Respuesta :
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".
[tex]\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[/tex]
[tex]\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}}[/tex]
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.
The integers are -3 and -4.
Let's represent
The larger consecutive integer as: x
The smaller consecutive integer as: y
Consecutive integers are represented as:
x , x + 1
where,
x = x + 1
y = x
The larger of two consecutive integers is 13 greater than four times the smaller. This is expressed algebraically as:
x = 4y + 13
Substitute x + 1 for x and x for y in Equation 1
x + 1 = 4x + 13
Collect like terms
x - 4x = 13 - 1
-3x = 12
Divide both sides by -3
-3x/-3 = 12 /-3
x = -4
Therefore, the smaller integer is -4
Solving for x
x + 1
-4 + 1
= -3
Therefore, the larger integer is -3
The correct option is Option B
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