The u-drive rent-a-truck company plans to spend $15 million on 310 new vehicles. each commercial van will cost $35 comma 000 , each small truck $70 comma 000 , and each large truck $60 comma 000 . past experience shows that they need twice as many vans as small trucks. how many of each type of vehicle can they buy?

Let x, y, and z be number of vans, small trucks and large trucks respectively. Therefore,
x+y+z = 310 (1)

But, the company will spend $15 million. Then,
35000x+ 70000y +60000z = 15,000,000
 35x + 70y + 60z = 15000 (2)

Additionally, twice as many van as small trucks will be required. Then,
x = 2y (3).

Substituting for x in (1) and (2);
2y +y +z = 310 => 3y +z =310 (4)
35(2y) + 70y + 60z = 15000 => 70y+70y+60z = 15000 => 140y + 60z = 15000 (5)

Solving (4) and (5)
From (4), z = 310 -3y
Then, using (5)
140y +60(310-3y) = 15000
140y + 18600 - 180y = 15000
-40y = -3600
y = 3600/40= 90
Then, z= 310 -3y = 310 -3(90) = 310 -270 = 40
And, x = 2y = 2(90) = 180

Therefore, the number will be:
Vans (x) = 180
Small trucks (y) = 90
Large  trucks (z) = 40

Basic algebra equation. first set up your equation: divide out 9.and now you are left with 2x + 4 = 8.