If you add the digits in a two-digit number and multiply the sum by 7, you get the original number. If you reverse the digits in
the two-digit number, the new number is 18 more than the sum of its two digits. What is the original number? 42
1 answer:
The digits are x and y
multiply sum of digits by 7
7(x+y) you get the original number which would be 10x+y
so
7x+7y=10x+y
if reverse digits, new number is18 more than sum
10y+x=18+x+y
so we got
7x+7y=10x+y
10y+x=18+x+y
we can minus x+y from both sides in all the equations to obtain
6x+6y=9x
9y=18
last one, divide both sides by 9
y=2
sub back
6x+6y=9x
6x+6(2)=9x
6x+12=9x
minus 6x both sides
12=3x
divide both sides by 3
4=x
the number is 42
why did you write it down there, oh well
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