Problem 37
Question
For Problems 37-46, solve each equation for the indicated variable. \(2 x-5 y=7\) for \(x\)
Step-by-Step Solution
Verified Answer
The solution for \(x\) is \(x = \frac{5y + 7}{2}\).
1Step 1: Isolate the Term Containing the Variable
Start with the equation \(2x - 5y = 7\). To solve for \(x\), first add \(5y\) to both sides of the equation to isolate the term with \(x\): \[2x = 5y + 7\].
2Step 2: Solve for the Variable
Now, divide both sides of the equation by 2 in order to solve for \(x\): \[x = \frac{5y + 7}{2}\].
Key Concepts
Linear EquationsAlgebraic ManipulationIsolation of Variables
Linear Equations
Linear equations are equations where variables appear only to the first power, without any exponents. These equations form a straight line when graphed. They typically take the general form:
The beauty of linear equations lies in their simplicity, making them a foundational concept in algebra. They are used to solve practical problems involving relationships between quantities. Understanding how to manipulate and solve these equations is key to mastering more complex algebraic problems.
- For one variable: \(ax + b = c\)
- For two variables: \(ax + by = c\)
The beauty of linear equations lies in their simplicity, making them a foundational concept in algebra. They are used to solve practical problems involving relationships between quantities. Understanding how to manipulate and solve these equations is key to mastering more complex algebraic problems.
Algebraic Manipulation
Algebraic manipulation involves using mathematical operations to rearrange equations. The goal is to simplify an expression or solve an equation.
Common techniques include:
For example, in the equation \(2x - 5y = 7\), adding \(5y\) to both sides is an effective move to bring the equation closer to a form where \(x\) can be easily isolated. Remember, the key is to perform operations equally on both sides of the equation to maintain equality.
Common techniques include:
- Adding or subtracting the same term from both sides
- Multiplying or dividing both sides by the same non-zero number
For example, in the equation \(2x - 5y = 7\), adding \(5y\) to both sides is an effective move to bring the equation closer to a form where \(x\) can be easily isolated. Remember, the key is to perform operations equally on both sides of the equation to maintain equality.
Isolation of Variables
The isolation of variables is a method used to solve an equation for a particular variable. This often involves multiple steps, as seen in the original exercise.
The process usually follows these patterns:
In the example \(2x - 5y = 7\) when solving for \(x\), we first move \(5y\) to the other side to leave \(2x\) alone. Then, we simply divide by 2 to isolate \(x\) completely: \( x = \frac{5y + 7}{2}\)
Mastering isolation techniques is vital for systematically solving equations and understanding algebraic expressions.
The process usually follows these patterns:
- Move all terms containing the variable of interest to one side of the equation
- Other terms should be shifted to the opposite side
In the example \(2x - 5y = 7\) when solving for \(x\), we first move \(5y\) to the other side to leave \(2x\) alone. Then, we simply divide by 2 to isolate \(x\) completely: \( x = \frac{5y + 7}{2}\)
Mastering isolation techniques is vital for systematically solving equations and understanding algebraic expressions.
Other exercises in this chapter
Problem 37
Solve each equation and inequality. \(\left|x-\frac{3}{4}\right|=\frac{2}{3}\)
View solution Problem 37
Solve each compound inequality and graph the solution sets. Express the solution sets in interval notation. \(x+23\)
View solution Problem 37
Solve each equation. If a pair of shoes costs a retailer \(\$ 24\), and he sells them for \(\$ 39.60\), what is his rate of profit based on the cost?
View solution Problem 37
Solve each equation. \(\frac{1}{2}(2 x-1)-\frac{1}{3}(5 x+2)=3\)
View solution