Respuesta :
Let's represent the cherry pies and apple pies with variables
[tex]\begin{gathered} \text{cherry pies=x} \\ \text{apple pies=y} \end{gathered}[/tex]The equation of julio's sales will be
[tex]\begin{gathered} 3\times x+14\times y=\text{ \$300} \\ 3x+14y=\text{ \$300 }\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots.(Eqn\text{ 1)} \end{gathered}[/tex]The equation of Jisel's sales will be
[tex]\begin{gathered} 3\times x+9\times y=\text{ \$210} \\ 3x+9y=\text{ \$210}\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots..(Eqn\text{ 2)} \end{gathered}[/tex]Let's combine Equations (1 )and (2 )and solve them simultaneously
[tex]\begin{gathered} 3x+14y=\text{ \$300}\ldots\ldots\ldots\ldots\ldots\text{.}(\text{Eqn 1)} \\ 3x+9y=\text{ \$210}\ldots\ldots\ldots\ldots\ldots.(Eqn\text{ 2)} \end{gathered}[/tex]Subtract Eqn (2) from Eqn (1) we will have
[tex]\begin{gathered} 3x+14y-(3x+9y)=\text{ \$300- \$210} \\ 3x+14y-3x-9y=\text{ \$90} \\ 3x-3x+14y-9y\text{ =\$90} \\ 5y=\text{ \$90} \\ \frac{5y}{5}=\text{ }\frac{90}{5} \\ y=\text{ \$18} \end{gathered}[/tex]Substitute y= $18 in Eqn 1 we will have
[tex]\begin{gathered} 3x+14y=\text{ \$300} \\ 3x+14(\text{ \$18)= \$300} \\ 3x+\text{ \$252= \$300} \\ 3x=\text{ \$300- \$252} \\ 3x=\text{ \$48} \\ \frac{3x}{3}=\frac{48}{3} \\ x=\text{ \$16} \end{gathered}[/tex]Therefore,
The equation for Julio's sales is 3x+14y= $300
The equation for Jisel's sales is 3x+9y= $210
The cost of one cherry pie is x= $16
The cost of one apple pie is y= $18
