Question

Solve for \(x.\) Simple Literal Equations. $$c(x-1)=5$$

Short answer

\(x = \frac{5}{c} + 1\)

Step-by-step solution

Isolate the variable term

To solve the equation for x, start by isolating the variable term which is the entire expression within the parentheses. To do this, divide both sides of the equation by c.

Divide both sides by c

Divide both sides of the equation by the coefficient c to get the term with x by itself. The equation then becomes \(x - 1 = \frac{5}{c}\).

Add 1 to both sides of the equation

Now that the term containing x is isolated, add 1 to both sides of the equation to solve for x. Therefore, \(x = \frac{5}{c} + 1\).

Key concepts

Isolate the Variable

Understanding how to isolate the variable is crucial in solving literal equations. The goal is to rearrange the equation so that the variable we're solving for is on one side of the equation all by itself. Do this by performing operations that 'undo' what's being done to the variable. For instance, if the variable is being multiplied by a number, we divide by that number to neutralize the multiplication. Similarly, if it is inside parentheses with a number subtracting or adding to it, we reverse this by doing the opposite operation to both sides of the equation. This allows us to 'isolate' the variable and make it the subject of the formula.

In the given exercise, with the equation \(c(x-1)=5\), the variable \(x\) is not isolated because it is both inside the parentheses and being multiplied by \(c\). The first step of division by \(c\) eliminates the multiplication, leading us towards having \(x\) by itself.

Literal Equations Step by Step

Tackling literal equations step by step ensures clarity and avoids common mistakes. Here's how you apply this method:

Identify the Variable

First, identify the variable to be isolated, which in our case is \(x\).

Undo Multiplication or Division

Next, if the variable is multiplied by or divided into a number (as \(x\) is multiplied by \(c\) in our exercise), you need to perform the opposite operation. We divided both sides by \(c\) to neutralize the multiplication.

Undo Addition or Subtraction

Finally, if the variable has numbers added to or subtracted from it (as \(1\) is subtracted from \(x\) in our exercise), do the opposite operation to both sides. We added \(1\) to both sides of our equation to isolate \(x\). By following these steps systematically, you can solve for the variable without confusion.

Algebraic Manipulation

Algebraic manipulation involves the use of mathematical operations to rearrange and simplify algebraic equations. It's a core skill in solving literal equations as it requires familiarity with algebraic properties, such as the distributive property, the commutative property, and the property of inverses.

For successful algebraic manipulation:

• Ensure you perform the same operation on both sides of the equation to maintain balance.
• Be comfortable with inverse operations: addition and subtraction are inverses, as are multiplication and division.
• Keep track of the equation as it transforms from one form to another, so you know at each stage what has been done and what still needs to be done.

Using algebraic manipulation, one can navigate through the complexities of literal equations with ease. In our exercise, after dividing both sides by \(c\), we employed addition to move the \(1\) over to the other side, bringing us one step closer to the solution, \(x\).