Question

My classesExample 5: The number of calories in a piece of pie is 20 less thanthree times the number of calories in a scoop of ice cream. The pieand ice cream together have 500 calories. How many calories are ineach?x = pie, y = ice creamMULTIPLE CHOICE QUESTIONx= 3y - 20x + y = 500How many calories are in a scoop of icecream?1090370250130

Answer 1

We will investigate the number of calories in ice cream and pie. We will define the variables as such:

`\begin{gathered} x\text{ = Number of calories in pie} \\ y\text{ = Number of calories in ice-cream} \end{gathered}`

From the given data a relationhsip between the number of calories in a pie ( x ) and the number of calories in an ice-cream scoop are given by the following statement:

" The number of calories in a piece of pie is 20 less than three times the number of calories in a scoop of ice cream. "

We will go ahead and decrypt the above given statement and express it mathematically in terms of declared variables ( x and y ) as follows:

`*\text{ = 3y - 20 }\ldots\textcolor{#FF7968}{Eq1}`

The next statement gives us the combined calories in both ice-cream scoop and a pie:

" The pie and ice cream together have 500 calories. "

We will go ahead and decrypt the above given statement and express it mathematically in terms of declared variables ( x and y ) as follows:

`x\text{ + y = 500 }\ldots\textcolor{#FF7968}{Eq2}`

We have two equations ( derived ) and two variables ( x and y ):

`\begin{gathered} *\text{ - 3y = -20 }\ldots Eq1\text{ } \\ x\text{ + y = 500 }\ldots Eq2 \end{gathered}`

We will solve the two equations ( Eq1 and Eq2 ) simultaneously using elimation method as follows:

`\begin{gathered} x\text{ - 3y = -20 --- > }\text{\textcolor{#FF7968}{-x}}\text{ + 3y = 20} \\ \textcolor{#FF7968}{x}\text{ + y = 500} \\ =================== \\ 4y\text{ = 520 } \\ y\text{ = }(520)/(4) \\ \textcolor{#FF7968}{y}\text{\textcolor{#FF7968}{ = 130 calories}} \end{gathered}`

We first multiplied the ( Eq1 ) by ( -1 ) then added the modified ( Eq1 ) to ( Eq2 ) whilst eliminating ( x ) and solved for the number of calories in an ice-cream scoop ( y ) i.e 130 calories.

Then we will back substitute the value of ( y ) into ( Eq2 ) and solve for ( x ):

`\begin{gathered} x\text{ + y = 500} \\ x\text{ + 130 = 500} \\ \textcolor{#FF7968}{x}\text{\textcolor{#FF7968}{ = 370 calories}} \end{gathered}`

Therefore, the number of calories in each confectionary is as such:

`\begin{gathered} \textcolor{#FF7968}{x=370}\text{\textcolor{#FF7968}{ calories in pie}} \\ \textcolor{#FF7968}{y=130}\text{\textcolor{#FF7968}{ calories in an ice-cream scoop}} \end{gathered}`