Solve the following systems of inequalities for all `x in R` `5(2x-7)-3(2x+3)le0` and `2x+19le6x+45`.

Answer

To solve the system of inequalities given by:

1. \( 5(2x - 7) - 3(2x + 3) \leq 0 \)
2. \( 2x + 19 \leq 6x + 45 \)

we will solve each inequality step by step.

### Step 1: Solve the first inequality

Start with the first inequality:

\[ 5(2x - 7) - 3(2x + 3) \leq 0 \]

Distributing the terms:

\[ 10x - 35 - 6x - 9 \leq 0 \]

Combine like terms:

\[ (10x - 6x) - (35 + 9) \leq 0 \]
\[ 4x - 44 \leq 0 \]

Now, isolate \(x\):

\[ 4x \leq 44 \]

Dividing both sides by 4:

\[ x \leq 11 \]

### Step 2: Solve the second inequality

Now, consider the second inequality:

\[ 2x + 19 \leq 6x + 45 \]

Rearranging the terms:

\[ 19 - 45 \leq 6x - 2x \]
\[ -26 \leq 4x \]

Now, isolate \(x\):

\[ \frac{-26}{4} \leq x \]

Simplifying:

\[ -\frac{13}{2} \leq x \quad \text{or} \quad x \geq -\frac{13}{2} \]

### Step 3: Combine the results

Now we have two inequalities:

1. \( x \leq 11 \)
2. \( x \geq -\frac{13}{2} \)

Combining these gives us:

\[ -\frac{13}{2} \leq x \leq 11 \]

### Final Answer

Thus, the solution for the system of inequalities is:

\[ x \in \left[-\frac{13}{2}, 11\right] \]