Help with algebraic inequalities containing fractions: −3 ≤ (3 – 5x)/3 < 3/8

[staceymeggs]

Help with algebraic inequalities containing fractions: −3 ≤ (3 – 5x)/3 < 3/8

I've suffered brain damage, and I have trouble remembering things, and I get mixed up a lot. I've been wracking my brain trying to figure out which rule in fractions I'm breaking that's causing my to get a very wrong answer. Can someone please help me? I know the answer I got is wrong, and even if my work was right, my answer runs backwards. Please help me figure out what I'm doing wrong so I can fix it and do better with future problems.


Question:
1. Solve the following inequalities and give your answer in interval notation. Hint: Either both the numerator and the denominator are positive, or both are negative; consider both cases.

−3 ≤ (3 – 5x)/3 < 3/8

My Work:
First half of equation:
-3/1 ≤ (3 – 5x)/3


-3/1(-3) ≤ (3 – 5x)/3 (1)


9/3 ≤ (3 – 5x)/3


3 ≤ -15x


Second half of equation:
(3 – 5x)/3 < 3/8


(3 – 5x)/3 (6) < 3/8 (3)


(18-30x)/24 < 9/24


(18-30x)/24(24) < 9/24(24)


432/576 < 216/576
0.75 < 1/8

ANSWER:
3 ≤ (-15x = 0.75) < 1/8

 

[JeffM]

I've suffered brain damage, and I have trouble remembering things, and I get mixed up a lot. I've been wracking my brain trying to figure out which rule in fractions I'm breaking that's causing my to get a very wrong answer. Can someone please help me? I know the answer I got is wrong, and even if my work was right, my answer runs backwards. Please help me figure out what I'm doing wrong so I can fix it and do better with future problems.


Question:
1. Solve the following inequalities and give your answer in interval notation. Hint: Either both the numerator and the denominator are positive, or both are negative; consider both cases.

−3 ≤ (3 – 5x)/3 < 3/8

My Work:
First half of equation:
-3/1 ≤ (3 – 5x)/3


-3/1(-3) ≤ (3 – 5x)/3 (1) MISTAKE I


9/3 ≤ (3 – 5x)/3


3 ≤ -15x MISTAKE II


Second half of equation:
(3 – 5x)/3 < 3/8


(3 – 5x)/3 (6) < 3/8 (3) MISTAKE III


(18-30x)/24 < 9/24


(18-30x)/24(24) < 9/24(24)


432/576 < 216/576 MISTAKE IV
0.75 < 1/8 MISTAKE V

ANSWER:
3 ≤ (-15x = 0.75) < 1/8

I am not going to talk about mistakes II, IV, and V because you were already on the wrong path, but you should try to find them on your own.

Your first mistake is huge. It appears that you are sort of trying to get common denominators, but that is not necessary in this problem. You want to get rid of denominators altogether. You can do that by multiplying each side of the inequation by the least common multiple of the denominators. Remember in equations, whatever you do to change the value of one side of the equation you must do to the other side of the equation.

\(\displaystyle -\ 3 \le \dfrac{3 - 5x}{3} \implies \dfrac{-\ 3}{1} \le \dfrac{3 - 5x}{3} \implies\)

\(\displaystyle 3 * \dfrac{-\ 3}{1} \le 3 * \dfrac{3 - 5x}{3} \implies 3 - 5x = \dfrac{-\ 9}{1} = -\ 9.\)

Now proceed from here carefully, and see what you get.

Mistake III is similar to mistake 1.

What is the least common multiple of the two denominators?

What do you get when you multiply both sides by that least common multiple?

See what you get from that.

 

[Steven G]

I've suffered brain damage, and I have trouble remembering things, and I get mixed up a lot. I've been wracking my brain trying to figure out which rule in fractions I'm breaking that's causing my to get a very wrong answer. Can someone please help me? I know the answer I got is wrong, and even if my work was right, my answer runs backwards. Please help me figure out what I'm doing wrong so I can fix it and do better with future problems.


Question:
1. Solve the following inequalities and give your answer in interval notation. Hint: Either both the numerator and the denominator are positive, or both are negative; consider both cases.

−3 ≤ (3 – 5x)/3 < 3/8

My Work:
First half of equation:
-3/1 ≤ (3 – 5x)/3


-3/1(-3) ≤ (3 – 5x)/3 (1) You are denoting that you are multiplying the left side by -3 and the right side by 1. You need to multiply both sides by the same amount


9/3 ≤ (3 – 5x)/3 When you multiply a fraction by -3 you only multiply the numerator by -3, NOT the denominator. Also, when multiply the 1 by -3 (which you should have not done) you got the wrong result as 1*(-3) = -3 NOT 3


3 ≤ -15x above you had (3 – 5x)/3, so how did you get -15x from this


Second half of equation:
(3 – 5x)/3 < 3/8


(3 – 5x)/3 (6) < 3/8 (3) You are denoting that you are multiplying the left side by 6 and the right side by 3. You need to multiply both sides by the same amount


(18-30x)/24 < 9/24 9/24 reduces back to 3/8! Why? Because you did NOT multiply 3/8 by 3, but rather by 3/3 which is 1! When you multiply a fraction by 3, you just multiply the numerator by 3.
Also 3*6=18 NOT 24. Again when you multiplied (3 – 5x)/3 by 6 you multiplied by 6/6, ie you multiplied the numerator and denominator by 6. Agai, you multiplied both sides by different numbers


(18-30x)/24(24) < 9/24(24)


432/576 < 216/576
0.75 < 1/8

ANSWER:
3 ≤ (-15x = 0.75) < 1/8

−3 ≤ (3 – 5x)/3 < 3/8 . You want to get x alone in the middle. You have three sides in this equation, the left hand side, the middle and the right hand side. Just concentrate on the middle. How would you get x alone. Well first you multiply by 3 (from all three sides!), then you'll just have 3-5x in the middle. Next you thould subtract 3 from all sides giving you -5x in the middle. Then divide all sides by -5 since -5x/-5 =x. Remember that whenever you multiply or divide an inequality by a negative number (like -5), you MUST switch the direction of the inequality symbol(s).

Please try again and show us your work.

 

[mmm4444bot]

… (3 – 5x)/3 … first you can subtract 3 …

Did you forget about the denominator?

 

[Steven G]

Did you forget about the denominator?

Yes, thanks for pointing that out!!

 

[mmm4444bot]

Hi staceymeggs. You can solve for x by working on the entire inequality at once, instead of breaking it in half.

I agree that clearing denominators is a good first step. Here's a similar example shown with steps.


\(\displaystyle -1 \; < \; \dfrac{-5x + 2}{3} \; \le \; \dfrac{7}{4}\)


12 is the LCM of 3 and 4, so multiplication by 12 will cause each denominator to cancel.


\(\displaystyle (12)(-1) \; < \; \dfrac{12}{1} \cdot \dfrac{-5x + 2}{3} \; \le \; \dfrac{12}{1} \cdot \dfrac{7}{4}\\
\;\\
\;\\
-12 \; < \; (4)(-5x + 2) \; \le \; 21\\
\;\\
\;\\
-12 \; < -20x + 8 \; \le \; 21\)


Look at -20x + 8. To solve for x, we need to first subtract 8 and then divide by -20. Here's the subtraction:


\(\displaystyle -12 - 8 \; < -20x + 8 - 8 \; \le \; 21 - 8\\
\;\\
\;\\
-20 \; < \; -20x \; < \; 13\)


Important Note: When we multiply (or divide) an inequality by a negative amount, we must change the direction of the inequality symbols. That's a rule we must remember. Here's the division by -20 (each inequality symbol has changed direction):


\(\displaystyle \dfrac{-20}{-20} > \dfrac{-20x}{-20} \; \ge \dfrac{13}{-20}\\
\;\\
\;\\
1 \; > \; x \; \ge \; -\dfrac{13}{20}\)


That's a correct answer, but most teachers want to see compound inequalities written in smallest-number to largest-number order, like this:

\(\displaystyle -\dfrac{13}{20} \; \le x \; < \; 1\)


If you have any more questions, let us know. :cool: