Answer:
[tex]\large \boxed{\text{15 hockey cards and 5 baseball cards}}[/tex]
Step-by-step explanation:
Let h = the number of hockey
and b = the number of baseball cards
You have two conditions:
[tex]\begin{array}{lrcl}(1) & h + b & = & 20\\(2) & h & = & b + 10 \\\end{array}[/tex]
Solve the equations for h and b
[tex]\begin{array}{rcll}(3) \; b + 10 + b & = & 20 & \text{Substituted (2) into (1)}\\2b + 10& = &20 &\text{Simplified}\\2b & = &10 &\text{Subtracted 10 from each side}\\(4)\qquad \qquad b & = &\mathbf{5}& \text{Divided each side by 2}\\h + 5 & = & 20 & \text{Substituted (4) into (1)}\\h & = &\mathbf{15}&\text{Divided each side by 2} \\\end{array}\\\text{Kirk has $\large \boxed{\textbf{15 hockey cards and 5 baseball cards}}$}[/tex]}
Check:
[tex]\begin{array}{ccc}15 + 5 = 20 & \qquad & 5 + 15 = 20\\20 = 20 & \qquad & 20 = 20\\\end{array}[/tex]
OK.
