Question

Which type of compound inequality is 3y +71 – 16 >5?

Answer 1

The type of compound inequality represented by
`\(3y + 71 - 16 > 5\)` is a "greater than" compound inequality.

Step-by-step explanation:

Certainly! Let's go through the detailed calculations and explanation for the given compound inequality:

The original inequality is
`\(3y + 71 - 16 > 5\).`

Step 1: Combine like terms on the left side of the inequality:

`\[3y + 71 - 16 > 5\]`

Combine the constants (71 and -16):

`\[3y + 55 > 5\]`

Step 2: Subtract 55 from both sides to isolate the
`\(3y\)` term:

`\[3y > -50\]`

Step 3: Divide both sides by 3 to solve for
`\(y\):`

`\[y > -(50)/(3)\]`

Now, let's interpret the result:

The solution
`\(y > -(50)/(3)\)` indicates that any real number greater than
`\(-(50)/(3)\)` will satisfy the original inequality. To visualize this on a number line, we mark a circle at
`\(-(50)/(3)\)` and draw an arrow to the right, indicating all values greater than this point.

In conclusion, the compound inequality
`\(3y + 71 - 16 > 5\)` represents a "greater than" relationship, and the solution set for
`\(y\) is \(y > -(50)/(3)\).`