PARENTALIGUARDIAN PERMISSION $\newline$ FOR JUVENILE VOLUNTEERS $\newline$ Math Analysis Unit $7$ Demonstrate Mastery $\newline$ Below are problems that represent Unit $7$ . Problems are worth $2 \%$ back on the Unit $7$ T problems you could earn $8 \%$ back. With a maximum of $\mathbf{8 7 \%}$ . If you only choose to do you could earn $2 \%$ back. You may choose to do as many or as few as you would like. credit. In order to receive credit you must have the proper equations, work and solutio The worksheet with all solutions is due no later than Wednesday, April $17$ th at $3$ : $15$ submit to me in my classroom, to my mailbox in the office or submit it on my viewp you to come in and work with me on these problems during I block as well as after $\newline$ $1$ ) Solve the inequality. Write your answer in interval notation. $\newline$ $\frac{x+3}{x-4} \leq 2$

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Grade 7

One-step inequalities: word problems

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Q. PARENTALIGUARDIAN PERMISSION

\newline

FOR JUVENILE VOLUNTEERS

\newline

Math Analysis Unit

7

Demonstrate Mastery

\newline

Below are problems that represent Unit

7

. Problems are worth

2 \%

back on the Unit

7

T problems you could earn

8 \%

back. With a maximum of

\mathbf{8 7 \%}

. If you only choose to do you could earn

2 \%

back. You may choose to do as many or as few as you would like. credit. In order to receive credit you must have the proper equations, work and solutio The worksheet with all solutions is due no later than Wednesday, April

17

th at

3

15

submit to me in my classroom, to my mailbox in the office or submit it on my viewp you to come in and work with me on these problems during I block as well as after

\newline

1

) Solve the inequality. Write your answer in interval notation.

\newline

\frac{x+3}{x-4} \leq 2

Get Terms Together: First, we need to get all terms on one side of the inequality to compare it to $0$ . $\newline$ $(x + 3)/(x - 4) - 2 \leq 0$ $\newline$ Now, find a common denominator and subtract $2$ from both sides. $\newline$ $(x + 3)/(x - 4) - 2(x - 4)/(x - 4) \leq 0$
Find Common Denominator: Combine the terms over the common denominator. $\newline$ (x + \(3 - $2$ x + $8$ )/(x - $4$ ) \leq $0$ $\newline$ Simplify the numerator. $\newline$ (-x + \(11)/(x - $4$ ) \leq $0$
Combine Terms: Now, we need to find the critical points where the numerator or denominator is zero. $\newline$ The numerator is zero when $-x + 11 = 0$ , which means $x = 11$ . $\newline$ The denominator is zero when $x - 4 = 0$ , which means $x = 4$ . $\newline$ These are our critical points.
Find Critical Points: We'll test intervals around our critical points to see where the inequality holds true. $\newline$ The intervals are $(-\infty, 4$ ), $(4, 11)$ , and $(11, \infty)$ . $\newline$ Pick test points from each interval, like $x = 3$ , $x = 5$ , and $x = 12$ , and plug them into the inequality.
Test Intervals: Test $x = 3$ : $\frac{-3 + 11}{3 - 4} \leq 0$ $\frac{8}{-1} \leq 0$ $-8 \leq 0$ is true.
Test $x = 3$ : Test $x = 5$ : $\frac{-5 + 11}{5 - 4} \leq 0$ $\frac{6}{1} \leq 0$ $6 \leq 0$ is false.
Test $x = 5$ : Test $x = 12$ : $\frac{-12 + 11}{12 - 4} \leq 0$ $\frac{-1}{8} \leq 0$ $\frac{-1}{8} \leq 0$ is true.
Test $x = 12$ : From the tests, we see that the inequality holds true for $x$ in $(-\infty, 4)$ and $(11, \infty)$ . However, we must exclude $x = 4$ because the original inequality is undefined at $x = 4$ .
Final Solution: Write the solution in interval notation. $\newline$ The solution is $(-\infty, 4) \cup (11, \infty)$ .