Answer:
We can express the number of baseball Linda has in two ways.
Either using variable c and y or using their values as follows,
[tex]\Rightarrow y=5+\frac{1}{3}c[/tex]
or [tex]\Rightarrow 38=5+\frac{1}{3} \times 99[/tex]
Step-by-step explanation:
Given : Jerrod has c baseball cards and Linda has 38 baseball cards.
Also, number of baseball cards Linda has is 5 more than one-third the number Jerrod has.
Mathematically, it can be stated as ,
[tex]\text{Number of baseball cards Linda has}=5+\frac{1}{3}c[/tex]
Let Linda has y number of baseball cards, then
[tex]\Rightarrow y=5+\frac{1}{3}c[/tex]
Also, it is given Linda has 38 baseball cards, so y = 38
[tex]\Rightarrow 38=5+\frac{1}{3}c[/tex]
From here we can solve for c,
[tex]38-5=\frac{1}{3}c[/tex]
[tex]33=\frac{1}{3}c[/tex]
[tex]c=33 \times 3[/tex]
[tex]\Rightarrow c=99[/tex]
Thus, Jerrod has 99 baseball cards.
We can express the number of baseball Linda has in two ways.
Either using variable c and y or using their values as follows,
[tex]\Rightarrow y=5+\frac{1}{3}c[/tex]
or [tex]\Rightarrow 38=5+\frac{1}{3} \times 99[/tex]
