The answer: 3 ⁵/⏨ = 7/2 ; There are 7 (seven) “halves” in 3 ⁵/⏨ .
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Explanation:
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Method 1): 3 ⁵/⏨ x/2; solve for "x".
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→ 3 ⁵/⏨ = 3 ½ = [ [(2*3) + 1] / 2 = (6+1)/2 = 7/2 = x/2 ; → x = 7;
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→ {Note: We know that "3 ⁵/⏨ = 3 ½ " because "⁵/⏨" = "½" ;
The fraction, "⁵/⏨"; can be simplified (reduced) to "½", since:
→ ⁵/⏨ = 5/10 = (5÷5) / (10÷5) = 1/2 = ½ ; → So, 3 ⁵/⏨ = 3 ½ .}.
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→ x = 7 ; → 3 ⁵/⏨ = 7/2 ; → There are 7 (seven) “halves” in 3 ⁵/⏨.
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Method 2): → 3 ⁵/⏨ = x/2; solve for "x".
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→3 ⁵/⏨ = 3.5 = x/2 ; {Note: "3 ⁵/⏨" can be written in decimal form as: "3.5".}.
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→ (3.5)*2 = x ↔ x = (3.5)*2 = 7 ↔ x = 7 ;
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→ 3 ⁵/⏨ = 7/2 ; → There are 7 (seven) “halves” in 3 ⁵/⏨ .
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Method 3): → 3 ⁵/⏨ = x/2; solve for "x".
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→ 3 ⁵/⏨ = 35/10 = x/2 ;
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→{We change the "3 ⁵/⏨" to an "improper fraction", which is "35/10";
→We do this as follows: "3 ⁵/⏨ = [(10*3) + 5] / 10 = (30 + 5) / 10 = 35/10"}.
→We have: 35/10 = x/2
→ Change the "10" to a "5"; and change the "2" to a "1" ;
→ Since "5*2 = 10"; and since, "10÷2 = 5"; and "2÷2 =1" ;
→ Rewrite as: 35/5 = x ; & solve: → 7 = x ↔ x = 7 ;
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→ 3 ⁵/⏨ = 7/2 ; → There are 7 (seven) “halves” in 3 ⁵/⏨ .
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Method 4): (variant of "Method 3"): → 3 ⁵/⏨ = x/2; solve for "x".
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→ 3 ⁵/⏨ = 35/10 = x / 2 ;
→ {Note: We write: "3 ⁵/⏨" as "35/10", the "improper fraction" form.
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Here is the variant: We shall continue WITHOUT simplifying the "10" and the "2" to a "5" and a "1"; rather, we will simply "cross-multiply":
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→ We have: → 35/10 = x/2 ; → Cross-multiply, to get:
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→ 10x = 35*2 ; From this point we have a few options:
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Option 1): We have: → 10x = 35*2 ;
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→ Multiply "(35*2)" ; → to get "70" ; and rewrite the equation:
→ 10x = (35*2) ; → 10x = 70 ;
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Divide EACH side of the equation by "10"; to isolate "x" on one side of the equation; &to solve for "x":
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→ 10x = 70 ; → 10x / 10 = 70 / 10 ; → x = 7 ;
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→ 3 ⁵/⏨ = 7/2 ; → There are 7 (seven) “halves” in 3 ⁵/⏨ .
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Option 2): We have: → 10x = 35 * 2 ;
→Divide each side of the equation by "10" to isolate "x" on one side of the equation (and ultimately, to solve for "x");
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→ Rewrite, as follows: 10x / 10 =
:
→ Now, on the "right-hand side of the equation, "cancel out" the "2" (that is, change the "2" to a "1");& change the "10" to a "5" ; (since "2/10", or "2 ÷ 10", = "1/5", or, "⅕"). The "1" can be eliminated, since anything multiplied by "1" is equal to that same value.
→ We now have "x" =
= 7; → x = 7 ;
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→ 3 ⁵/⏨ = 7/2 ; → There are 7 (seven) “halves” in 3 ⁵/⏨ .
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Option 2): → We have: 10x = 35*2 ;
→ Divide EACH side of the equation by "(35*2)" ;
→ 10x / (35*2) = (35*2) / (35*2) ;
→ Rewrite as: 10x / (35*2) = 1 ;
{Note: The "right-hand side" of the equation; "(35 * 2) / (35 *2)" ; or,
"(35 * 2) ÷ (35 *2)" = "1" ;
→ since any [non-zero value], divided by itself, is equal to: "1".}.
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→ So, we have:
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→ 10x / (35*2) = 1 ; ↔ 10x ÷ (35*2) = 1 ;
→ 1*(35*2) = 10x; → 35*2 = 10x ; → 70 = 10x ↔ 10x = 70 ;
→ We have: 10x = 70; → Now, divide EACH side of the equation by "10", to isolate "x" on one side of the equation; & to solve for "x" ;
→ 10x/10 = 70/10 ; → x = 7 ;
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→ 3 ⁵/⏨ = 7/2 ; → There are 7 (seven) “halves” in 3 ⁵/⏨ .
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Option 3): very similar to {option 2}.
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We have: → 10x = 35*2 ;
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Here is the variant from "Option 2" :
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We multiply the: (35*2) on the right-hand side of our equation:
→ 35*2 = 70 ; & rewrite our equation as:
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→ 10x = 70 ; → Now, we divide each side of the equation by "70";
→ 10x /70 = 70/70 ; → to get: 10x/70 = 1;
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WIth 10x /70 = 1; we can EITHER:
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1) Reduce the "10x/70" on the left-hand side of the equation, to:
→ "1x/7", or "x/7 " ; [Since (10x÷10) / (70÷10) = 1x/7 = x/7] ;
→ AND THEN; Rewrite the equation as:
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→ x/7 = 1; → Then solve for "x" ;
→ 1*7 = x → 7 = x ↔ x = 7 ;
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→ 3 ⁵/⏨ = 7/2 ; → There are 7 (seven) “halves” in 3 ⁵/⏨ .
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OR: 2) Keep the equation as: 10x/70 = 1 ; and solve for "x"
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→ 10x /70 = 1 ; → 10x = 1*70 ;
→ to get: → 10x = 70 ; Solve for "x" by dividing each side of the equation by "10"; to isolate "x" on one side of the equation; & to solve for "x" ;
→ 10x = 70 ; → 10x/10 = 70/10 ; → x = 7 ;
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→ 3 ⁵/⏨ = 7/2 ; → There are 7 (seven) “halves” in 3 ⁵/⏨ .
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