Question

1. solve $-12 geq 3x$ and graph the solution.
2. solve $\frac{t}{-2}+4 leq 6$ and graph the solution.
3. graph the solution of $2x + 7 geq -5$ and $2x - 7 < 9$

Explanation:

Step1: Solve the first inequality

Given \( - 12\geq3x\). Divide both sides by \(3\).
\(\frac{-12}{3}\geq\frac{3x}{3}\)
\(-4\geq x\) or \(x\leq - 4\)

Step2: Solve the second inequality

Given \(\frac{t}{-2}+4\leq6\). Subtract \(4\) from both sides:
\(\frac{t}{-2}+4 - 4\leq6 - 4\)
\(\frac{t}{-2}\leq2\). Multiply both sides by \(-2\) and reverse the inequality sign (since multiplying by a negative number).
\(t\geq - 4\)

Step3: Solve the compound - inequality in problem 5

First, solve \(2x + 7\geq - 5\). Subtract \(7\) from both sides:
\(2x+7 - 7\geq - 5 - 7\)
\(2x\geq - 12\). Divide by \(2\): \(x\geq - 6\)
Second, solve \(2x - 7\lt9\). Add \(7\) to both sides:
\(2x-7 + 7\lt9 + 7\)
\(2x\lt16\). Divide by \(2\): \(x\lt8\)

Answer:

For problem 3: \(x\leq - 4\). Graph: A closed circle at \(-4\) and a line extending to the left on the number - line.
For problem 4: \(t\geq - 4\). Graph: A closed circle at \(-4\) and a line extending to the right on the number - line.
For problem 5: \(-6\leq x\lt8\). Graph: A closed circle at \(-6\), an open circle at \(8\), and a line connecting them on the number - line.