The school athletic director had a budget of $500 to purchase 164 items for the soccer team. She purchased vests for $2.50 each, soccer balls for $9.25 each, and cones for $0.75 each. She purchased 40 more cones than soccer balls.

The question is incomplete:

The school athletic director had a budget of $500 to purchase 164 items for the soccer team. She purchased vests for $2.50 each, soccer balls for $9.25 each, and cones for $0.75 each. She purchased 40 more cones than soccer balls. How many of each item can she purchase?

She can purchase 60 vests, 32 soccer balls and 72 cones.

Step-by-step explanation:

You can write the following equations:

x+y+z=164 (1)

2.50x+9.25y+0.75z=500 (2)

y=z-40 (3), where:

x is the amount of vests

y is the amount of soccer balls

z is the amount of cones

First, you have to replace (3) in (1) and (2):

x+z-40+z=164

x+2z=204 (4)

2.50x+9.25(z-40)+0.75z=500

2.50x+9.25z-370+0.75z=500

2.50x+10z=870 (5)

Now, you have two equations:

x+2z=204 (4)

2.50x+10z=870 (5)

Then, you have to isolate x in (4)

x=204-2z (6)

After this, you can replace (6) in (5):

2.50(204-2z)+10z=870

510-5z+10z=870

5z=870-510

5z=360

z=360/5

z=72

Next, you can replace the value of z in (6):

x=204-2z

x=204-2(72)

x=204-144

x=60

Finally, you can replace the value z in (3):

y=z-40

y=72-40

y=32

According to this, the answer is that she can purchase 60 vests, 32 soccer balls and 72 cones.

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