When solving a system of equations, use the elimination method or the substitution method.

Since the second equation can be easily solved for aa, the substitution method is the most appropriate method for this system.

Solving a+5b=22 for aa gives a=22−5b.

Now, substitute a=22−5b into the first equation of 3a+5b=26 and solve for b:

3a+5b=26 3(22-5b)+5b=26 66-15b+5b=26 66-10b=26 -10b=-40 b=4

To find aa, substitute b=4 into a=22−5b. This gives a=22−5(4)=2.

The solution of the system is then (a,b)=(2,4).

To find aa, substitute b=4b=4 into a=22−5b. This gives a=22−5(4)=2.

The solution of the system is then (a,b)=(2,4).

Since the second equation can be easily solved for aa, the substitution method is the most appropriate method for this system.

Solving a+5b=22 for aa gives a=22−5b.

Now, substitute a=22−5b into the first equation of 3a+5b=26 and solve for b:

3a+5b=26 3(22-5b)+5b=26 66-15b+5b=26 66-10b=26 -10b=-40 b=4

To find aa, substitute b=4 into a=22−5b. This gives a=22−5(4)=2.

The solution of the system is then (a,b)=(2,4).

To find aa, substitute b=4b=4 into a=22−5b. This gives a=22−5(4)=2.

The solution of the system is then (a,b)=(2,4).