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Question
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the sum of the digits of a three-digit number is 13. the tens digit, $t$, is 1 more than the hundreds digit, $h$. the units digit, $u$, is 3 more than the sum of the tens and hundreds digits. which system of equations can be used to find each digit?
a. $h + t + u = 13$
$h - t = 1$
$h + t - u = 3$
b. $h + t + u = 13$
$-h + u = 1$
$h - t + u = 3$
c. $h + t + u = 13$
$-h + t = 1$
$-h - t + u = 3$
d. $h + t + u = 13$
$h + u = 1$
$-h - t = 3$
Step1: Translate sum of digits
$h + t + u = 13$
Step2: Translate tens-hundreds relation
$t = h + 1 \implies -h + t = 1$
Step3: Translate units-hundreds/tens relation
$u = h + t + 3 \implies -h - t + u = 3$
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C. $h + t + u = 13$
$-h + t = 1$
$-h - t + u = 3$