[tex]\huge \bf༆ Answer ༄[/tex]
Let the capacity of bus be x students
And van be y students, now ;
From the given statements we get two equations ~
[tex]{ \qquad{ \sf{ \dashrightarrow}}} \: \: \sf \:2x + 10y = 260 \: \: \: \: \: (1)[/tex]
[tex]{ \qquad{ \sf{ \dashrightarrow}}} \: \: \sf \:13x + 5y = 670 \: \: \: \: \: (2)[/tex]
multiply the equation (2) with 2 [ it won't change the values ]
[tex]{ \qquad{ \sf{ \dashrightarrow}}} \: \: \sf \:2x + 10y = 260 \: \: \: \: \: \: \: \: \: (1)[/tex]
[tex]{ \qquad{ \sf{ \dashrightarrow}}} \: \: \sf \:26x + 10y = 1340 \: \: \: \: \: (3)[/tex]
Now, deduct equation (1) from equation (3)
[tex]{ \qquad{ \sf{ \dashrightarrow}}} \: \: \sf \:26x + 10y - 2x - 10y = 1340 - 260[/tex]
[tex]{ \qquad{ \sf{ \dashrightarrow}}} \: \: \sf \:24x = 1080[/tex]
[tex]{ \qquad{ \sf{ \dashrightarrow}}} \: \: \sf \:x = 1080 \div 24[/tex]
[tex]{ \qquad{ \sf{ \dashrightarrow}}} \: \: \sf \:x = 45[/tex]
Therefore each bus can carry (x) = 45 students
Now, plug the value of x in equation (1) to find y ~
[tex]{ \qquad{ \sf{ \dashrightarrow}}} \: \: \sf \:2x + 10y = 260[/tex]
[tex]{ \qquad{ \sf{ \dashrightarrow}}} \: \: \sf \:(2 \times 45) + 10y = 260[/tex]
[tex]{ \qquad{ \sf{ \dashrightarrow}}} \: \: \sf \:90 + 10y = 260[/tex]
[tex]{ \qquad{ \sf{ \dashrightarrow}}} \: \: \sf \:10y = 260 - 90[/tex]
[tex]{ \qquad{ \sf{ \dashrightarrow}}} \: \: \sf \:10y = 170[/tex]
[tex]{ \qquad{ \sf{ \dashrightarrow}}} \: \: \sf \:y = 170 \div 10[/tex]
[tex]{ \qquad{ \sf{ \dashrightarrow}}} \: \: \sf \:y = 17[/tex]
Hence, each van can carry (y) = 17 students in total.
