QUESTION IMAGE
Question
below are two inequalities and the graphs of their lines without the shading. by imagining where the shading should be, identify which point would satisfy both inequalities.
$y < -\frac{2}{5}x + 6$
$y < 5x - 6$
To solve this, we need to find a point that satisfies both inequalities \( y < -\frac{2}{5}x + 6 \) and \( y < 5x - 6 \). Let's assume we test a point, say \((2, 0)\):
Step 1: Test the first inequality \( y < -\frac{2}{5}x + 6 \)
Substitute \( x = 2 \) and \( y = 0 \):
\( 0 < -\frac{2}{5}(2) + 6 \)
\( 0 < -\frac{4}{5} + 6 \)
\( 0 < \frac{26}{5} \) (which is true, since \( \frac{26}{5} = 5.2 \))
Step 2: Test the second inequality \( y < 5x - 6 \)
Substitute \( x = 2 \) and \( y = 0 \):
\( 0 < 5(2) - 6 \)
\( 0 < 10 - 6 \)
\( 0 < 4 \) (which is also true)
Let's check another point, say \((3, 5)\):
Step 1: First inequality
\( 5 < -\frac{2}{5}(3) + 6 \)
\( 5 < -\frac{6}{5} + 6 \)
\( 5 < \frac{24}{5} \) (which is false, since \( \frac{24}{5} = 4.8 \))
Now, let's check \((2, 0)\) again. Wait, maybe a better approach is to find the intersection of the lines and then check the region. The lines are \( y = -\frac{2}{5}x + 6 \) and \( y = 5x - 6 \). Let's find their intersection:
Set \( -\frac{2}{5}x + 6 = 5x - 6 \)
Multiply both sides by 5: \( -2x + 30 = 25x - 30 \)
\( 30 + 30 = 25x + 2x \)
\( 60 = 27x \)
\( x = \frac{60}{27} = \frac{20}{9} \approx 2.22 \)
Then \( y = 5(\frac{20}{9}) - 6 = \frac{100}{9} - \frac{54}{9} = \frac{46}{9} \approx 5.11 \)
Now, the region for \( y < -\frac{2}{5}x + 6 \) is below the first line, and \( y < 5x - 6 \) is below the second line. Let's check a point in the overlapping region. Let's take \( (3, 4) \):
Step 1: First inequality
\( 4 < -\frac{2}{5}(3) + 6 \)
\( 4 < -\frac{6}{5} + 6 \)
\( 4 < \frac{24}{5} \) (false, since \( \frac{24}{5} = 4.8 \), but 4 < 4.8 is true? Wait, 4 is less than 4.8, so that's true. Wait, my mistake earlier. \( 4 < 4.8 \) is true.
Step 2: Second inequality
\( 4 < 5(3) - 6 \)
\( 4 < 15 - 6 \)
\( 4 < 9 \) (true)
Wait, maybe I made a mistake with \((3, 5)\). Let's recheck \((3, 5)\) in the first inequality: \( 5 < -\frac{6}{5} + 6 = \frac{24}{5} = 4.8 \)? No, 5 is not less than 4.8, so that's false.
Now, let's check \((2, 0)\) again. It works for both. Let's check \((1, -1)\):
Step 1: First inequality
\( -1 < -\frac{2}{5}(1) + 6 \)
\( -1 < -\frac{2}{5} + 6 \)
\( -1 < \frac{28}{5} \) (true)
Step 2: Second inequality
\( -1 < 5(1) - 6 \)
\( -1 < -1 \) (false, since it's not less than, it's equal)
So, \((2, 0)\) works. Let's confirm with the graph. The dashed lines mean the inequalities are strict (<, not ≤). The line \( y = 5x - 6 \) passes through (1, -1) and (2, 4), (3, 9), etc. The line \( y = -\frac{2}{5}x + 6 \) passes through (0, 6), (5, 4), (10, 2), etc. The region below both lines would be where \( y \) is less than both. For \( x = 2 \), \( y < -\frac{4}{5} + 6 = 5.2 \) and \( y < 10 - 6 = 4 \). So the stricter condition is \( y < 4 \) when \( x = 2 \). So \( (2, 0) \) is in that region.
Another point: (3, 4) – wait, \( y = 4 \) for \( x = 3 \) in the second inequality: \( 4 < 5(3) - 6 = 9 \) (true), and in the first inequality: \( 4 < -\frac{6}{5} + 6 = 4.8 \) (true). Wait, but earlier I thought (3, 5) was false, which it is. So (3, 4) is also a solution. But let's check the graph. The dashed lines: the line with negative slope is \( y = -\frac{2}{5}x + 6 \) (dashed, since it's <), and the line with positive slope is \( y = 5x - 6 \) (dashed). The region below both would be to the right of the intersection point? Wait, no. Let's see the slopes: the first line has slope -2/5 (negative), the second has slope 5 (positive). So the first line is decreasing, the second is increasing. Their intersection is at (20/9, 46/9) ≈ (2.22, 5.11). So below…
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A. (2, 0) (assuming this is one of the options; if options differ, adjust accordingly)