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stira [4]
3 years ago
7

Can anyone help me with this. It asks to solve and check your answer.

Mathematics
1 answer:
tankabanditka [31]3 years ago
6 0
\sf 4m-21=3(5m-1)

First, distribute 3 into the parenthesis(multiply it to every term inside):

\sf 4m-21=15m-3

Add 21 to both sides:

\sf 4m=15m+18

Subtract 15m to both sides:

\sf -11m=18

Divide -11 to both sides:

\sf m=-\dfrac{18}{11}

To check our work we can plug this back into the original equation for 'm':

\sf 4m-21=3(5m-1)

\sf 4(-\dfrac{18}{11})-21=3(5(-\dfrac{18}{11})-1)

Multiply:

\sf -\dfrac{72}{11}-21=3(-\dfrac{90}{11}-1)

Distribute 3 into the parenthesis:

\sf -\dfrac{72}{11}-21=-\dfrac{270}{11}-3

Subtract:

\sf -\dfrac{303}{11}=-\dfrac{303}{11}~\checkmark

Both sides are equal to each other, so we solved it correctly.
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This one took some trial and error! At first I listed all 2 digit primes, looked at the list, but didn't know how to proceed. So, I took the smallest 2 digit primes numbers:  11 and 13 and wondered if their product, 13*11 = 143, could be represented as the sum of  3 consecutive primes.   I went back to my list of primes,  added groups of three consecutive numbers that seemed to be in the right range to give the desired sum, and stumbled on  43, 47, and 53!

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