Question 1051170
{{{A+B+C+D=59}}}
{{{D=4A}}}
{{{B=C=A+5}}}
Substituting into the first equation,
{{{A+(A+5)+(A+5)+4A=59}}}
{{{7A+10=59}}}
{{{7A=49}}}
{{{A=7}}}
So then,
{{{B=C=7+5=12}}}
{{{D=4(7)=28}}}