Solve for F 2/3(F+G)=D

The problem presents an equation in which the quantity $F$ is related to another quantity $G$ and to a third quantity $D$ through a mathematical relationship. Specifically, the equation states that two-thirds of the sum of $F$ and $G$ is equal to $D$ . The challenge posed by the problem is to isolate and solve for $F$ to find its value in terms of the other two known quantities, $G$ and $D$ . It requires the application of algebraic manipulation skills, such as distributing, combining like terms, and solving for a variable.

$\frac{2}{3} (F + G) = D$

Answer

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Solution:

Step 1:

Isolate the term involving F by multiplying both sides by $\frac{3}{2}$ to eliminate the fraction.

$\frac{3}{2} \times \frac{2}{3} (F + G) = \frac{3}{2} \times D$

Step 2:

Proceed to simplify the equation.

Step 2.1:

Begin with the left-hand side of the equation.

Step 2.1.1:

First, simplify the expression $\frac{3}{2} \times \frac{2}{3} (F + G)$ .

Step 2.1.1.1:

Utilize the distributive property to expand $\frac{3}{2} \times (\frac{2}{3} F + \frac{2}{3} G) = \frac{3}{2} D$ .

Step 2.1.1.2:

Combine like terms by multiplying $\frac{2}{3}$ with F to get $\frac{3}{2} \times (\frac{2 F}{3} + \frac{2}{3} G) = \frac{3}{2} D$ .

Step 2.1.1.3:

Similarly, combine $\frac{2}{3}$ with G to obtain $\frac{3}{2} \times (\frac{2 F}{3} + \frac{2 G}{3}) = \frac{3}{2} D$ .

Step 2.1.1.4:

Apply the distributive property again to get $\frac{3}{2} \times \frac{2 F}{3} + \frac{3}{2} \times \frac{2 G}{3} = \frac{3}{2} D$ .

Step 2.1.1.5:

Eliminate the common factor of 3.

Step 2.1.1.5.1:

Remove the common factor to simplify the expression to $\frac{1}{2} \times 2 F + \frac{3}{2} \times \frac{2 G}{3} = \frac{3}{2} D$ .

Step 2.1.1.5.2:

Rewrite the equation as $\frac{1}{2} \times 2 F + \frac{3}{2} \times \frac{2 G}{3} = \frac{3}{2} D$ .

Step 2.1.1.6:

Eliminate the common factor of 2.

Step 2.1.1.6.1:

Extract the factor of 2 from 2F to get $\frac{1}{2} \times (2 \times F) + \frac{3}{2} \times \frac{2 G}{3} = \frac{3}{2} D$ .

Step 2.1.1.6.2:

Remove the common factor to simplify further to $F + \frac{3}{2} \times \frac{2 G}{3} = \frac{3}{2} D$ .

Step 2.1.1.6.3:

Rewrite the expression as $F + \frac{3}{2} \times \frac{2 G}{3} = \frac{3}{2} D$ .

Step 2.1.1.7:

Eliminate the common factor of 3.

Step 2.1.1.7.1:

Remove the common factor to get $F + \frac{1}{2} \times 2 G = \frac{3}{2} D$ .

Step 2.1.1.7.2:

Rewrite the equation as $F + \frac{1}{2} \times 2 G = \frac{3}{2} D$ .

Step 2.1.1.8:

Eliminate the common factor of 2.

Step 2.1.1.8.1:

Extract the factor of 2 from 2G to get $F + \frac{1}{2} \times (2 \times G) = \frac{3}{2} D$ .

Step 2.1.1.8.2:

Remove the common factor to simplify to $F + G = \frac{3}{2} D$ .

Step 2.1.1.8.3:

Rewrite the final expression as $F + G = \frac{3}{2} D$ .

Step 2.2:

Now, simplify the right-hand side of the equation.

Step 2.2.1:

Combine $\frac{3}{2}$ with D to get $F + G = \frac{3 D}{2}$ .

Step 3:

Finally, isolate F by subtracting G from both sides.

$F = \frac{3 D}{2} - G$

Knowledge Notes:

To solve the equation $2 / 3 (F + G) = D$ , we follow a systematic approach:

Multiplication by the Reciprocal: To eliminate the fraction $2 / 3$ , we multiply both sides of the equation by its reciprocal, which is $3 / 2$ . This step simplifies the equation by getting rid of the fraction.
Simplification: We simplify the equation step by step, starting with the left side and then the right side. This involves expanding and combining like terms.
Distributive Property: This property states that $a (b + c) = a b + a c$ . We use this property to expand expressions like $\frac{3}{2} \times \frac{2}{3} (F + G)$ .
Canceling Common Factors: When we have common factors in the numerator and the denominator, we can cancel them out to simplify the expression further.
Isolating the Variable: The final step is to isolate the variable we are solving for, in this case, F. We do this by moving all other terms to the opposite side of the equation.

Throughout this process, we ensure that each step is mathematically valid and brings us closer to the simplest form of the equation, from which we can easily find the solution for F.