Answer:
Step-by-step explanation:
Remember that it doesn't matter how many numbers are after the tenths place of a decimal point what matters is the tenths place itself. If you have a decimal that has dozens of numbers but it's tenth place is a 3 and then you have a decimal with 5 numbers in whole but it's tenths place is 4, the 4 is greater then the 3 simply because it's greater in tenths.
Hope this helps.
Answer: " a = 21 " .
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<u>Step-by-step explanation</u>:
Given: 3a/4-2a/3 = 7/4 ; Solve for "a" ;
Rewrite as: (3a/4) - (2a/3) = (7/4) ;
Now, for each of the three (3) "denominator values" in "fraction form" within the equation given:
→ Find the "LCD" ["Least Common Denominator"]:
The denominators are: 4, 3, and 4 ;
that is: "3" and "4" ;
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To find the LCD: First; multiply the denominators: "4 * 3 = 12" .
So; the value "12" could be the LCD; so, the value for the LCD is no greater than "12" ; however, there <u><em>could </em></u>be a smaller value.
To determine the LCD:
List the multiples of the given denominators:
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3: 3, 6, 9, <u><em>12</em></u>, 15 .... ;
4: 4, 8, <u><em>12</em></u>, 16... ;
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We find that "12" is, in fact, the LCD of "3" and 4:
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We can multiply each side of the equation by "12" ; to eliminate the "fractional values" :
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<u>Note the</u><u> "</u><u>distributive property</u><u>"</u><u> of multiplication</u>:
→ a(b+c) = ab + ac ;
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As such:
;
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Let us start with the "left-hand side" of the equation:
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= ;
=
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Note: " " ;
;
→ The "12" cancels to a "3" ; and the "4" cancels to a "1" ;
since: "12÷4 = 3" ; and since: "4÷4 = 1" ;
→ and we can rewrite the "left-hand-side" expression as:
→ " " ; which we can simplify as:
→ "3 * 3a" ; which we can simplify as: " 9a " .
then we have: " [12 * ] " ;
which equals:
= " " ;
<u>Note</u>: The "12" cancels out to a "4"; & the "3" cancels out to a "1" ;
→ {since: "(12 ÷ 3 = 4)"; & since: "(3 ÷ 3 = 1)" ;
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→ And we can rewrite the expression as:
→ " " ; which we can simplify as:
→ " 4 * 2a " ; which we can simplify/calculation as: " 8a " ;
Now, we can rewrite the expression of the "left-hand side"
of the equation as:
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→ " [9a] − [8a] " ; (don't forget to carry down the "minus sign"!) ;
which we can simplify/calculate to get:
→ " [9a − 8a] " ; which we can further simplify/calculate;
→ to get:
→ " 1a " ; or: "a" —the value for which we wish to solve!
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Now, let us examine the "right-hand side" of the equation:
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→ " " ;
<u>Note</u>: The "12" cancels out to a "3" ; & the "4" cancels out to a "1" ;
→ {Since: "12 ÷4 = 3 " ; & since: "4 ÷ 4 = 1 "} ;
And we can rewrite the expression as:
→ " " ; which we can simplify as:
→ " 3 * 7 " ; which can simplify/calculate to get: " 21" ;
⇒ Now, let us rewrite the equation; by using our simplified values for both the "left-hand side" and the "right-hand side" of the equation; to solve for "a" :
⇒ a = 21 ;
→ which is the correct answer:
→ " a = 21 " .
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Hope this helps!
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