Question

1. determine what solutions make this situation true. choose 3 answers. 3(2y - 4) + 5y ≤ 20 0 4 -4 2 6

Explanation:

Step1: Simplify the inequality

First, expand the left - hand side of the inequality \(3(2y - 4)+5y\leq20\).
Using the distributive property \(a(b + c)=ab+ac\), we have \(3\times2y-3\times4 + 5y\leq20\), which simplifies to \(6y-12 + 5y\leq20\).
Combine like terms: \((6y + 5y)-12\leq20\), so \(11y-12\leq20\).

Step2: Solve for y

Add 12 to both sides of the inequality: \(11y-12 + 12\leq20 + 12\), which gives \(11y\leq32\).
Then divide both sides by 11: \(y\leq\frac{32}{11}\approx2.91\).

Step3: Check each option

• For \(y = 0\): Since \(0\leq\frac{32}{11}\), substitute \(y = 0\) into the original inequality: \(3(2\times0 - 4)+5\times0=3\times(- 4)+0=-12\leq20\), which is true.
• For \(y = 4\): Since \(4>\frac{32}{11}\), substitute \(y = 4\) into the original inequality: \(3(2\times4 - 4)+5\times4=3\times4 + 20 = 12 + 20=32>20\), which is false.
• For \(y=-4\): Since \(-4\leq\frac{32}{11}\), substitute \(y = - 4\) into the original inequality: \(3(2\times(-4)-4)+5\times(-4)=3\times(-8 - 4)-20=3\times(-12)-20=-36-20=-56\leq20\), which is true.
• For \(y = 2\): Since \(2\leq\frac{32}{11}\), substitute \(y = 2\) into the original inequality: \(3(2\times2 - 4)+5\times2=3\times(4 - 4)+10=0 + 10 = 10\leq20\), which is true.
• For \(y = 6\): Since \(6>\frac{32}{11}\), substitute \(y = 6\) into the original inequality: \(3(2\times6 - 4)+5\times6=3\times(12 - 4)+30=3\times8+30=24 + 30 = 54>20\), which is false.

Answer:

The three values that satisfy the inequality are 0, - 4, and 2. So the correct options are:

• 0
• - 4
• 2