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

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

A. p = 2(l + w)p/2 - l = ww = p/2 - l b. while isolating the variable w in the equation, i first divided both sides by 2 and then i subtracted l from both sides. i hope this .

paul's with a y-intercept of $ 2856