line region includes \((0,0)\):
- For \(y>\frac{2}{5}x - 2\), the origin satisfies this, so the shaded region is above the line? No, correction: Wait, original inequality \(2x-5y<10\) can be tested directly: substitute (0,0): 0 - 0 <10 is true. Now check which dashed region includes (0,0): the second region (dashed line, shaded left/below) includes (0,0), while the fourth (dashed line, shaded right/above) does not. Wait, no: when we rewrite to \(y > \frac{2}{5}x -2\), (0,0) gives 0 > -2, which is true, so the region is above the line? No, wait the line \(2x-5y=10\) has x-intercept (5,0) and y-intercept (0,-2). The region \(2x-5y<10\) is equivalent to all points where the value of \(2x-5y\) is less than 10. For a point below the line (like (0,-3)), \(2(0)-5(-3)=15>10\) (false). For (0,0), 0<10 (true), which is above the line? Wait no, the line goes from (5,0) down to (0,-2). So (0,0) is above the line, and \(2x-5y<10\) is true there. But wait the second region is shaded below the dashed line, which would be points like (0,-3) which gives 15>10, which violates the inequality. Wait I made a mistake in rearrangement:
\(2x -5y <10\)
\(-5y < -2x +10\)
Divide by -5 (flip inequality): \(y > \frac{2}{5}x -2\). So the region is **above** the line. But the fourth region is shaded above the dashed line? No, looking at the diagrams:
First: solid line, shaded below
Second: dashed line, shaded below
Third: solid line, shaded above
Fourth: dashed line, shaded above
Wait testing (0,0) in \(y > \frac{2}{5}x -2\): 0 > -2, which is true, so (0,0) is in the solution region, which is above the line. But wait \(2x-5y <10\) at (0,0) is 0<10, true. At (6,0): 12-0=12>10, which is false, so (6,0) is not in the region. The line has x-intercept (5,0), so (6,0) is to the right of the line, which is above the line? No, the line goes from (5,0) to (0,-2), so to the right of (5,0) is below the line? Wait no, slope is 2/5, so as x increases, y increases. So (6,0): y=0, the line at x=6 has y=(2/5)(6)-2=12/5-2=2/5=0.4, so (6,0) is below the line (0 < 0.4). And \(2(6)-5(0)=12>10\), which violates the inequality, so points below the line do not satisfy the inequality. Points above the line, like (0,0): line at x=0 has y=-2, 0 > -2, and 0<10, which is true. So the solution region is above the dashed line, which is the fourth region? Wait no, wait the question's inequality is \(2x-5y <10\). Let's use the test point (5,1): 2(5)-5(1)=10-5=5<10, which is true. (5,1) is above the line (line at x=5 is y=0, 1>0). (5,-1): 10 - (-5)=15>10, false, which is below the line. So the solution is above the dashed line, which is the fourth region? Wait no, the second region is dashed line, shaded below, which is false. The fourth is dashed line, shaded above, which is correct. Wait I messed up earlier. Let's recheck:
Step 1: Strict inequality \(<\) means dashed line (eliminate solid line options 1 and 3).
Step 2: Test (0,0) in \(2x-5y <10\): 0 <10, true. So (0,0) must be in the shaded region. Looking at the dashed line options:
- Second region: shaded below the dashed line, (0,0) is above the line, so not in the shaded area.
- Fourth region: shaded above the dashed line, (0,0) is in this shaded area.
Wait this contradicts my earlier mistake. So correct answer is the fourth region? No, wait the line \(2x-5y=10\): when x=0, y=-2; when y=0, x=5. So the line goes from (5,0) to (0,-2). The region \(2x-5y <10\) is all points where \(2x-5y\) is less than 10. For (0,0): 0<10, true, so (0,0) is in the region, which is above the line (since the line at x=0 is y=-2, 0 is above -2). So the shaded region should be above the dashed line, which is the fourth diagram. Wait but let's check the second diagram: shaded below the dashed line, which would include (0,-3): 0 -5(-3)=15>10, which is false, so that region does not satisfy the inequality. So the correct answer is the fourth region (dashed line, shaded above the line, including (0,0)).

Wait no, wait I think I messed up the direction of the inequality when rearranging. Let's do it again:
\(2x -5y <10\)
Subtract 2x: \(-5y < -2x +10\)
Divide by -5, reverse inequality: \(y > \frac{2}{5}x - 2\)
So y is greater than the line, which is above the line, since the line has a positive slope. So the region above the dashed line is the solution, which is the fourth diagram.

Wait but the question's diagrams:
1. Solid line, shaded below
2. Dashed line, shaded below
3. Solid line, shaded above
4. Dashed line, shaded above

Yes, so the fourth one is correct. Wait but earlier I thought the second, that was a mistake. So correcting:

Step1: Identify line type

Strict inequality \(<\) → dashed line. Eliminates solid line options 1 and 3.

Step2: Test origin in inequality

Substitute \((0,0)\) into \(2x-5y<10\):
\(2(0)-5(0) <10 \implies 0<10\), which is true. So \((0,0)\) is in the solution region.

Step3: Match region to test point

The fourth diagram (dashed line, shaded above the line) includes \((0,0)\), while the second (dashed line, shaded below) does not.

Final Answer: The fourth (bottom-most) shaded region (dashed boundary line \(2x-5y=10\), shaded area above the line)

Answer:

The second (top-to-bottom) shaded region (dashed line \(2x-5y=10\), shaded area below/left of the line)