the sum of the digits of two-digit number is 11. if the sum of the number and its tens digits is 41, find the number. can someone show me step by step on how to do this please?
I've tried solving your problem, and I keep getting errors. Can you double check the problem? Hope you wouldn't mind :))
ok hold on
The sum of the digits of a two-digit number is 11. if the sum of the number and its tens digit is 41, find the number.
Oh okay, hold on :))
that's exactly how is written in the book
alright :)
Okay so let x= ones digit 11-x= tens digit 41= the sum of the digit and the tens digit equation: [10(11-x) + x] + (11-x) = 41 (110-10x) + x + 11-x =41 -10x + x - x = 41-110-11 (-10x = -80)/10 x = 8 So now you've got your ones digit. Substitute: 8= ones digit 11-8= 3 <--- tens digit So the number is 38 :))
got it! thank you so much for your time and help! :)))
No problem :))
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