Questions: 7(z+2)+4>3(z+4)-7 and -2z-5>2(z+2)+5

7(z+2)+4>3(z+4)-7 and -2z-5>2(z+2)+5
Transcript text: $7(z+2)+4>3(z+4)-7$ and $-2 z-5>2(z+2)+5$
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Solution

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Solution Steps

To solve the given inequalities, we need to simplify each inequality separately and then find the solution set that satisfies both inequalities simultaneously. This involves isolating the variable \( z \) in each inequality and determining the range of values for \( z \) that satisfy both conditions.

Step 1: Simplify the First Inequality

The first inequality is given by: \[ 7(z + 2) + 4 > 3(z + 4) - 7 \]

Simplifying both sides, we have: \[ 7z + 14 + 4 > 3z + 12 - 7 \] \[ 7z + 18 > 3z + 5 \]

Subtract \(3z\) from both sides: \[ 4z + 18 > 5 \]

Subtract 18 from both sides: \[ 4z > -13 \]

Divide by 4: \[ z > -\frac{13}{4} \]

Step 2: Simplify the Second Inequality

The second inequality is: \[ -2z - 5 > 2(z + 2) + 5 \]

Simplifying both sides, we have: \[ -2z - 5 > 2z + 4 + 5 \] \[ -2z - 5 > 2z + 9 \]

Add \(2z\) to both sides: \[ -5 > 4z + 9 \]

Subtract 9 from both sides: \[ -14 > 4z \]

Divide by 4: \[ z < -\frac{7}{2} \]

Step 3: Find the Intersection of the Solution Sets

The solution to the first inequality is: \[ z > -\frac{13}{4} \]

The solution to the second inequality is: \[ z < -\frac{7}{2} \]

The intersection of these solutions is: \[ -\frac{13}{4} < z < -\frac{7}{2} \]

Final Answer

The solution set for the inequalities is: \[ \boxed{\left(-\frac{13}{4}, -\frac{7}{2}\right)} \]

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