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Sagot :
We are given that high school A filled 3 vans and 8 buses with 322 students. Let "x" be the number of students in a van and "y" the number of students in the bus then this can be written mathematically as:
[tex]3x+8y=322,(1)[/tex]High school B filled 8 vans and 6 buses with 322. Since each van and bus has the same number of students then we can write this as:
[tex]8x+6y=322,(2)[/tex]We get 2 equations and 2 variables. To determine the solution we will solve for "x" in equation (1). To do that we will subtract "8y" from both sides:
[tex]3x=322-8y[/tex]Now, we divide both sides by 3:
[tex]x=\frac{322-8y}{3}[/tex]Now, we substitute the value of "x" in equation (2):
[tex]8(\frac{322-8y}{3})+6y=322[/tex]Now, we apply the distributive law on the parenthesis:
[tex]\frac{2576-64y}{3}+6y=322[/tex]Now, we multiply both sides by 3:
[tex]2576-64y+18y=966[/tex]Now, we add like terms:
[tex]2576-46y=966[/tex]Now, we subtract 2576 from both sides:
[tex]\begin{gathered} -46y=966-2576 \\ -46y=-1610 \end{gathered}[/tex]Now, we divide both sides by -46:
[tex]y=-\frac{1610}{-46}=35[/tex]Now, we substitute the value of "y" in equation (1):
[tex]3x+8(35)=322[/tex]Solving the product:
[tex]3x+280=322[/tex]Now, we subtract 280:
[tex]\begin{gathered} 3x=322-280 \\ 3x=42 \end{gathered}[/tex]Dividing by 3:
[tex]x=\frac{42}{3}=14[/tex]Therefore, there are 14 students in the van and 35 in the bus.
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