Did I Simplify These Precalculus Problems Correctly?

In summary, the conversation discusses simplifying two expressions and finding the correct answer. The first expression is simplified to 2x^4 + 9/x^7 and the person confirms that it is correct. The second expression is simplified to (8x^3)/(y^15), but the other person points out that it is incorrect and provides steps to solve it correctly.
  • #1
MaximumPhysics
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Just joined this website for math help like right now :)

Background info: I was sick for 2 years before going to university so my level in math is horrible; though I will try my best in all questions which I would ask.1. Simplify 2x^-3 + 9x^-7

Here's what I did

2/x^3 + 9/x^7
find common denominator
2/x^3 multiplied top and bottom by x^4

2x^4/x^7+9/x^7= 2x^4+9/x^7

Did I get the correct answer?

2. Simplify ((5x^4+y^-8)/(10x^5y^3))^-3

I first multiplied everything by ^-3

so (5^-3 x^-12+ y^-24)/ (10^-3 x^-15 y^-9)

(1/5^3 x^-12 + y^-24 ) / (1/10^3 x^-15 y^-9)

1/125 multiplied by 1000/1 (flipped this one)

I get 1000/125 = 8 so 8 on top

rest I can use the property of subtracting variables with exponents when it is multiplication...anyways fast forward and I get.

(8x^3)/(y^15)
 
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  • #2
tenthfire said:
Just joined this website for math help like right now :)

Background info: I was sick for 2 years before going to university so my level in math is horrible; though I will try my best in all questions which I would ask.1. Simplify 2x^-3 + 9x^-7

Here's what I did

2/x^3 + 9/x^7
find common denominator
2/x^3 multiplied top and bottom by x^4

2x^4/x^7+9/x^7= 2x^4+9/x^7

Did I get the correct answer? ... Yes

2. Simplify ((5x^4+y^-8)/(10x^5y^3))^-3

...

(8x^3)/(y^15)

Good morning,

your 2nd result is wrong.

\(\displaystyle \left( \frac{5x^4+\frac1{y^8}}{10x^5 \cdot y^3} \right)^{-3}\)

1. step:

\(\displaystyle \left( \frac{5x^4+\frac1{y^8}}{10x^5 \cdot y^3} \right)^{-3} = \left( \frac{10x^5 \cdot y^3} {5x^4+\frac1{y^8}}\right)^{3}\)

2. step:

\(\displaystyle \left( \frac{10x^5 \cdot y^3} {5x^4+\frac1{y^8}}\right)^{3}= \left( \frac{10x^5 \cdot y^3} {\frac{5x^4 y^8+1}{y^8}}\right)^{3}\)

3. I'll leave the rest for you.
 

1. What does it mean to simplify in precalculus?

Simplifying in precalculus means to manipulate an expression or equation to make it easier to solve or understand. This can involve combining like terms, factoring, or using algebraic properties to rewrite the expression in a simpler form.

2. Why is simplifying important in precalculus?

Simplifying is important in precalculus because it allows us to solve equations and expressions more efficiently. It also helps us better understand the underlying concepts and relationships within the problem.

3. What are some common techniques for simplifying in precalculus?

Some common techniques for simplifying in precalculus include combining like terms, factoring, using the distributive property, and applying algebraic rules such as the associative and commutative properties. In some cases, trigonometric identities may also be used to simplify expressions.

4. How can I tell if an expression or equation is simplified?

An expression or equation is considered simplified when it cannot be further reduced or rewritten in a simpler form. This means that all like terms have been combined, all factors have been factored, and any unnecessary parentheses have been removed.

5. Can simplifying lead to incorrect solutions in precalculus?

No, simplifying should not lead to incorrect solutions in precalculus. As long as the simplification is done correctly and all steps are shown, the final solution should be equivalent to the original expression or equation. However, it is always important to check your work and make sure all simplification steps are accurate.

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