First problem: let [tex]b[/tex] and [tex]h[/tex] be the base and height of the rectangle, respectively. We know that [tex]b=11+h[/tex] (the base if 11 longer than the height), and that [tex]2(b+h)=58[/tex] (the perimeter is 58 centimeters).
So, we have the system
[tex]\begin{cases}b=11+h\\2(b+h)=58\end{cases}\iff\begin{cases}b=11+h\\b+h=29\end{cases}[/tex]
Use the first equation to substitute into the second:
[tex]b+h=11+h+h=2h+11=29 \iff 2h=18 \iff h=9[/tex]
And since the base is 11 centimeters longer, we have
[tex]b=11+h=11+9=20[/tex]
Second problem: Let [tex]m,n,o[/tex] be the number of cards owned by Mark, Norm and Oscar, respectively. We know that:
[tex]m+n+o=810[/tex] (they have 810 cards in total)
[tex]n=m+30[/tex] (Norm has 30 more than Mark)
[tex]o=2m[/tex] (Oscar has twice as much as Mark)
The second and third equation express [tex]n[/tex] and [tex]o[/tex] in terms of [tex]m[/tex], and substituting those expressions in the first equation we have
[tex]m+n+o=m+(m+30)+2m=810 \iff 4m+30=810 \iff 4m=780 \iff m=195[/tex]
And then we can use again the second and third equations:
[tex]n=m+30=195+30=225[/tex]
[tex]o=2m=2\cdot 195=390[/tex]
