a = hours worked by the first mechanic
b = hours worked by the second mechanic.
since the first mechanic charges $55 per hour, then for "a" hours that'd be a total of 55*a or 55a, likewise, for the second mechanic that'd be a total charge of 80*b or 80b.
we know all hours combined are 15, so then a + b = 15.
we also know that all charges combined are $950, so 55a + 80b = 950.
[tex]\bf \begin{cases} a+b=15\\ \boxed{b}=15-a\\ \cline{1-1} 55a+80b=950 \end{cases}\qquad \qquad \stackrel{\textit{substituting on the 2nd equation}}{55a+80\left( \boxed{15-a} \right)=950} \\\\\\ 55a+1200-80a=950\implies -25a+1200=950\implies -25a=-250 \\\\\\ a=\cfrac{-250}{-25}\implies \blacktriangleright a = 10\blacktriangleleft \\\\\\ \stackrel{\textit{since we know that}}{b=15-a\implies }b=15-10\implies \blacktriangleright b=5 \blacktriangleleft[/tex]
