1. For the first inequality, Luz can buy less than 5 computers (since
represents the number of computers and must be less than 5).
2. For the second set of inequalities,
must be greater than -6 and less than 0.
The image shows a mathematical problem with two inequalities:
1.

2.

To solve these inequalities step by step, we'll perform the following calculations:
For the first inequality:
![\[ 70x < 350 \]](https://img.qammunity.org/2021/formulas/mathematics/middle-school/vxpr19e7auu9gavldn484f5hawfv2oi5pe.png)
1. Divide both sides by 70 to solve for x:
![\[ x < (350)/(70) \]](https://img.qammunity.org/2021/formulas/mathematics/middle-school/jrcdbr1673ocqlzmrak8cm48y76da2aca2.png)
For the second set of inequalities:
![\[ x + 3 < 3 \] and \( x + 3 > -3 \)](https://img.qammunity.org/2021/formulas/mathematics/middle-school/in23deyap2fpzfib71p96upustpq0mi0fi.png)
1. Subtract 3 from all parts of the compound inequality:
and

and

Now let's calculate the values.
Solution for the inequalities:
First inequality:
![\[ 70x < 350 \]](https://img.qammunity.org/2021/formulas/mathematics/middle-school/vxpr19e7auu9gavldn484f5hawfv2oi5pe.png)
After dividing both sides by 70, we get:
![\[ x < 5 \]](https://img.qammunity.org/2021/formulas/mathematics/middle-school/zeh204ssjpjooywe50fzgr55o59lcpi55z.png)
Second set of inequalities:
and

After subtracting 3 from all parts of the compound inequality, we get:
and

Thus, the solution to the inequalities are:
1. For the first inequality, Luz can buy less than 5 computers (since
represents the number of computers and must be less than 5).
2. For the second set of inequalities,
must be greater than -6 and less than 0.
If we interpret
as the number of items Luz can buy, the negative values don't make sense in a real-world context, as you cannot buy a negative number of items. Therefore, the practical solution for the second inequality is that Luz cannot buy any items since
must be less than 0 and whole numbers (non-negative integers) are required for the count of items.