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!