Respuesta :
The answer is: "x = 10 , y = 30" ; or; write as: " (10, 30) " .
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Explanation:
___________________________________________________________
Given:
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" x + y = 40 ";
and: " 0.08x + 0.03y = 1.7 " ;
Find the solutions set (the "x-value" and the "y-value"} for this system of equations ; {which can be written as: " (x, y) " ; as the "solution set" for the system of equations.}.
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NOTE:
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For the second equation given:
" 0.08x + 0.03y = 1.7 " ;
→ Multiply EACH SIDE of the equation by "100" ;
to get rid of the "decimal values" ;
→ 100 * {0.08x + 0.03y} = 100* {1.7} ;
to get: " 8x + 3y = 170 " ;
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So, we have:
" 8x + 3y = 170 " ; AND:
" x + y = 40" ;
___________________________________________________
Let us take: "x + y = 40 " ; and solve for "x" (in terms of "y"); and then plug that obtained value {i.e. "substitution"} for "x" in the other equation:
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x + y = 40 ;
Subtract "y" from each side of the equation ;
x + y − y = 40 − y ;
To get:
x = 40 − y ;
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Now, take the "other" equation in the system:
→ " 8x + 3y = 170 " ;
And substitute: "(40 − y)" for "x" ; & then solve for "x" ;
→ 8(40 − y) + 3y = 170 ;
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Note: We have the expression: "8(40− y)" ;
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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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So; " 8(40 − y) " = (8*40) − (8*y) = "320 − 8y" ;
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And rewrite: " 8(40 − y) + 3y = 170 " ;
→ {by substituting: "320 − 8y" for: "8(40 − y)" ; as follows:
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→ 320 − 8y + 3y = 170 ;
Combine the "like terms" on the "left-hand side" of the equation:
-8y + 3y = -5y ;
And rewrite the equation:
320 − 5y = 170 ;
Subtract "320" from each side of the equation:
320 − 5y − 320 = 170 − 320 ;
to get:
-5y = -150 ;
Divide EACH SIDE of the equation by "-5" ;
to isolate "y" on one side of the equation; & to solve for "y" ;
-5y / -5 = -150 / -5 ;
to get:
y = 30 ;
To solve for "x";
x + y = 40 ;
x = 40 − y ;
Plug in: "30" for "y" ; to solve for "x" ;
x = 40 − (30) ;
x = 10 ;
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So; the answer is: "x = 10, y = 30": ; or, write as: " (10, 30)" .
_____________________________________________________
To confirm our answer:
_____________________________________________________
Let us plug in our values for "x" and 'y" ; into the equation:
" (0.08)x + (0.03)y = 1.7 " ; to see if the equation holds true;
→ (0.08)*(10) + (0.03)(30) = ? 0.9 ? ;
→ 0.8 + 0.9 = ? 1.7 ? YES!
______________________________________________________
___________________________________________________________
Explanation:
___________________________________________________________
Given:
___________________________________________________________
" x + y = 40 ";
and: " 0.08x + 0.03y = 1.7 " ;
Find the solutions set (the "x-value" and the "y-value"} for this system of equations ; {which can be written as: " (x, y) " ; as the "solution set" for the system of equations.}.
_________________________________________________________
NOTE:
_________________________________________________________
For the second equation given:
" 0.08x + 0.03y = 1.7 " ;
→ Multiply EACH SIDE of the equation by "100" ;
to get rid of the "decimal values" ;
→ 100 * {0.08x + 0.03y} = 100* {1.7} ;
to get: " 8x + 3y = 170 " ;
_______________________________________________________
So, we have:
" 8x + 3y = 170 " ; AND:
" x + y = 40" ;
___________________________________________________
Let us take: "x + y = 40 " ; and solve for "x" (in terms of "y"); and then plug that obtained value {i.e. "substitution"} for "x" in the other equation:
___________________________________________________
x + y = 40 ;
Subtract "y" from each side of the equation ;
x + y − y = 40 − y ;
To get:
x = 40 − y ;
____________________________________________
Now, take the "other" equation in the system:
→ " 8x + 3y = 170 " ;
And substitute: "(40 − y)" for "x" ; & then solve for "x" ;
→ 8(40 − y) + 3y = 170 ;
_________________________________________________
Note: We have the expression: "8(40− y)" ;
_________________________________________________
Note the "distributive property of multiplication:
_________________________________________________
a(b + c) = ab + ac ;
a(b − c) = ab − ac ;
_________________________________________________
So; " 8(40 − y) " = (8*40) − (8*y) = "320 − 8y" ;
_________________________________________________
And rewrite: " 8(40 − y) + 3y = 170 " ;
→ {by substituting: "320 − 8y" for: "8(40 − y)" ; as follows:
_________________________________________________
→ 320 − 8y + 3y = 170 ;
Combine the "like terms" on the "left-hand side" of the equation:
-8y + 3y = -5y ;
And rewrite the equation:
320 − 5y = 170 ;
Subtract "320" from each side of the equation:
320 − 5y − 320 = 170 − 320 ;
to get:
-5y = -150 ;
Divide EACH SIDE of the equation by "-5" ;
to isolate "y" on one side of the equation; & to solve for "y" ;
-5y / -5 = -150 / -5 ;
to get:
y = 30 ;
To solve for "x";
x + y = 40 ;
x = 40 − y ;
Plug in: "30" for "y" ; to solve for "x" ;
x = 40 − (30) ;
x = 10 ;
__________________________
So; the answer is: "x = 10, y = 30": ; or, write as: " (10, 30)" .
_____________________________________________________
To confirm our answer:
_____________________________________________________
Let us plug in our values for "x" and 'y" ; into the equation:
" (0.08)x + (0.03)y = 1.7 " ; to see if the equation holds true;
→ (0.08)*(10) + (0.03)(30) = ? 0.9 ? ;
→ 0.8 + 0.9 = ? 1.7 ? YES!
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