Question

the sum of three numbers is 61. the second number is 6 times the first. and the third number is 5 more than the second. find the numbers

Answer 1

$x-the\ first\ number\\6x-the\ second\ number\\6x+5-the\ third\ number\\\\(x)+(6x)+(6x+5)=61\\x+6x+6x+5=61\\13x+5=61\ \ \ \ |subtract\ 5\ from\ both\ sides\\13x=56\ \ \ \ |divide\ both\ sides\ by\ 13\\\boxed{x=\frac{56}{13}\to x=4\frac{4}{13}-the\ first\ number}\\\boxed{6x=6\cdot\frac{56}{13}=\frac{336}{13}=25\frac{11}{13}-the\ second\ number}\\\boxed{6x+5=25\frac{11}{13}+5=30\frac{11}{13}-the\ third\ number}$