1/8 of pizza cost 2 dollars. In an hour, 6 1/4 pizzas were sold. How much money was made at the food booth in that hour

1 answer:

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First of all, you would make 6 1/4 into an improper fraction. This makes it 25/4, then you multiply it by 2 to get 50/8 and have common denominators. Divide that by 1/8, which would be 50. Then since each 1/8 costs 2 dollars you would multiply 50 by 2. So, the answer would be that they made 100 dollars.

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The sum of twice a first number and five times a second number is 78. If the second number is subtracted from five times the fir

ss7ja [257]

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.
_____________________________________________________
→ Rewrite the equation as:
_____________________________________________________
→ " 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" .
_____________________________________________________
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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Igoryamba

425+650=1075 1075x3=3225

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