Solve for D C=1-5/2D

Solution:

Step 1

Reformulate the equation to $1-\frac{5}{2}D=C$.

Step 2

Simplify the terms in the equation.

Step 2.1

Combine the variable $D$ with the coefficient $\frac{5}{2}$ to get $1-\frac{5D}{2}=C$.

Step 2.2

Position the number $5$ before the variable $D$ to maintain the expression $1-\frac{5D}{2}=C$.

Step 3

Isolate the term containing $D$ by subtracting $1$ from both sides, resulting in $-\frac{5D}{2}=C-1$.

Step 4

To solve for $D$, multiply both sides by $-\frac{2}{5}$, yielding $-\frac{2}{5}(-\frac{5D}{2})=-\frac{2}{5}(C-1)$.

Step 5

Proceed to simplify both sides of the equation.

Step 5.1

First, simplify the left side of the equation.

Step 5.1.1

Start by simplifying $-\frac{2}{5}(-\frac{5D}{2})$.

Step 5.1.1.1

Cancel out the common factor of $2$.

Step 5.1.1.1.1

Shift the negative sign in $-\frac{2}{5}$ to the numerator to get $\frac{-2}{5}(-\frac{5D}{2})=-\frac{2}{5}(C-1)$.

Step 5.1.1.1.2

Move the negative sign in $-\frac{5D}{2}$ to the numerator, resulting in $\frac{-2}{5}\cdot \frac{-5D}{2}=-\frac{2}{5}(C-1)$.

Step 5.1.1.1.3

Extract the factor of $2$ from $-2$ to get $\frac{2(-1)}{5}\cdot \frac{-5D}{2}=-\frac{2}{5}(C-1)$.

Step 5.1.1.1.4

Eliminate the common factor to simplify to $\frac{\cancel{2}\cdot -1}{5}\cdot \frac{-5D}{\cancel{2}}=-\frac{2}{5}(C-1)$.

Step 5.1.1.1.5

Rewrite the simplified expression as $\frac{-1}{5}(-5D)=-\frac{2}{5}(C-1)$.

Step 5.1.1.2

Cancel out the common factor of $5$.

Step 5.1.1.2.1

Extract the factor of $5$ from $-5D$ to get $\frac{-1}{5}(5(-D))=-\frac{2}{5}(C-1)$.

Step 5.1.1.2.2

Eliminate the common factor to simplify to $\frac{-1}{\cancel{5}}(\cancel{5}(-D))=-\frac{2}{5}(C-1)$.

Step 5.1.1.2.3

Rewrite the simplified expression as $D=-\frac{2}{5}(C-1)$.

Step 5.1.1.3

Perform the multiplication.

Step 5.1.1.3.1

Multiply $-1$ by $-D$ to get $D=-\frac{2}{5}(C-1)$.

Step 5.1.1.3.2

Multiply $D$ by $1$ to maintain $D=-\frac{2}{5}(C-1)$.

Step 5.2

Next, simplify the right side of the equation.

Step 5.2.1

Begin by simplifying $-\frac{2}{5}(C-1)$.

Step 5.2.1.1

Apply the distributive property to get $D=-\frac{2}{5}C+\frac{2}{5}(-1)$.

Step 5.2.1.2

Combine the variable $C$ with the coefficient $\frac{2}{5}$ to form $D=-\frac{2C}{5}+\frac{2}{5}(-1)$.

Step 5.2.1.3

Multiply $-\frac{2}{5}$ by $-1$.

Step 5.2.1.3.1

Multiply $-1$ by $-1$ to get $D=-\frac{2C}{5}+\frac{2}{5}$.

Step 5.2.1.3.2

Multiply $\frac{2}{5}$ by $1$ to maintain $D=-\frac{2C}{5}+\frac{2}{5}$.

Step 5.2.1.4

Position the number $2$ before the variable $C$ to finalize $D=-\frac{2C}{5}+\frac{2}{5}$.

Knowledge Notes:

This problem involves solving a linear equation in one variable, which is a fundamental concept in algebra. The steps taken to solve the equation include:

1. Rearranging the Equation: The equation is rewritten to isolate the terms involving the variable on one side and the constants on the other side.

2. Simplifying Terms: Combining like terms and simplifying expressions are essential to reduce the equation to a simpler form.

3. Isolating the Variable: Operations such as addition, subtraction, multiplication, and division are used to isolate the variable on one side of the equation.

4. Cancellation: Common factors in the numerator and denominator of fractions are canceled to simplify the equation further.

5. Distributive Property: This property is used to expand expressions like $a(b+c)$ into $ab+ac$.

6. Multiplication and Division: These operations are used to solve for the variable after all terms have been simplified.

In the context of this problem, the solution process involves manipulating the equation to isolate $D$ and then performing arithmetic operations to solve for $D$. The use of Latex formatting ensures that the mathematical expressions are presented clearly and accurately.