Answer :
Let x be the length of a fire truck and y the length of the ambulance.
The total length of 3 fire trucks and 4 ambulances will be:
[tex]3x+4y[/tex]To this we have to add the 1 foot space for each vehicle, in this case we have seven vehicles then the total length in the first case is:
[tex]3x+4y+7[/tex]and this can be parked in a 152 lane, tha means that:
[tex]3x+4y+7=152[/tex]Now, the total length of 2 fire trucks and 5 ambulances is:
[tex]2x+5y[/tex]To this, once again, we have to add the foot between them and since we have 7 vehicles, then:
[tex]2x+5y+7[/tex]but this can be achieved in a 163 long lane, then:
[tex]2x+5y+7=136[/tex]Hence we have the system of equations:
[tex]\begin{gathered} 3x+4y+7=152 \\ 2x+5y+7=136 \end{gathered}[/tex]Solving this system will give us the lengths of the fire trucks and ambulance. To solve the system we first write it as:
[tex]\begin{gathered} 3x+4y=145 \\ 2x+5y=129 \end{gathered}[/tex]Now we muñltiply the first equation by 2 and the second by -3, then we have:
[tex]\begin{gathered} 6x+8y=290 \\ -6x-15y=-387 \end{gathered}[/tex]now, we add the equations. This gives us a equation of only the y variable:
[tex]\begin{gathered} -7y=-97 \\ y=\frac{97}{7} \end{gathered}[/tex]Then the length of the anbulances is 97/7 feet.
Once we have this value we plug it in the first equation and solve for x:
[tex]\begin{gathered} 3x+4(\frac{97}{7})=145 \\ 3x=145-4\cdot\frac{97}{7} \\ 3x=\frac{627}{7} \\ x=\frac{209}{7} \end{gathered}[/tex]Hence, the length of the fire trucks is 209/7 feet.
Now that we have the lengths of each type of vehicle we can know how much we need for 1 fire truck and 5 ambulances (rememeber that we have to add 1 foot per vehicle, in this case 6 in total):
[tex]\frac{209}{7}+5(\frac{97}{7})+6=\frac{736}{7}[/tex]Therefore, we need a lane of 736/7 feet (or approximately 105.14 feet).