Question

Simplify u^2+3u/u^2-9
A.u/u-3, =/ -3, and u=/3
B. u/u-3, u=/-3

Answer 1

Answer:a

Step-by-step explanation:

Answer 2

The correct answer is: Answer choice: [A]:
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→ "
`(u)/(u-3)` " ; " { u
`\\eq` ± 3 } " ;

→ or, write as: " u / (u − 3) " ; {" u ≠ 3 "} AND: {" u ≠ -3 "} ;
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Step-by-step explanation:
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We are asked to simplify:


`((u^2+3u))/((u^2-9))` ;


Note that the "numerator" —which is: "(u² + 3u)" — can be factored into:
→ " u(u + 3) " ;

And that the "denominator" —which is: "(u² − 9)" — can be factored into:
→ "(u − 3) (u + 3)" ;
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Let us rewrite as:
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→
`(u(u+3))/((u-3)(u+3))` ;

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→ We can simplify by "canceling out" BOTH the "(u + 3)" values; in BOTH the "numerator" AND the "denominator" ; since:

"
`((u+3))/((u+3)) = 1` " ;

→ And we have:
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→ "
`(u)/(u-3)` " ; that is: " u / (u − 3) " ; { u
`\\eq 3` } .
and: { u
`\\eq-3` } .

→ which is: "Answer choice: [A] " .
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NOTE: The "denominator" cannot equal "0" ; since one cannot "divide by "0" ;

and if the denominator is "(u − 3)" ; the denominator equals "0" when "u = -3" ; as such:

"u
`\\eq`3" ;

→ Note: To solve: "u + 3 = 0" ;

Subtract "3" from each side of the equation;

→ " u + 3 − 3 = 0 − 3 " ;

→ u = -3 (when the "denominator" equals "0") ;

→ As such: " u
`\\eq` -3 " ;

Furthermore, consider the initial (unsimplified) given expression:

→
`((u^2+3u))/((u^2-9))` ;

Note: The denominator is: "(u² − 9)" .

The "denominator" cannot be "0" ; because one cannot "divide" by "0" ;

As such, solve for values of "u" when the "denominator" equals "0" ; that is:
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→ " u² − 9 = 0 " ;

→ Add "9" to each side of the equation ;

→ u² − 9 + 9 = 0 + 9 ;

→ u² = 9 ;

Take the square root of each side of the equation;
to isolate "u" on one side of the equation; & to solve for ALL VALUES of "u" ;

→ √(u²) = √9 ;

→ | u | = 3 ;

→ " u = 3" ; AND; "u = -3 " ;

We already have: "u = -3" (a value at which the "denominator equals "0") ;

We now have "u = 3" ; as a value at which the "denominator equals "0");

→ As such: " u
`\\eq 3`" ; "u
`\\eq` -3 " ;

or, write as: " { u
`\\eq` ± 3 } " .

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