Solve by substitution method 6x + 5y=3 and  7x + 8y= 10

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.

Learn how to solve the system of linear equations 6x + 5y = 3 and 7x + 8y = 10 using the substitution method with a step-by-step solution. This problem involves solving linear equations, applying the substitution method, and verifying solutions. Suitable for students in Grades 7-9.

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