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
ANSWERED
asked 2021-08-02
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.

Expert Answers (1)

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
\(\displaystyle{\left({x}-{7}\right)}={7}-{7}={0}\)
\(\displaystyle{x}={7}\)
\(\displaystyle{x}+{7}={7}+{7}={14}\)
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