The numbers are: "30" and "32" .
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The two (2) consecutive even numbers are: "x" and "(x+2)" .
We are asked to find these numbers.
If we can solve for "x" (one of these numbers); we can solve for "(x+2)" by adding "2" to "x".
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→ (⅔) x + (¼) (x + 2) = 28 ; Solve for "x" ;
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Note the distributive property of multiplication:
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a(b+c) = ab + ac ;
a(b − c) = ab − ac ;
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As such: " + (¼) (x + 2) = (¼) x + (¼) (2) = (¼) x + ½ ;
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Rewrite the equation: → (⅔) x + (¼) (x + 2) = 28 ;
as: " → (⅔) x + (¼) x + ½ = 28 ;
Multiply the entire equation by "12" ; to get rid of the fractions:
→ 12 * { (⅔) x + (¼) x + ½ = 28 } ;
→ (24/3) x + (12/4)x + (12/2) = 28*(12) ;
→ 8x + 3x + 6 = 336
→ 11x + 6 = 336 ;
Subtract "6" from each side of the equation:
→ 11x + 6 − 6 = 336 − 6 ;
→ 11x = 330 ;
Divide each side of the equation by "11" ;
to isolate "x" on one side of the equation; & to solve for "x" ;
→ 11x / 11 = 330 / 11 ;
→ x = 30 ;
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The next consecutive even integer is:
"(x + 2)" ; which equals: "30 + 2 = 32" .
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The numbers are: "30" and "32" .
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Let us check our answer, by plugging the value for "x" in the original equation; to see if it holds true:
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→ (⅔) x + (¼) (x + 2) = 28 ;
→ (⅔) (30) + (¼) (30 + 2) = ? 28 ? ;
→ (60/3) + (¼)*(32) = ? 28 ? ;
→ 20 + 8 = ? 28 ? ; Yes!
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