Question

The larger of two consecutive integers is 13 greater than four times the smaller. Find the integers

Answer 1

now, let's keep in mind that, the integers are consecutrive, therefore, if one is say 2, the other is either, 3 or 1, because 1,2,3  <--- notice, 1 is before 2 and 3 is after 2.

so say, our first integer is "a", then the next one can just be "a+1".

$\bf \begin{cases} a&=small\\ a+1&=\stackrel{consecutive}{large}\\ \end{cases}\qquad \begin{cases} \stackrel{\textit{4 times the small}}{4a}\\ \stackrel{\textit{13 greater than that}}{4a + 13} \end{cases}~\hfill \stackrel{\textit{large}}{a+1}=\stackrel{\stackrel{\textit{13 more than }}{\textit{4 times the small}}}{4a+13} \\\\[-0.35em] ~\dotfill$

$\bf a+1=4a+13\implies 1=3a+13\implies -12=3a\implies \cfrac{-12}{3}=a \\\\\\ \blacktriangleright -4=a \blacktriangleleft\qquad \qquad \stackrel{\textit{consecutive and large}~\hfill }{a+1\implies -4+1\implies \blacktriangleright -3 \blacktriangleleft}~\hfill \boxed{\stackrel{\textit{small}}{-4}~~,~~\stackrel{\textit{large}}{-3}}$

let's recall that, on the negative side in a number line, the closer to 0, the larger the number, therefore, -1,000,000 is a much smaller number than -1, because -1 is closer to 0, thus is larger.