What is the simplified form of x minus 5 over x squared minus 3x minus 10 ⋅ x plus 4 over x squared plus x minus 12? (6 points)
2 answers:
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Answer: This took me a long time to type, hope you read it and find it helpful and easy to understand...
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Step-by-step explanation:
1.
To make it easier to add / subtract fractions, we are looking for common denominators. I can see that if I move the fraction on the left with a denominator 3x to the right, I'll be able to add two equations with like denominator.
Let's add
Add the numerators and keep the same denominator.
To get rid of the denominator x on the left side, we can multiply by x.
Now multiply by 3.
Subtract 2.
The value of x is 1.
Proof.
I can't show the proof to the following problems because I have a 5000 character limit.
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2.
Again, we have like denominators, therefore, we can simply subtract the numerators and keep the same denominator.
Again, we can multiply by x to get rid of the x in the denominator.
Do not distribute the x in the numerator yet because we're gonna eliminate it later.
Factor the denominator.
Simplify x's.
Multiply by x-4 to get rid of the denominator.
Add 10
Subtract x
Combine like terms;
Divide by 4.
Simplify by 2.
6/2 = 3
4/2 = 2
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3.
Same denominator, add numerators.
Multiply by x to get rid of the x.
Subtract
Combine like terms;
Subtract 11.
Divide by -11
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4.
It's easier if you move the operation with denominator x+6 to the right side with negative sign and bring the operation on the right side to the left side with negative sign as well.
Now, since we have the same denominators, we can simply subtract numerators.
Distribute the negative sign.
Combine like terms;
Factor.
I can see that I can factor a 5 on both numerator and numerator. This will allow me to simplify them.
Simplify.
Multiply by x-6
This will simplify the denominators.
Add x and 6.
Combine like terms;
Divide by 5.
Hi there,
This is the original inequality equation:
So, we first need to find the critical points of equality, and we can do that by switching the less than sign to an equal sign.
Now, we multiply both sides by x + 1:
Then, we multiply both sides by x - 1:
Next, we subtract x² from both sides:
After that, we solve for x. We do this by adding -x to both sides and dividing by 2. Doing so gives us x = 0, which is our first critical point. We need to find a few more critical points by testing x = -1 and x = 1. Here is how we do that:
<span>x = <span>−1 </span></span>(Makes left denominator equal to 0)<span>x = 1 </span>(Makes right denominator equal to 0)Check intervals in between critical points. (Test values in the intervals to see if they work.)<span>x <<span>−1 </span></span>(Doesn't work in original inequality)<span><span><span>−1 </span>< x </span><0 </span>(Works in original inequality)<span><span>0 < x </span>< 1 </span>(Doesn't work in original inequality)<span>x > 1 </span><span>(Works in original inequality)
Therefore, the answer to your query is -1 < x < 0 or x > 1. Hope this helps and have a phenomenal day!</span>
This is the original inequality equation:
So, we first need to find the critical points of equality, and we can do that by switching the less than sign to an equal sign.
Now, we multiply both sides by x + 1:
Then, we multiply both sides by x - 1:
Next, we subtract x² from both sides:
After that, we solve for x. We do this by adding -x to both sides and dividing by 2. Doing so gives us x = 0, which is our first critical point. We need to find a few more critical points by testing x = -1 and x = 1. Here is how we do that:
<span>x = <span>−1 </span></span>(Makes left denominator equal to 0)<span>x = 1 </span>(Makes right denominator equal to 0)Check intervals in between critical points. (Test values in the intervals to see if they work.)<span>x <<span>−1 </span></span>(Doesn't work in original inequality)<span><span><span>−1 </span>< x </span><0 </span>(Works in original inequality)<span><span>0 < x </span>< 1 </span>(Doesn't work in original inequality)<span>x > 1 </span><span>(Works in original inequality)
Therefore, the answer to your query is -1 < x < 0 or x > 1. Hope this helps and have a phenomenal day!</span>