Problem 50
Question
Solve each equation for \(x\). $$5 x-(2 a+c)=4(x+c)$$
Step-by-Step Solution
Verified Answer
The solution for x is \( x = 2a + 5c \).
1Step 1 - Distribute terms on the right side
Distribute the 4 on the right side of the equation: \[ 5x - (2a + c) = 4x + 4c \]
2Step 2 - Move all terms involving x to one side
Subtract 4x from both sides of the equation to isolate the terms involving x on one side: \[ 5x - 4x - (2a + c) = 4c \]This simplifies to: \[ x - (2a + c) = 4c \]
3Step 3 - Move constant terms to the other side
Add (2a + c) to both sides to isolate x: \[ x = 4c + 2a + c \]Combine like terms on the right side: \[ x = 2a + 5c \]
Key Concepts
linear equationsalgebraic manipulationdistributive propertyisolating variables
linear equations
Linear equations are fundamental in algebra. They are equations of the first order, meaning they involve variables raised to the power of one. These equations take the form ax + b = c , where 'a,' 'b,' and 'c' are constants, and 'x' is the variable. The goal in solving linear equations is to find the value of 'x' that makes the equation true. Let's explore simple steps and techniques to solve linear equations.
algebraic manipulation
Algebraic manipulation involves rearranging equations to simplify and solve for a variable. Here are some key techniques:
- Adding or subtracting the same value from both sides of the equation.
- Multiplying or dividing both sides of the equation by the same nonzero value.
distributive property
The distributive property is an important algebraic property used to simplify expressions. It states that: a(b + c) = ab + ac . In our given equation 5 x-(2 a+c)=4(x+c) , applying the distributive property helps us deal with parentheses. For example, 4(x+c) becomes 4x + 4c when distributed. Distributing terms efficiently can clarify complex equations and make them more manageable.
isolating variables
Isolating the variable means getting the variable (x) alone on one side of the equation. This is typically done in stages:
- Move all terms containing the variable to one side of the equation.
- Move all constant terms to the opposite side.
Other exercises in this chapter
Problem 50
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