To solve the system of linear equations using the substitution method, we'll follow these steps:

Given Equations:

1. $6x + 5y = 3$ $\quad \text{(Equation 1)}$
2. $7x + 8y = 10$ $\quad \text{(Equation 2)}$

Step 1: Solve one equation for one variable.

Let's solve Equation 1 for $x$:

$6x + 5y = 3$ $6x = 3 - 5y$ $x = \frac{3 - 5y}{6}$ $x = \frac{1}{2} - \frac{5y}{6}$

Step 2: Substitute the expression for $x$ into the other equation.

Substitute $x = \frac{1}{2} - \frac{5y}{6}$ into Equation 2:

$7\left(\frac{1}{2} - \frac{5y}{6}\right) + 8y = 10$

Step 3: Solve for $y$.

Distribute and simplify:

$\frac{7}{2} - \frac{35y}{6} + 8y = 10$

To eliminate the fractions, multiply the entire equation by 6:

$6\left(\frac{7}{2} - \frac{35y}{6} + 8y\right) = 60$

$21 - 35y + 48y = 60$

Combine like terms:

$13y + 21 = 60$

Subtract 21 from both sides:

$13y = 39$

Now, solve for $y$:

$y = \frac{39}{13} = 3$

Step 4: Substitute $y = 3$ back into the expression for $x$.

Now that we have $y = 3$, substitute it into the expression for $x$:

$x = \frac{1}{2} - \frac{5(3)}{6}$

Simplify:

$x = \frac{1}{2} - \frac{15}{6}$

Convert $\frac{15}{6}$ to $\frac{5}{2}$:

$x = \frac{1}{2} - \frac{5}{2} = \frac{1 - 5}{2} = \frac{-4}{2} = -2$

Final Answer:

The solution to the system of equations is: $x = -2, \quad y = 3$

Would you like more details or have any questions?

Here are some related questions for further practice:

1. How do you solve a system of equations using the elimination method?
2. What are the benefits of using the substitution method over other methods?
3. How can you verify the solution of a system of equations graphically?
4. Can a system of linear equations have no solution or infinite solutions? If so, when?
5. How do you interpret the solution of a system of equations in a real-world context?

Tip: Always substitute the solutions back into the original equations to check if they satisfy both equations.