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Bas_tet [7]
3 years ago
7

the sum of 2 consecutive integers is at most the difference between nine times the smaller and 5 times the larger

Mathematics
1 answer:
Reil [10]3 years ago
6 0
<h2>Answer:</h2>

x\geq 3 \ and \ x+1 \geq 4

<h2>Step-by-step explanation:</h2>

The question in this problem is:

<em>The sum of 2 consecutive integers is at most the difference between nine times the smaller and 5 times the larger. What are the numbers?</em>

<em />

First of all, let's name the first variable x which is the smaller number. Accordingly, the lager number will be x+1 given that those numbers are consecutive. On the other hand<em> at most </em>conveys the idea of an inequality, which is:

\leq \\ which \ means \ less \ than

So:

1. The sum of 2 consecutive integers can be written as:

v+(v+1)

2. Nine times the smaller and 5 times the larger can be written as:

9v-5(v+1)

Finally, the whole statement:

The sum of 2 consecutive integers is at most the difference between nine times the smaller and 5 times the larger:

x+(x+1) \leq 9x-5(x+1) \\ \\ x+x+1\leq 9x-5x-5 \\ \\ 2x+1 \leq 4x-5 \\ \\ 6 \leq 2x \\ \\

x+(x+1) \leq 9x-5(x+1) \\ \\ x+x+1\leq 9x-5x-5 \\ \\ 2x+1 \leq 4x-5 \\ \\ 6 \leq 2x \\ \\ \frac{6}{2} \leq \frac{2x}{2} \\ \\ 3 \leq x \\ \\ x\geq 3 \\ \\ and \\ \\ x+1 \geq 4

The two numbers are:

x\geq 3 \ and \ x+1 \geq 4

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