Answer:
Option no C
Step-by-step explanation:
Given:
Price of crackers $ 9 for 3 boxes
Price of Granola Bars $6 for 3 boxes
Total bought $ 20 worth of snacks
To find the equation
Solution:
Price of cracker $ 9 for 3 boxes
price of one box of a cracker = $ 9 / 3
=$ 3 per box
Price of granola bars $ 6 for 3 boxes
price of one box of a granola bars = $ 6 / 3
=$ 2 per box
Let total crackers bought are x
price of x crackers will be 3x
Let granola bars bought are y
price of y granola bars will be 2y
as total price is 20
so price of x crackes + y granolas = 20
which will be
3x + 2y = 20
which will be option no C
Answer:
I can't see it the picture is not clean or close enough but it's in the bottom left
Answer:
2x
Step-by-step explanation:
We can first pull the x out of every term:
Then, simplify:
So, the answer is 2x.
The numbers are: "9" and "12" .
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Explanation:
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Let: "x" be the "first number" ; AND:
Let: "y" be the "second number" .
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From the question/problem, we are given:
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2x + 5y = 78 ; → "the first equation" ; AND:
5x − y = 33 ; → "the second equation" .
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From "the second equation" ; which is:
" 5x − y = 33" ;
→ Add "y" to EACH side of the equation;
5x − y + y = 33 + y ;
to get: 5x = 33 + y ;
Now, subtract: "33" from each side of the equation; to isolate "y" on one side of the equation ; and to solve for "y" (in term of "x");
5x − 33 = 33 + y − 33 ;
to get: " 5x − 33 = y " ; ↔ " y = 5x − 33 " .
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Note: We choose "the second equation"; because "the second equation"; that is; "5x − y = 33" ; already has a "y" value with no "coefficient" ; & it is easier to solve for one of our numbers (variables); that is, "x" or "y"; in terms of the other one; & then substitute that value into "the first equation".
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Now, let us take "the first equation" ; which is:
" 2x + 5y = 78 " ;
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We have our obtained value; " y = 5x − 33 " .
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We shall take our obtained value for "y" ; which is: "(5x− 33") ; and plug this value into the "y" value in the "first equation"; and solve for "x" ;
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Take the "first equation":
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→ " 2x + 5y = 78 " ; and write as:
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→ " 2x + 5(5x − 33) = 78 " ;
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Note the "distributive property of multiplication" :
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a(b + c) = ab + ac ; AND:
a(b − c) = ab − ac .
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So; using the "distributive property of multiplication:
→ +5(5x − 33) = (5*5x) − (5*33) = +25x − 165 .
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So we can rewrite our equation:
→ " 2x + 5(5x − 33) = 78 " ;
by substituting the: "+ 5(5x − 33) " ; with: "+25x − 165" ; as follows:
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→ " 2x + 25x − 165 = 78 " ;
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→ Now, combine the "like terms" on the "left-hand side" of the equation:
+2x + 25x = +27x ;
Note: There are no "like terms" on the "right-hand side" of the equation.
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→ Rewrite the equation as:
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→ " 27x − 165 = 78 " ;
Now, add "165" to EACH SIDE of the equation; as follows:
→ 27x − 165 + 165 = 78 + 165 ;
→ to get: 27x = 243 ;
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Now, divide EACH SIDE of the equation by "27" ; to isolate "x" on one side of the equation ; and to solve for "x" ;
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27x / 27 = 243 / 27 ;
→ to get: x = 9 ; which is "the first number" .
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Now; Let's go back to our "first equation" and "second equation" to solve for "y" (our "second number"):
2x + 5y = 78 ; (first equation);
5x − y = 33 ; (second equation);
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Start with our "second equation"; to solve for "y"; plug in "9" for "x" ;
→ 5(9) − y = 33 ;
45 − y = 33;
Add "y" to each side of the equation:
45 − y + y = 33 + y ; to get:
45 = 33 + y ;
↔ y + 33 = 45 ; Subtract "33" from each side of the equation; to isolate "y" on one side of the equation ; & to solve for "y" ;
→ y + 33 − 33 = 45 − 33 ;
to get: y = 12 ;
So; x = 9 ; and y = 12 . The numbers are: "9" and "12" .
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To check our work:
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1) Let us plug these values into the original "second equation" ; to see if the equation holds true (with "x = 9" ; and "y = 12") ;
→ 5x − y = 33 ; → 5(9) − 12 =? 33 ?? ; → 45 − 12 =? 33 ?? ; Yes!
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2) Let us plug these values into the original "second equation" ; to see if the equation holds true (with "x = 9" ; and "y = 12") ;
→ 2x + 5y = 78 ; → 2(9) + 5(12) =? 78?? ; → 18 + 60 =? 78?? ; Yes!
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So, these answers do make sense!
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