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Chapter 4: Problem 54
To prepare for Section \(4.2,\) review solving equations and formulas (Sections2.2 and 2.3 ). Solve. [ 2.2] $$ 6 y-3(5-2 y)=4 $$
Short Answer
Expert verified
y = \frac{19}{12}
Step by step solution
01
- Distribute
Distribute the \( -3 \) through the parentheses: \[ 6y - 3(5 - 2y) = 4 \Rightarrow 6y - 3 \cdot 5 + 3 \cdot 2y = 4 \Rightarrow 6y - 15 + 6y = 4 \]
02
- Combine Like Terms
Combine the \( 6y \) terms: \[ 6y + 6y - 15 = 4 \Rightarrow 12y - 15 = 4 \]
03
- Isolate the Variable
Add \( 15 \) to both sides to isolate the term with \( y \): \[ 12y - 15 + 15 = 4 + 15 \Rightarrow 12y = 19 \]
04
- Solve for y
Divide both sides by \( 12 \): \[ y = \frac{19}{12} \]
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
distributive property
The distributive property is a fundamental algebraic rule that allows us to simplify expressions. It states that multiplying a sum by a number gives the same result as multiplying each addend by the number and then adding the products. The property can be written in two ways: \ \ The exercise begins by applying the distributive property to the equation \(6y - 3(5 - 2y) = 4\): \ \ Here’s what happens: \ \ When you distribute the \(-3\) through the parentheses, you multiply \(-3\) by every term inside the parentheses: \ \[ 6y - 3(5) + 3(2y) = 4 \ \rightarrow 6y - 15 + 6y = 4 \] \ \ This step simplifies the expression, making it easier to solve. The key to mastering this concept is to remember to multiply every term inside the parentheses by the term outside.
combining like terms
Combining like terms is an essential step in simplifying equations and expressions. Like terms are terms that contain the same variable raised to the same power. For example, \(6y\) and \(6y\) are like terms because they both have the variable 'y'. \ \ In the example from the exercise, we have: \ \[ 6y - 15 + 6y = 4 \ \rightarrow 6y + 6y - 15 = 4 \] \ \ By combining the like terms, you add or subtract the coefficients of the terms. In this case, adding \(6y\) and \(6y\) gives \(12y\): \ \[ 12y - 15 = 4 \] \ \ Now, the equation is simpler and ready for the next step. This process is crucial because it reduces the number of terms and makes the equation easier to solve.
isolating the variable
Isolating the variable involves rearranging the equation to get the variable by itself on one side of the equation. This is typically done using addition, subtraction, multiplication, or division. \ \ In our example, we need to isolate the term with \(y\): \ \[ 12y - 15 = 4 \] \ \ Start by adding 15 to both sides to get rid of the constant term on the left: \ \[ 12y - 15 + 15 = 4 + 15 \rightarrow 12y = 19 \] \ \ Now, the equation is \(12y = 19\). To solve for \(y\), divide both sides by 12: \ \[ y = \frac{19}{12} \] \ \ And that's it! By isolating the variable, you have found that \(y = \frac{19}{12}\). This method can be applied to any linear equation, making it a valuable tool in algebra.
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