Solve for a and `b:2(a+b)-(a-b)=6,4(a-b)=2(a+b)-9`

Options

A. `a=3/4,b=1 3/4`
B. `a=1/2,b=1 1/2`
C. `a=1 3/4, b=3/4`
D. `a=3/2, b=3/4`

Answer

To solve the system of equations given by:

1. \( 2(a + b) - (a - b) = 6 \)
2. \( 4(a - b) = 2(a + b) - 9 \)

we will follow these steps:

### Step 1: Simplify the first equation

Starting with the first equation:

\[ 2(a + b) - (a - b) = 6 \]

Distributing the terms:

\[ 2a + 2b - a + b = 6 \]

Combining like terms:

\[ (2a - a) + (2b + b) = 6 \]

This simplifies to:

\[ a + 3b = 6 \]

Let's label this as Equation (1):

\[ a + 3b = 6 \quad \text{(1)} \]

### Step 2: Simplify the second equation

Now, let's simplify the second equation:

\[ 4(a - b) = 2(a + b) - 9 \]

Distributing the terms:

\[ 4a - 4b = 2a + 2b - 9 \]

Rearranging the equation by moving all terms to one side:

\[ 4a - 4b - 2a - 2b + 9 = 0 \]

Combining like terms:

\[ (4a - 2a) + (-4b - 2b) + 9 = 0 \]

This simplifies to:

\[ 2a - 6b + 9 = 0 \]

Let's label this as Equation (2):

\[ 2a - 6b + 9 = 0 \quad \text{(2)} \]

### Step 3: Solve the system of equations

Now we have the two equations:

1. \( a + 3b = 6 \) (1)
2. \( 2a - 6b + 9 = 0 \) (2)

From Equation (1), we can express \( a \) in terms of \( b \):

\[ a = 6 - 3b \]

Now, substitute \( a \) in Equation (2):

\[ 2(6 - 3b) - 6b + 9 = 0 \]

Distributing the 2:

\[ 12 - 6b - 6b + 9 = 0 \]

Combining like terms:

\[ 12 + 9 - 12b = 0 \]

This simplifies to:

\[ 21 - 12b = 0 \]

Now, solving for \( b \):

\[ 12b = 21 \implies b = \frac{21}{12} = \frac{7}{4} \]

### Step 4: Find the value of \( a \)

Now that we have \( b \), we can find \( a \) using Equation (1):

\[ a = 6 - 3b = 6 - 3\left(\frac{7}{4}\right) \]

Calculating \( 3 \times \frac{7}{4} \):

\[ 3 \times \frac{7}{4} = \frac{21}{4} \]

Now substituting back:

\[ a = 6 - \frac{21}{4} \]

Converting 6 into a fraction with a denominator of 4:

\[ 6 = \frac{24}{4} \]

Now, substituting:

\[ a = \frac{24}{4} - \frac{21}{4} = \frac{3}{4} \]

### Final Solution

Thus, the values of \( a \) and \( b \) are:

\[ a = \frac{3}{4}, \quad b = \frac{7}{4} \]