question 10 is: a pen and a pencil cost RM2.40 and RM0.30 respectively. The price of an eraser is one third of the price of the
pen. Adam buys one of each item. Find the total amount of money Adam has to pay.
1 answer:
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The correct answer is: Answer choice: [A]:
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→ "
" ; " { u 
± 3 } " ;
→ or, write as: " u / (u − 3) " ; {" u ≠ 3 "} AND: {" u ≠ -3 "} ;
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Explanation:
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We are asked to simplify:
;
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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→
;
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→ We can simplify by "canceling out" BOTH the "(u + 3)" values; in BOTH the "numerator" AND the "denominator" ; since:
"
" ;
→ And we have:
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→ " " ; that is: " u / (u − 3) " ; { u
} .
and: { u
} .
→ 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:
"u3" ;
→ 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 -3 " ;
Furthermore, consider the initial (unsimplified) given expression:
→ ;
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" ; "u -3 " ;
or, write as: " { u ± 3 } " .
_________________________________________________________
__________________________________________________________
→ "
→ or, write as: " u / (u − 3) " ; {" u ≠ 3 "} AND: {" u ≠ -3 "} ;
__________________________________________________________
Explanation:
__________________________________________________________
We are asked to simplify:
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)" ;
___________________________________________________________
Let us rewrite as:
___________________________________________________________
→
___________________________________________________________
→ We can simplify by "canceling out" BOTH the "(u + 3)" values; in BOTH the "numerator" AND the "denominator" ; since:
"
→ And we have:
_________________________________________________________
→ "
and: { u
→ which is: "Answer choice: [A] " .
_________________________________________________________
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
→ 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
Furthermore, consider the initial (unsimplified) given expression:
→
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:
_______________________________________________________
→ " 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
or, write as: " { u
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