I really need help setting this problem up

Question

In a basketball game, Team A defeated Team B with a score of 92. Team A won by scoring a combination of two-point baskets, three-point baskets, and one-point free throws. The number of two-point baskets was four more than the number of free throws. The number of free throws was four more than five times the number of three-point baskets. What combination of scoring accounted for the Team A's 92 points?

 

two-point baskets =

 

three-point baskets= 

 

free throws=

Answer 1

Hi Blake, the first tendency is to set up an equation something like this:

 

Let x equal the 3-point baskets

 

Then 5x + 4 equals the 1-point baskets

 

And (5x + 4) + 4 equals the 2 point baskets

 

The equation would be: x + (5x + 4) + ((5x + 4) + 4) = 92

 

The problem with this equation is it does not take into account the value of each of the three types of baskets, 1-point, 2-points, or 3-points.

 

The correct equation:

 

The correct equation takes into consideration the point value of each type of basket:

 

Three point      One point               two point                 Total

baskets            baskets                 baskets                    points

3 * x +            1 * (5x + 4) +       2 * ((5x + 4) + 4) = 92

 

This becomes:

 

3x + 5x + 4 + 10x + 16 = 92

 

18x + 20 = 92

 

Subtract 20 from both sides gives:

 

18x = 72

 

Divide both sides by 18 gives:

 

x = 4, or 4 3-point baskets were made

 

5x + 4, or 24 1-point baskets were made

 

(5x + 4) + 4, or 28 2-point baskets were made.

 

Solution:

 

24 1-point baskets

 

28 2-point baskets

 

 4 3-point baskets

 

Questions?

 

 

 

 

Answer 2

Hello Blake,

 

First we need to set up just a few equations and assign variables to the unknowns. 

 

Let a = free throw; b = 2 pointers; and c = 3 pointers 

 

All of these scoring types will total to be 92 points.

 

The number of free throw plus the number of 2 pointers times the value of two plus the number of 3 pointers times the value of 3. The equation looks like the following:

 

a + 2b + 3c = 92 

 

We will need to manipulate this equation based off of the provided information: 

 

b = a + 4 

 

a = 5c + 4 

 

We would need to solve each of these equations with respect to free throws, as we have information with regards to free throws (a). 

 

a + 2(a+4) + 3(a-4/5) = 92 

 

Try solving for (a) yourself. You should get 28 And be able to solve for the number of 2 pointers and 3 pointers from this. 

 

Hope this helps! 

 

Answer 3

Let A = 3-point baskets

Let B = 2-point baskets

Let C = 1-point free throw baskets

 

The total points of these baskets give a score of 92 points.  We can set up equations that represent what the problem is saying.

 

A + B + C = 92                   eq1

B = C + 4                           eq2

C = 5A + 4                          eq3

 

 

We have here a system of equations.  What we want to do is use substitution/elimination methods, to reduce the number of equations to work with.  This way, it will be easier to solve.  We want to get an equation with only one variable.

 

Substitute eq2 into eq1.  Keep eq3.

 

A + (C + 4) + C = 92           

A + 2C + 4 = 92                   

A + 2C = 88                        new eq1

 

-5A + C = 4                           eq3

 

 

We now have 2 equations with 2 variables.

 

  A + 2C = 88             new eq1

-5A + C = 4                     eq3

 

 

Multiply new eq1 by 5.

 

5A + 10C = 440               new eq1

-5A + C = 4                        eq3

 

 

Add the equations to eliminate the A terms.

 

11C = 444

 

    C = 444 / 11

 

 

But wait.  This value of C is NOT a whole number.  The number of each type of basket made by Team A must be whole numbers.  Therefore, there is an error in the problem you posted.