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
balls. How many of each item can purchase?

She purchased 72 cones, 32 balls, and 60 vests.

Step-by-step explanation:

The total number of items is 164.

The price of the vests is $2.50 each.

The price of the soccer balls is $9.25 each.

The price of the cones is $0.75 each.

The total budget was  $500.

If V is the number of vests, B the number of balls, and C the number of cones, we have that:

V + B + C = 164.

V*$2.50 + B*$9.25 + C*$0.75 = $500.

and:

"She purchased 40 more cones than  balls."

C = B + 40.

Then we have a system of 3 equations and 3 variables:

V + B + C = 164.

V*$2.50 + B*$9.25 + C*$0.75 = $500.

C = B + 40.

Here the first step is to replace the third equation into the other two, in this way we eliminate one variable and one equation:

V + B + (B + 40) = 164

V*$2.50 + B*$9.25 + (B + 40)*$0.75 = $500.

Now we want to isolate one variable in one of the equations, i will isolate V in the first equation:

V = 164 - 2*B - 40 = 124 - 2*B

Now i can replace this into the other equation:

(124 - 2*B)*$2.50 + B*$9.25 + (B + 40)*$0.75 = $500

Now let's solve this for B.

$310 - $5*B + B*$9.25 + B*$0.75 + $30 = $500

B*( $9.25 - $5 + $0.75) = $500 - $310 - $30

B*$5 = $160

B = 160/5 =  32

Now we know the value of B.

And by the equations above:

V = 124 -2*B = 124 - 2*32 = 60

C = B + 40 = 32 + 40 = 72.

Then:

She purchased 72 cones, 32 balls and 60 vests.

its a straight line parallel to the time axis

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step-by-step explanation: