Lesson Topics Discussion Quiz: Class Homework |
Steps-3 |
Title: Linear inequalities in one or two variables |
Grade Lesson s5-l4 |
Explanation: Hello Students, time to practice and review the steps for the problem. |
Quiz: Discussion Step
Id | Type | Name | Note |
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1 |
Problem |
Solve the inequality \$ | 3x - 2 | + | 2x + 1 | < 5 \$. |
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2 |
Step |
THe given inequality is |
\$ | 3x - 2 | + | 2x + 1 | < 5 \$ |
3 |
Clue |
To solve this, we need to consider the critical points where the expressions inside the absolute values change signs. These points are: \$3x - 2 = 0 => x = 2/3\$ \$2x + 1 = 0 => x = -1/2\$ |
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4 |
Step |
These points divide the number line into three intervals: |
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5 |
Step |
Let’s analyze each interval: |
\$x < -1/2\$ \$3x - 2 < 0 => | 3x - 2 | = -(3x - 2) = -3x + 2\$ \$2x + 1 < 0 => | 2x + 1 | = -(2x + 1) = -2x - 1\$ |
6 |
Step |
The inequality becomes: |
\$(-3x + 2) + (-2x - 1) < 5\$ \$-5x + 1 < 5\$ \$-5x < 4\$ \$x > -4/5\$ |
7 |
Step |
Since we are considering \$x < -1/2\$, we need to find the intersection of \$x < -1/2 and x > -4/5\$ |
\$-4/5 = -0.8\$ \$-1/2 = -0.5\$ So, \$-4/5 < x < -1/2\$ |
8 |
Step |
Next Interval |
\$-1/2 ≤ x < 2/3\$ \$3x - 2 < 0 => | 3x - 2 | = -(3x - 2) = -3x + 2\$ \$2x + 1 ≥ 0 => | 2x + 1 | = 2x + 1\$ The inequality becomes: (-3x + 2) + (2x + 1) < 5 -x + 3 < 5 -x < 2 x > -2 |
9 |
Step |
Since we are considering \$-1/2 ≤ x < 2/3\$, we need to find the intersection of \$x > -2 and -1/2 ≤ x < 2/3\$. Therefore, \$-1/2 ≤ x < 2/3\$. |
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10 |
Step |
Final Interval |
\$x ≥ 2/3\$ \$3x - 2 ≥ 0 => | 3x - 2 | = 3x - 2\$ \$2x + 1 > 0 => |2x + 1 | = 2x + 1\$ The inequality becomes: (3x - 2) + (2x + 1) < 5 5x - 1 < 5 5x < 6 \$x < 6/5\$ |
11 |
Step |
Since we are considering \$x ≥ 2/3\$, we need to find the intersection of \$x ≥ 2/3 and x < 6/5\$ |
\$2/3 = 0.666...\$ \$6/5 = 1.2\$ Therefore, \$2/3 ≤ x < 6/5\$ |
12 |
Clue |
Now we combine the solutions from each interval: \$-4/5 < x < -1/2\$ \$-1/2 ≤ x < 2/3\$ \$2/3 ≤ x < 6/5\$ Combining these intervals, we get: \$-4/5 < x < 6/5\$. |
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13 |
Solution |
The solution to the inequality \$| 3x - 2 | + | 2x + 1 | < 5 is -4/5 < x < 6/5\$. |
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14 |
Sumup |
Please summarize steps |
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Choices |
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15 |
Choice-A |
This option is incorrect. It suggests that x is greater than \$-4/5\$ and less than \$-6/5\$. However, \$-4/5\$ is greater than \$-6/5\$, so no number can satisfy both conditions simultaneously. This is an empty set |
Wrong \$(-4/5) < x < -6/5\$ |
16 |
Choice-B |
This option is a possible range, but it is incorrect because it misses the negative portion of the solution |
Wrong \$(4/5) < x < 6/5\$ |
17 |
Choice-C |
This option has the correct upper bound of \$6/5\$, but the lower bound of \$-4/7\$ is incorrect |
Wrong \$(-4/7) < x < 6/5\$ |
18 |
Choice-D |
This option is the correct solution. It accurately represents the range of x values that satisfy the inequality |
Correct \$(-4/5) < x < 6/5\$ |
19 |
Answer |
Option |
D |
20 |
Sumup |
Please summarize choices |
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