Practice Problems

Problems:

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Problem 1

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Problem 2

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Problem 3

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Problem 4

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Problem 5

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Solutions:

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Problem 1

Step 1: Get all the $x$ terms on one side.

• We move the $x$ terms to one side by subtracting $2x$ from both sides: $3x - 2x - 4 > 2x - 2x + 5$ Simplifying: $x - 4 > 5$ Step 2: Isolate the variable $x$.

• Add $4$ to both sides to solve for $x$: $x - 4 + 4 > 5 + 4$ Simplifying: $x > 9$ Solution: The solution is $x > 9$. This means $x$ can be any number greater than $9$.

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Problem 2

Step 1: Get all the $x$ terms on one side.

• Add $3x$ to both sides to move the $x$ terms together: $-5x + 3x + 7 \leq 2 - 3x + 3x$ Simplifying: $-2x + 7 \leq 2$ Step 2: Isolate the variable $x$.

• Subtract $7$ from both sides: $-2x + 7 - 7 \leq 2 - 7$ Simplifying: $-2x \leq -5$ Step 3: Solve for $x$.

• Divide both sides by $-2$ and reverse the inequality sign: $\frac{-2x}{-2} \geq \frac{-5}{-2}$ Simplifying: $x \geq \frac{5}{2}$ Solution: The solution is $x \geq \frac{5}{2}$. This means $x$ can be any number greater than or equal to $\frac{5}{2}$.

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Problem 3

Step 1: Distribute the $4$ on the left side.

• Distribute (multiply) the $4$: $4 \times 2x - 4 \times 3 > 5x + 1$ Simplifying: $8x - 12 > 5x + 1$ Step 2: Get all the $x$ terms on one side.

• Subtract $5x$ from both sides: $8x - 5x - 12 > 5x - 5x + 1$ Simplifying: $3x - 12 > 1$ Step 3: Isolate the variable $x$.

• Add $12$ to both sides: $3x - 12 + 12 > 1 + 12$ Simplifying: $3x > 13$ Step 4: Solve for $x$.

• Divide by $3$: $x > \frac{13}{3}$ Solution: The solution is $x > \frac{13}{3}$. This means $x$ can be any number greater than $\frac{13}{3}$.

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Problem 4

Step 1: Clear the fractions.

• Multiply both sides by the least common multiple (LCM) of the denominators, which is $6$: $6 \times \frac{2x - 1}{3} \geq 6 \times \frac{5x + 4}{2}$ Simplifying: $2(2x - 1) \geq 3(5x + 4)$ Step 2: Distribute the numbers on both sides.

• Distribute the $2$ and the $3$: $4x - 2 \geq 15x + 12$ Step 3: Get all the $x$ terms on one side.

• Subtract $4x$ from both sides: $4x - 4x - 2 \geq 15x - 4x + 12$ Simplifying: $-2 \geq 11x + 12$ Step 4: Isolate the variable $x$.

• Subtract $12$ from both sides: $-2 - 12 \geq 11x$ Simplifying: $-14 \geq 11x$ Step 5: Solve for $x$.

• Divide by $11$: $\frac{-14}{11} \geq x \quad \text{or} \quad x \leq \frac{-14}{11}$ Solution: The solution is $x \leq \frac{-14}{11}$. This means $x$ can be any number less than or equal to $\frac{-14}{11}$.

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Problem 5

Step 1: Distribute the $2$ on the left side.

• Distribute (multiply) the $2$: $2 \times x - 2 \times 4 - 3 \geq 4x + 1$ Simplifying: $2x - 8 - 3 \geq 4x + 1$ Step 2: Combine like terms.

• Combine $-8$ and $-3$ on the left side: $2x - 11 \geq 4x + 1$ Step 3: Get all the $x$ terms on one side.

• Subtract $2x$ from both sides: $2x - 2x - 11 \geq 4x - 2x + 1$ Simplifying: $-11 \geq 2x + 1$ Step 4: Isolate the variable $x$.

• Subtract $1$ from both sides: $-11 - 1 \geq 2x$ Simplifying: $-12 \geq 2x$ Step 5: Solve for $x$.

• Divide by $2$: $\frac{-12}{2} \geq x \quad \text{or} \quad x \leq -6$ Solution: The solution is $x \leq -6$. This means $x$ can be any number less than or equal to $-6$.