Question

Which set expresses the solution to the compound inequality −2≤2x+1/3 −4<1?(1 point)Responses{x:5/2≤x<1}left brace x colon Start Fraction 5 over 2 End Fraction less-than-or-equal-to x less than 1 right brace{x:5/2≤x<7}left brace x colon Start Fraction 5 over 2 End Fraction less-than-or-equal-to x less than 7 right brace{x:3≤x<7}left brace x colon 3 less-than-or-equal-to x less than 7 right brace{x:−3/2≤x<1}left brace x colon Start Fraction negative 3 over 2 End Fraction less-than-or-equal-to x less than 1 right brace

Answer 1

To solve the compound inequality $-2\le 2x+\frac{1}{3}-4<1$, we can break it into two parts:

1. $-2\le 2x+\frac{1}{3}-4$
2. $2x+\frac{1}{3}-4<1$

Step 1: Solve the first part:

$$-2\le 2x+\frac{1}{3}-4$$

First, simplify $\frac{1}{3}-4$:

$$\frac{1}{3}-4=\frac{1}{3}-\frac{12}{3}=\frac{1-12}{3}=\frac{-11}{3}$$

So the first inequality becomes:

$$-2\le 2x-\frac{11}{3}$$

Add $\frac{11}{3}$ to both sides:

$$-2+\frac{11}{3}\ge 2x$$

Convert $-2$ to a fraction:

$$-\frac{6}{3}+\frac{11}{3}=\frac{5}{3}$$

This gives us:

$$\frac{5}{3}\ge 2x\text{or}2x\le \frac{5}{3}$$

Now, divide both sides by 2:

$$x\le \frac{5}{6}$$

Step 2: Solve the second part:

$$2x+\frac{1}{3}-4<1$$

Using the same simplification as before:

$$2x-\frac{11}{3}<1$$

Add $\frac{11}{3}$ to both sides:

$$2x<1+\frac{11}{3}$$

Convert $1$ to a fraction:

$$1=\frac{3}{3}$$

So, this becomes:

$$2x<\frac{3}{3}+\frac{11}{3}=\frac{14}{3}$$

Now, divide both sides by 2:

$$x<\frac{14}{6}=\frac{7}{3}$$

Step 3: Combine the results:

From the first part, we have:

$$x\le \frac{5}{6}$$

From the second part, we have:

$$x<\frac{7}{3}$$

Since $\frac{5}{6}\approx 0.833$ is less than $\frac{7}{3}\approx 2.333$, the solution set for the compound inequality can be written as:

$$x\le \frac{5}{6}$$

Thus, the correct answer for the set that expresses the solution to the compound inequality is:

$$x:-\frac{3}{2}\le x<1$$

So the correct option is:

$x:-\frac{3}{2}\le x<1$