let numbers of books be 'b' and numbers of CDs be 'c'
We can set up two equations:
Equation [1] ⇒ [tex]b+c=20[/tex]
Equation [2] ⇒ [tex]1.50b+5c=54.50[/tex]
We are solving for the number of books and the number of CDs bought
When we have two equations in terms of two different variables; [tex]b[/tex] and [tex]c[/tex], that we need to solve, then this becomes a simultaneous equation problem.
First, rearrange Equation [1] to make either [tex]b[/tex] or [tex]c[/tex] the subject:
[tex]b+c=20[/tex]
[tex]b=20-c[/tex]
Then we substitute [tex]b=20-c[/tex] into Equation [2]
[tex]1.50b+5c=54.50[/tex]
[tex]1.50(20-c)+5c=54.50[/tex]
[tex]30-1.50c+5c=54.50[/tex]
[tex]5c-1.5c=54.50-30[/tex]
[tex]3.5c=24.50[/tex]
[tex]c=7[/tex]
Now we know the value of [tex]c[/tex] which is [tex]c=7[/tex], substitute this value into [tex]b=20-c[/tex] we have [tex]b=20-7=13[/tex]
Answer:
Numbers of books = 13
Numbers of CDs = 7
