find a number which when multiplied by 7 and then reduced by 3 is equal to 47 .
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r={8}
PREMISES
y=(7r-3=53)
CALCULATIONS
y=(7r-3=53)
7r-3=53
7r/7-(3–3)=(53+3)/7
1r-0=56/7
r=56/7 (Cancel the common factor 7 in the divisor then adjust the numerator)
r=8/1
r=
8
PROOF
If r=8, then the inverse of the mathematical phrase 7r-3 brings
y+3=7r
53+3=7(8) and
56=56 establishes proof of the root (zero) r=8 for the statement 7r-3=53
C.H.
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