[tex]x\geq 3 \ and \ x+1 \geq 4[/tex]
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 [tex]x[/tex] which is the smaller number. Accordingly, the lager number will be [tex]x+1[/tex] given that those numbers are consecutive. On the other hand at most conveys the idea of an inequality, which is:
[tex]\leq \\ which \ means \ less \ than[/tex]
So:
1. The sum of 2 consecutive integers can be written as:
[tex]v+(v+1)[/tex]
2. Nine times the smaller and 5 times the larger can be written as:
[tex]9v-5(v+1)[/tex]
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:
[tex]x+(x+1) \leq 9x-5(x+1) \\ \\ x+x+1\leq 9x-5x-5 \\ \\ 2x+1 \leq 4x-5 \\ \\ 6 \leq 2x \\ \\ [/tex]
[tex]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[/tex]
The two numbers are:
[tex]x\geq 3 \ and \ x+1 \geq 4[/tex]
