Question

# I am looking for 3 consecutive multiples of 7. The first multiple of 7 is (x-7), the 2nd multiple of 7 is (x), and the 3rd multiple of 7 (x+7). Find the numbers.

Factors and multiples
I am looking for 3 consecutive multiples of 7. The sum of the squares of the 2 smallest is 147 less than the square of the largest. The first multiple of 7 is $$\displaystyle{\left({x}-{7}\right)}$$, the 2nd multiple of 7 is (x), and the 3rd multiple of $$\displaystyle{7}{\left({x}+{7}\right)}$$. Find the numbers.

2021-08-03
Step 1
I am looking for 3 consecutive multiples of 7. The sum of the squares of the 2 smallest is 147 less than the square of the largest. The first multiple of 7 is $$\displaystyle{\left({x}-{7}\right)}$$, the 2nd multiple of 7 is (x), and the 3rd multiple of $$\displaystyle{7}{\left({x}+{7}\right)}$$. Find the numbers.
According to question,
$$\displaystyle{\left({x}-{7}\right)}^{{{2}}}+{x}^{{{2}}}={\left({x}+{7}\right)}^{{{2}}}-{147}$$
$$\displaystyle\Rightarrow{x}^{{{2}}}+{49}-{14}{x}+{x}^{{{2}}}={x}^{{{2}}}+{49}+{14}{x}-{147}$$
$$\displaystyle\Rightarrow{x}^{{{2}}}-{14}{x}={14}{x}-{147}$$
$$\displaystyle\Rightarrow{x}^{{{2}}}-{28}{x}+{147}={0}$$
$$\displaystyle\Rightarrow{x}^{{{2}}}-{\left({21}+{7}\right)}{x}+{147}={0}$$
$$\displaystyle\Rightarrow{x}^{{{2}}}-{21}{x}-{7}{x}+{147}={0}$$
$$\displaystyle\Rightarrow{x}{\left({x}-{21}\right)}-{7}{\left({x}-{21}\right)}={0}$$
$$\displaystyle\Rightarrow{\left({x}-{21}\right)}-{\left({x}-{7}\right)}={0}$$
$$\displaystyle{x}-{21}={0},{x}-{7}={0}$$
$$\displaystyle\Rightarrow{x}={21},{x}={7}$$
Step 2
when $$\displaystyle{x}={21}$$
the consecutive multiples of 7 is
$$\displaystyle{\left({x}-{7}\right)}={21}-{7}={14}$$
$$\displaystyle{x}={21}$$
$$\displaystyle{x}+{7}={21}+{7}={28}$$
when $$\displaystyle{x}={7}$$
the consecutive multiples of 7 is
$$\displaystyle{\left({x}-{7}\right)}={7}-{7}={0}$$
$$\displaystyle{x}={7}$$
$$\displaystyle{x}+{7}={7}+{7}={14}$$