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Solve the inequality.

[tex]\[ 17x + 5 \ \textgreater \ 39 \][/tex]
[tex]\[ -13x - 6 \ \textgreater \ -45 \][/tex]

[tex]\[ \boxed{?} \ \textless \ x \ \textless \ \boxed{\square} \][/tex]


Sagot :

To solve the system of inequalities, we need to address each inequality separately and then find the common solution. Let's break this down step-by-step:

1. Solve the first inequality: [tex]\( 17x + 5 > 39 \)[/tex]

- Subtract 5 from both sides to isolate the term with [tex]\( x \)[/tex]:
[tex]\[ 17x + 5 - 5 > 39 - 5 \][/tex]
[tex]\[ 17x > 34 \][/tex]

- Divide both sides by 17 to solve for [tex]\( x \)[/tex]:
[tex]\[ x > \frac{34}{17} \][/tex]
[tex]\[ x > 2 \][/tex]

So, the solution to the first inequality is:
[tex]\[ x > 2 \][/tex]

2. Solve the second inequality: [tex]\( -13x - 6 > -45 \)[/tex]

- Add 6 to both sides to isolate the term with [tex]\( x \)[/tex]:
[tex]\[ -13x - 6 + 6 > -45 + 6 \][/tex]
[tex]\[ -13x > -39 \][/tex]

- Divide both sides by -13. Remember, when dividing by a negative number, the inequality sign reverses:
[tex]\[ x < \frac{-39}{-13} \][/tex]
[tex]\[ x < 3 \][/tex]

So, the solution to the second inequality is:
[tex]\[ x < 3 \][/tex]

3. Combine the solutions:

We have:
[tex]\[ x > 2 \][/tex]
and
[tex]\[ x < 3 \][/tex]

The common solution is the intersection of these two intervals. Therefore:
[tex]\[ 2 < x < 3 \][/tex]

So, the final solution to the system of inequalities [tex]\( 17x + 5 > 39 \)[/tex] and [tex]\( -13x - 6 > -45 \)[/tex] is:
[tex]\[ 2 < x < 3 \][/tex]
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