Question

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

Answer 1

Answer:

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

Step-by-step explanation:

The question in this problem is:

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?

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 at most 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:

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