Answer :
Sure, let's solve the inequality step by step:
We start with the given inequality:
[tex]\[ \frac{1}{3} x + \frac{5}{3} < -\frac{3}{4} x - \frac{1}{2} \][/tex]
First, we want to clear the fractions by finding a common multiple. In this case, the least common multiple (LCM) of the denominators (3 and 4) is 12. We multiply each term by 12:
[tex]\[ 12 \cdot \left( \frac{1}{3}x \right) + 12 \cdot \left( \frac{5}{3} \right) < 12 \cdot \left( -\frac{3}{4}x \right) + 12 \cdot \left( -\frac{1}{2} \right) \][/tex]
This simplifies to:
[tex]\[ 4x + 20 < -9x - 6 \][/tex]
Next, we want to isolate [tex]\( x \)[/tex]. Begin by getting all terms involving [tex]\( x \)[/tex] on one side of the inequality and the constant terms on the other side. To do this, add [tex]\( 9x \)[/tex] to both sides:
[tex]\[ 4x + 9x + 20 < -9x + 9x - 6 \][/tex]
This simplifies to:
[tex]\[ 13x + 20 < -6 \][/tex]
Now, isolate [tex]\( x \)[/tex] by subtracting 20 from both sides:
[tex]\[ 13x + 20 - 20 < -6 - 20 \][/tex]
This simplifies to:
[tex]\[ 13x < -26 \][/tex]
Now, solve for [tex]\( x \)[/tex] by dividing both sides by 13:
[tex]\[ x < -2 \][/tex]
Hence, the solution to the inequality [tex]\(\frac{1}{3} x+ \frac{5}{3} < -\frac{3}{4} x-\frac{1}{2}\)[/tex] is:
[tex]\[ x < -2 \][/tex]
In interval notation, this means:
[tex]\[ (-\infty, -2) \][/tex]
We start with the given inequality:
[tex]\[ \frac{1}{3} x + \frac{5}{3} < -\frac{3}{4} x - \frac{1}{2} \][/tex]
First, we want to clear the fractions by finding a common multiple. In this case, the least common multiple (LCM) of the denominators (3 and 4) is 12. We multiply each term by 12:
[tex]\[ 12 \cdot \left( \frac{1}{3}x \right) + 12 \cdot \left( \frac{5}{3} \right) < 12 \cdot \left( -\frac{3}{4}x \right) + 12 \cdot \left( -\frac{1}{2} \right) \][/tex]
This simplifies to:
[tex]\[ 4x + 20 < -9x - 6 \][/tex]
Next, we want to isolate [tex]\( x \)[/tex]. Begin by getting all terms involving [tex]\( x \)[/tex] on one side of the inequality and the constant terms on the other side. To do this, add [tex]\( 9x \)[/tex] to both sides:
[tex]\[ 4x + 9x + 20 < -9x + 9x - 6 \][/tex]
This simplifies to:
[tex]\[ 13x + 20 < -6 \][/tex]
Now, isolate [tex]\( x \)[/tex] by subtracting 20 from both sides:
[tex]\[ 13x + 20 - 20 < -6 - 20 \][/tex]
This simplifies to:
[tex]\[ 13x < -26 \][/tex]
Now, solve for [tex]\( x \)[/tex] by dividing both sides by 13:
[tex]\[ x < -2 \][/tex]
Hence, the solution to the inequality [tex]\(\frac{1}{3} x+ \frac{5}{3} < -\frac{3}{4} x-\frac{1}{2}\)[/tex] is:
[tex]\[ x < -2 \][/tex]
In interval notation, this means:
[tex]\[ (-\infty, -2) \][/tex]
