Problem 12
Question
Solve each equation. Check your solution. $$\frac{2}{3} p=-22$$
Step-by-Step Solution
Verified Answer
The solution is \( p = -33 \).
1Step 1: Understand the Equation
We start with the equation \( \frac{2}{3} p = -22 \). This means \( p \) has been multiplied by \( \frac{2}{3} \), and the product is \( -22 \). Our goal is to isolate \( p \) to solve for its value.
2Step 2: Isolate \( p \)
To solve for \( p \), we need to eliminate the fraction \( \frac{2}{3} \). We do this by multiplying both sides of the equation by the reciprocal of \( \frac{2}{3} \), which is \( \frac{3}{2} \). Thus, we multiply both sides:\[\frac{3}{2} \times \frac{2}{3} p = \frac{3}{2} \times (-22)\]This simplifies to:\[p = \frac{3}{2} \times (-22)\]
3Step 3: Simplify the Calculation
Now, perform the multiplication on the right side of the equation:\[p = \frac{3}{2} \times (-22) = -33\]Hence, we find that \( p = -33 \).
4Step 4: Verify the Solution
To verify this solution, substitute \( p = -33 \) back into the original equation:\[\frac{2}{3} \times (-33) = -22\]Simplifying the left-hand side:\[\frac{2}{3} imes (-33) = -\frac{66}{3} = -22\]Since both sides of the equation are equal, \( p = -33 \) is indeed the correct solution to the equation.
Key Concepts
PrealgebraFractionsEquation Verification
Prealgebra
Prealgebra lays the foundation for understanding algebra by dealing with basic mathematical concepts such as arithmetic operations and understanding variables. It is the gateway to more complex math topics and begins with simple equations that contain an unknown value referred to as a variable.
In the context of this problem, we are given the equation \( \frac{2}{3} p = -22 \). This equation tells us that "two-thirds of \( p \) equals -22." Our task in prealgebra is to solve such equations to find the value of this unknown variable, \( p \). Solving an equation means finding the particular value of the variable that makes the equation true.
This involves understanding how to manipulate the equation using various arithmetic operations to isolate the variable on one side, which ultimately tells us its value. This sets the stage for solving more complex equations in higher-level algebra.
In the context of this problem, we are given the equation \( \frac{2}{3} p = -22 \). This equation tells us that "two-thirds of \( p \) equals -22." Our task in prealgebra is to solve such equations to find the value of this unknown variable, \( p \). Solving an equation means finding the particular value of the variable that makes the equation true.
This involves understanding how to manipulate the equation using various arithmetic operations to isolate the variable on one side, which ultimately tells us its value. This sets the stage for solving more complex equations in higher-level algebra.
Fractions
Fractions are numbers that represent parts of a whole. They consist of a numerator (top number) and a denominator (bottom number). In the equation \( \frac{2}{3} p = -22 \), the fraction \( \frac{2}{3} \) indicates that \( p \) is being multiplied by two-thirds.
Dealing with fractions in equations often requires using their reciprocal. The reciprocal of a fraction is found by swapping its numerator and denominator. So the reciprocal of \( \frac{2}{3} \) is \( \frac{3}{2} \).
When solving equations involving fractions, we multiply both sides by the reciprocal. This "undoes" the fraction part of the equation by effectively cancelling it out, allowing the variable to be isolated. For our equation:
Dealing with fractions in equations often requires using their reciprocal. The reciprocal of a fraction is found by swapping its numerator and denominator. So the reciprocal of \( \frac{2}{3} \) is \( \frac{3}{2} \).
When solving equations involving fractions, we multiply both sides by the reciprocal. This "undoes" the fraction part of the equation by effectively cancelling it out, allowing the variable to be isolated. For our equation:
- Multiply both sides by \( \frac{3}{2} \): \( \frac{3}{2} \times \frac{2}{3} p = \frac{3}{2} \times (-22) \)
Equation Verification
Once you find a solution to an equation, it's crucial to verify it, ensuring it's correct. Verification involves substituting the found value back into the original equation to check if the equation holds true.
Let's verify \( p = -33 \) as the solution to our equation \( \frac{2}{3} p = -22 \):
Let's verify \( p = -33 \) as the solution to our equation \( \frac{2}{3} p = -22 \):
- Substitute \( -33 \) for \( p \): \( \frac{2}{3} \times (-33) \)
- Calculate: \( \frac{2}{3} \times (-33) = -\frac{66}{3} = -22 \)
Other exercises in this chapter
Problem 11
Find each quotient. Use an area model if necessary. $$2 \frac{1}{6} \div\left(-1 \frac{1}{5}\right)$$
View solution Problem 11
Replace each with \(,\) or \(=\) to make a true sentence. $$\frac{9}{10} \circ 0.90$$
View solution Problem 12
Find each product. Use an area model if necessary. $$-2 \frac{1}{2} \cdot 5 \frac{2}{3}$$
View solution Problem 12
Find the least common denominator (LCD) of each pair of fractions. $$\frac{5}{7} \cdot \frac{7}{9}$$
View solution