The numbers are: "3" and "9" .
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Explanation:
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Let "x" represent one of the two (2) numbers.
Let "y" represent the other one of the two (2) numbers.
x = 2y + 3 ;
x + y = 12 .
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Method 1)
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x = 12 − y ;
Plug this into "x" for "2y + 3 = x" ;
→ 2y + 3 = 12 − y ;
Add "y" to each side of the equation; & subtract "3" from each side of the equation ;
→ 2y + 3 + y − 3 = 12 − y + y − 3 ;
to get: 3y = 9 ;
Divide each side of the equation by "3" ;
to isolate "y" on one side of the equation; & to solve for "y" ;
3y / 3 = 9 / 3 ;
y = 3 .
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Now: x = 12 − y ; Plug in "3" for "y" ; to solve for "x" ;
→ x = 12 − 3 = 9
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So; x = 9, y = 3 .
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Method 2)
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When we have:
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x = 2y + 3 ;
x + y = 12 .
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→ y = 12 − x ;
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Substitute "(12−x)" for "y" in the equation:
" x = 2y + 3 " ;
→ x = 2(12 − x) + 3 ;
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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:
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→ 2(12 − x) = 2(12) − 2(x) = 24 − 2x ;
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So; rewrite:
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x = 2(12 − x) + 3 ;
as:
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→ x = 24 − 2x + 3 ;
→ x = 27 − 2x ;
Add "2x" to each side of the equation:
→ x + 2x = 27 − 2x + 2x ;
→ 3x = 27 ;
Divide each side of the equation by "3" ;
to isolate "x" on one side of the equation; & to solve for "x" ;
3x / 3 = 27 / 3 ;
x = 9 .
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Note: "y = 12 − x" ; Substitute "9" for "x" ; to solve for "y" ;
→ y = 12 − 9 = 3 ;
→ y = 3 .
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So, x = 9 ; and y = 3.
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The numbers are: "3" and "9" .
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To check our answers:
Let us plug these numbers into the original equations;
to see if the equations hold true ; (i.e. when, "x = 9" ; and "y = 3"
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→ x + y = 12 ;
→ 9 + 3 =? 12 ? Yes!
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→ x = 2y + 3 ;
→ 9 =? 2(3) + 3 ?? ;
→ 9 =? 6 + 3 ? Yes!!
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