Problem 70 - Intermediate Algebra

Question

Solve for the specified variable. $$ l=\frac{a-S+S r}{r} \quad \text { for } a $$

Step-by-Step Solution

Answer

The solution for \( a \) is: \( a = lr + S - Sr \).

Step 1: Multiply Both Sides by r

To eliminate the fraction, multiply both sides of the equation by \( r \):\[ l imes r = rac{a - S + Sr}{r} imes r \]This simplifies to:\[ lr = a - S + Sr \]

Step 2: Rearrange Terms

Move all terms involving \( a \) to one side and the other constants to the opposite side of the equation:\[ lr + S - Sr = a \]

Step 3: Isolate Variable a

We already have \( a \) isolated after the rearrangement in step 2:\[ a = lr + S - Sr \]

Step 4: Final Equation

Thus, the solution for \( a \) is:\[ a = lr + S - Sr \]

Key Concepts

Linear Equations

Linear equations are a fundamental part of algebra and appear frequently in mathematics. They are called "linear" because they represent lines when graphed in two dimensions. A standard linear equation usually appears in the form of \[ ax + b = c \] where "\( x \)" is the variable, and "\( a \)," "\( b \)," and "\( c \)" are constants. In a linear equation, the power of the variable is always one, which makes these types of equations easy to work with. You will often find yourself solving linear equations to find the value of one specific variable.

• Linear equation graphs form straight lines.
• These equations often model real-world situations.

Being comfortable with linear equations is an important step to mastering more complex algebraic concepts later on. Solving these equations typically involves simple operations such as addition, subtraction, multiplication, or division.

Variable Isolation

Variable isolation is a crucial step in solving equations as it allows you to find the value of the unknown variable. Essentially, isolating a variable means getting the variable by itself on one side of the equation, with everything else on the other side. This helps easily determine the value of the variable. In our example problem, the variable "\( a \)" was isolated step by step:

• We first eliminated any fractions by multiplying both sides by "\( r \)."
• We then rearranged the equation so that all terms involving "\( a \)" were isolated on one side.
• The remaining constants were grouped on the opposite side.

This involves applying inverse operations, which essentially undoes what is being done to the variable. By working step by step, it becomes easier to maintain balance and correctly isolate the variable.

Algebraic Manipulation

Algebraic manipulation involves using various mathematical operations to simplify and solve equations. These operations include addition, subtraction, multiplication, division, and rearranging terms to transform equations into simpler forms.In our step-by-step solution, we used algebraic manipulation:

• By multiplying both sides of the equation by "\( r \)," we eliminated the fraction.
• Then, we rearranged the terms to bring like variables together for simplification.
• Finally, by isolating "\( a \)," we found the solution as a function of other variables.

Understanding how to manipulate equations algebraically is essential. It not only helps solve equations but also aids in recognizing relationships between different components within an equation.