To solve the inequality \( 5(2k - 3) \geq -3(4 - 3k) \), we can follow these concise steps:

Question

1. **Distribute**:
- Left: \( 5(2k) - 5(3) = 10k - 15 \)
- Right: \( -3(4) + 9k = -12 + 9k \)

Rewrite the inequality:
\[
10k - 15 \geq -12 + 9k
\]

2. **Combine like terms**:
Subtract \( 9k \) from both sides:
\[
10k - 9k - 15 \geq -12
\]
Simplifying gives:
\[
k - 15 \geq -12
\]

3. **Isolate \( k \)**:
Add 15 to both sides:
\[
k \geq 3
\]

Thus, the solution is:
\[
k \geq 3
\]mkake htis like a 6th grader wrote it

Answer 1

Sure! Let's make it simpler for a 6th grader to understand.

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To solve the problem \( 5(2k - 3) \geq -3(4 - 3k) \):

1. Distribute the numbers:

   • On the left side, we do \( 5 \times 2k \) and \( 5 \times -3 \):

     • \( 5 \times 2k = 10k \)
     • \( 5 \times -3 = -15 \)
     • So, the left side becomes \( 10k - 15 \).

   • On the right side, we do \( -3 \times 4 \) and \( -3 \times -3k \):

     • \( -3 \times 4 = -12 \)
     • \( -3 \times -3k = +9k \) (because a negative times a negative is a positive)
     • So, the right side becomes \( -12 + 9k \).

Now our inequality looks like this: \[ 10k - 15 \geq -12 + 9k \]

2. Combine the like terms:

   • We want to get all the \( k \) terms on one side. So let's subtract \( 9k \) from both sides: \[ 10k - 9k - 15 \geq -12 \]
   • This simplifies to: \[ k - 15 \geq -12 \]

3. Isolate \( k \):

   • Now we add 15 to both sides to solve for \( k \): \[ k \geq -12 + 15 \]
   • Which simplifies to: \[ k \geq 3 \]

So, our final answer is: \[ k \geq 3 \]

That means any number that is 3 or bigger is a solution!