Three numbers

Find three numbers so that the second number is four times greater than the first and the third is lower by five than the second number. Their sum is 67.

Final Answer:

a =  8
b =  32
c =  27

Step-by-step explanation:

$$\begin{gathered} b=4a \\ c=b-5 \\ a+b+c=67 \\ a+4a+(b-5)=67 \\ a+4a+(4a-5)=67 \\ a+4\cdot a+(4\cdot a-5)=67 \\ 9a=72 \\ a=\frac{72}{9}=8 \\ a=8 \end{gathered}$$

$b=4\cdot a=4\cdot 8=32$

$$\begin{gathered} c=b-5=32-5=27 \\ \text{ Verifying Solution: } \\ s=a+b+c=8+32+27=67 \end{gathered}$$

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