If a + b + c = 7/12 3a - 4b + 5c = 3/4 and 7a - 11b - 13c = - 7/12

then what is the value of a + c = ?

We are given three equations:

1. \( a + b + c = \frac{7}{12} \)
2. \( 3a - 4b + 5c = \frac{3}{4} \)
3. \( 7a - 11b - 13c = -\frac{7}{12} \)

We need to find the value of \( a + c \).

Let's approach this step by step.

### Step 1: Express \( b \) in terms of \( a \) and \( c \)
From the first equation:
\[
a + b + c = \frac{7}{12} \implies b = \frac{7}{12} - a - c
\]

### Step 2: Substitute \( b \) into the other two equations
Now, substitute \( b = \frac{7}{12} - a - c \) into the second and third equations.

#### Substitute into the second equation:
\[
3a - 4\left( \frac{7}{12} - a - c \right) + 5c = \frac{3}{4}
\]
Simplify:
\[
3a - 4 \times \frac{7}{12} + 4a + 4c + 5c = \frac{3}{4}
\]
\[
3a - \frac{28}{12} + 4a + 9c = \frac{3}{4}
\]
\[
7a + 9c - \frac{7}{3} = \frac{3}{4}
\]
To eliminate fractions, multiply through by 12 (the least common denominator of 3 and 4):
\[
12 \times (7a + 9c - \frac{7}{3}) = 12 \times \frac{3}{4}
\]
\[
84a + 108c - 28 = 9
\]
\[
84a + 108c = 37 \quad \text{(Equation 4)}
\]

#### Substitute into the third equation:
\[
7a - 11\left( \frac{7}{12} - a - c \right) - 13c = -\frac{7}{12}
\]
Simplify:
\[
7a - 11 \times \frac{7}{12} + 11a + 11c - 13c = -\frac{7}{12}
\]
\[
7a - \frac{77}{12} + 11a - 2c = -\frac{7}{12}
\]
\[
18a - 2c - \frac{77}{12} = -\frac{7}{12}
\]
Multiply through by 12:
\[
12 \times (18a - 2c - \frac{77}{12}) = 12 \times -\frac{7}{12}
\]
\[
216a - 24c - 77 = -7
\]
\[
216a - 24c = 70 \quad \text{(Equation 5)}
\]

### Step 3: Solve the system of two equations
Now, we have the following system of two equations:
1. \( 84a + 108c = 37 \)
2. \( 216a - 24c = 70 \)

We can solve this system using elimination or substitution. Let's use elimination.

#### Multiply the first equation by 2 and the second equation by 9 to align the coefficients of \( c \):
\[
2 \times (84a + 108c) = 2 \times 37 \quad \text{(Equation 6)}
\]
\[
9 \times (216a - 24c) = 9 \times 70 \quad \text{(Equation 7)}
\]
This gives:
\[
168a + 216c = 74 \quad \text{(Equation 6)}
\]
\[
1944a - 216c = 630 \quad \text{(Equation 7)}
\]

#### Add the two equations:
\[
168a + 216c + 1944a - 216c = 74 + 630
\]
\[
2112a = 704
\]
\[
a = \frac{704}{2112} = \frac{1}{3}
\]

### Step 4: Substitute \( a = \frac{1}{3} \) into one of the original equations
Substitute \( a = \frac{1}{3} \) into the first equation \( a + b + c = \frac{7}{12} \):
\[
\frac{1}{3} + b + c = \frac{7}{12}
\]
\[
b + c = \frac{7}{12} - \frac{4}{12} = \frac{3}{12} = \frac{1}{4}
\]

Now we know that:
\[
b + c = \frac{1}{4}
\]

### Final Answer:
The value of \( a + c \) is:

\[
a + c = \frac{1}{3} + c
\]

Do you want any specific clarifications or more details on the steps? 
Here are 5 related questions for further practice:
1. How do you solve systems of linear equations using matrices?
2. What are the different methods to solve a system of linear equations?
3. How does substitution help in solving systems of equations?
4. Can you always solve three-variable systems using substitution?
5. What are some special cases where systems of equations have no solution?

**Tip**: When solving linear equations, aligning terms for easy elimination or substitution makes the process faster and more efficient!

If a + b + c = 7/12, 3a - 4b + 5c = 3/4, and 7a - 11b - 13c = -7/12, then what is the value of a + c = ?

Learn how to solve the system of linear equations a + b + c = 7/12, 3a - 4b + 5c = 3/4, and 7a - 11b - 13c = -7/12 using the substitution and elimination methods. Suitable for high school students in Grades 9-12.

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