Alright, let's solve the inequality:
[tex]\[
\frac{x-3}{2}+\frac{5}{4}<\frac{x}{12}+\frac{2x+9}{15}
\][/tex]
### Step-by-Step Solution:
1. Combine the left-hand side (LHS):
The LHS of the inequality is:
[tex]\[
\frac{x-3}{2} + \frac{5}{4}
\][/tex]
To combine these fractions, find a common denominator. The common denominator of 2 and 4 is 4.
Convert [tex]\(\frac{x-3}{2}\)[/tex] into a fraction with a denominator of 4:
[tex]\[
\frac{x-3}{2} = \frac{2(x-3)}{4} = \frac{2x-6}{4}
\][/tex]
Now, add [tex]\(\frac{2x-6}{4}\)[/tex] and [tex]\(\frac{5}{4}\)[/tex]:
[tex]\[
\frac{2x-6}{4} + \frac{5}{4} = \frac{2x-6+5}{4} = \frac{2x-1}{4}
\][/tex]
2. Combine the right-hand side (RHS):
The RHS of the inequality is:
[tex]\[
\frac{x}{12} + \frac{2x+9}{15}
\][/tex]
To combine these fractions, find a common denominator. The least common multiple of 12 and 15 is 60.
Convert [tex]\(\frac{x}{12}\)[/tex] and [tex]\(\frac{2x+9}{15}\)[/tex] into fractions with a denominator of 60:
[tex]\[
\frac{x}{12} = \frac{5x}{60}
\][/tex]
[tex]\[
\frac{2x+9}{15} = \frac{4(2x+9)}{60} = \frac{8x+36}{60}
\][/tex]
Now, add [tex]\(\frac{5x}{60}\)[/tex] and [tex]\(\frac{8x+36}{60}\)[/tex]:
[tex]\[
\frac{5x}{60} + \frac{8x+36}{60} = \frac{5x + 8x + 36}{60} = \frac{13x + 36}{60}
\][/tex]
3. Set up the new inequality:
[tex]\[
\frac{2x-1}{4} < \frac{13x + 36}{60}
\][/tex]
4. Clear the denominators by finding the least common multiple of 4 and 60, which is 60.
Multiply every term by 60:
[tex]\[
60 \cdot \frac{2x-1}{4} < 60 \cdot \frac{13x + 36}{60}
\][/tex]
Simplify:
[tex]\[
15(2x-1) < 13x + 36
\][/tex]
5. Distribute on the left:
[tex]\[
30x - 15 < 13x + 36
\][/tex]
6. Isolate [tex]\( x \)[/tex] on one side:
Subtract [tex]\(13x\)[/tex] from both sides:
[tex]\[
30x - 13x - 15 < 36
\][/tex]
Simplify:
[tex]\[
17x - 15 < 36
\][/tex]
Add 15 to both sides:
[tex]\[
17x < 51
\][/tex]
Divide both sides by 17:
[tex]\[
x < 3
\][/tex]
Therefore, the solution to the inequality is:
[tex]\[
-\infty < x < 3
\][/tex]
In interval notation, this can be written as:
[tex]\[
(-\infty, 3)
\][/tex]
So, the solution set for the inequality is:
[tex]\[
x \in (-\infty, 3)
\][/tex]
