Question 214345
Let x = the number of problems solved correctly and y = the number of problems solved incorrectly.
We can express the problem algebraically in the two equations as:
1). {{{50x-25y = 0}}}  "...50 centavos for each problem correctly solved (x) and ...fine him 25 centavos for each incorrect solution (y)."
2). {{{x+y = 30}}} The number of problems = 30. Rearrange this as: {{{x = 30-y}}} and substitute into the first equation.
1a). {{{50(30-y)-25y = 0}}} Simplify.
1b). {{{1500-50y-25y = 0}}}
1c). {{{1500-75y = 0}}} Add 75y to both sides.
1d). {{{1500 = 75y}}} Divide both sides by 75.
1e). {{{20 = y}}} and:
{{{x = 30-20}}}
{{{x = 10}}}
The boy solved 10 of the problems correctly.