How do you find the two consecutive integers whose sum is 223?

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Answer 1 · 2018-05-22T14:41:00

See a solution process below:

First, let's cal the first integer we are looking for: $n$ $n$

Then, because we are looking for consecutive integers the second integer we are looking for can be written as: $n + 1$ $n + 1$

We know these two integers sum to 223. Therefore, we can write this equation and solve for $n$ $n$ :

$n + (n + 1) = 223$ $n + (n + 1) = 223$

$n + n + 1 = 223$ $n + n + 1 = 223$

$1 n + 1 n + 1 = 223$ $1n + 1n + 1 = 223$

$(1 + 1) n + 1 = 223$ $(1 + 1)n + 1 = 223$

$2 n + 1 = 223$ $2n + 1 = 223$

$2 n + 1 - 1 = 223 - 1$ $2n + 1 - color(red)(1) = 223 - color(red)(1)$

$2 n + 0 = 222$ $2n + 0 = 222$

$2 n = 222$ $2n = 222$

$\frac{2 n}{2} = \frac{222}{2}$ $(2n)/color(red)(2) = 222/color(red)(2)$

$(color(red)(cancel(color(black)(2)))n)/cancel(color(red)(2)) = 111$

$n = 111$

Solution Check:

$111 + 112 = 223$