Greetings!
Let x represent the number of 2 pointers made.
Let y represent the number of 3 pointers made.
We have to create a system of equations to solve this problem:
[tex] \left \{ {{x+y=23} \atop {2x+3y=53}} \right. [/tex]
Isolate for [tex]x[/tex] in Equation #1:
[tex]x+y=23[/tex]
[tex]x=23-y[/tex]
Substitute the new Equation #1 into Equation #2:
[tex]2x+3y=53[/tex]
[tex]2(23-y)+3y=53[/tex]
Solve for [tex]y[/tex] :
[tex]2(23-y)+3y=53[/tex]
Distribute the Parenthesis:
[tex]46-2y+3y=53[/tex]
Combine like terms:
[tex]46+y=53[/tex]
Add -46 to both sides:
[tex](46+y)+(-46)=(53)+(-46)[/tex]
[tex]y=7[/tex]
Substitute this value into Equation #1:
[tex]x+y=23[/tex]
[tex]x+(7)=23[/tex]
Add -7 to both sides:
[tex](x+7)+(-7)=(23)+(-7)[/tex]
[tex]x=16[/tex]
The Solution Is:
[tex]\boxed{(16,7)}[/tex]
Jake made 16 two-pointers and 7 three-pointers!
I hope this helped!
-Benjamin
