Question 622631
Hi, there--
.
We can solve this problem using algebra.
.
Let x be the number the 1-point free throws.
let y be the number of 2-point shots.
let z be the numb of 3-point shots.
.
Now we need to write three equations using the information in the problem to model this situation.
.
The Lakers scored a total of 90 points, so the points from each type of shot must sum up to 90. Three-pointers are worth 3 times the number of shots, or 3z. Two-point baskets are worth two times the number of shots, or 2y. Free throws are worth 1 times the number of shots, or x. In algebra, we can write this as
.
{{{x+2y+3z=90}}}
.
There were 11 times as many two-point baskets as three-point baskets. In other words,
[the number of 2-point baskets] = [11] * [the number of 3-point baskets].
.
In algebra, we write this as
{{{y=11z}}}
.
There were five times as many free-throws as three-point baskets. In other words,
[the number of free throws] = [5] * [the number of 3-point baskets].
.
In algebra, we write this relationship as,
{{{x=5z}}}
.
Now we have a system of three equations and three variables. We will solve for x, y, and z.
{{{x+2y+3z=90}}}
{{{y=11z}}}
{{{x=5z}}}
.
We begin by using the substitution method. The third equation states that x=5z, so substitute 5z for x in the first equation.
{{{x+2y+3z=90}}}
{{{(5z)+2y+3z=90}}}
.
Simplify.
{{{2y+8z=90}}}
.
The second equation states that y=11z, so substitute 11z for y.
{{{2(11z)+8z=90}}}
.
Solve for z.
{{{22z+8z=90}}}
{{{30z=90}}}
{{{z=3}}}
.
The Lakers made 3 three-point shots. 
They made 11 times as many two-point shots, or 33 two-point shots.
They made 5 times as many free throws, or 15 free throws.
.
We need to make sure that this scoring combination gave them a total of 90 points:
3 three-pointers give 9 points.
33 two-point shots give 66 points.
15 free throws give 15 points.
9+66+15 does equal 90 points all together.
.
Tht's it. Feel free to email me if you have questions about any part of the explanation.
.
Ms.Figgy
math.in.the.vortex@gmail.com