solve for system of equations 2x+y=8 and x-2y=-3

We can solve the system of equations:

1. \( 2x + y = 8 \)
2. \( x - 2y = -3 \)

### Step 1: Solve one equation for a variable
We can solve the second equation for \( x \) in terms of \( y \):
\[
x = 2y - 3
\]

### Step 2: Substitute into the first equation
Now, substitute \( x = 2y - 3 \) into the first equation \( 2x + y = 8 \):
\[
2(2y - 3) + y = 8
\]
\[
4y - 6 + y = 8
\]
\[
5y - 6 = 8
\]
\[
5y = 14
\]
\[
y = \frac{14}{5}
\]

### Step 3: Substitute \( y \) back into the equation for \( x \)
Now, substitute \( y = \frac{14}{5} \) into the equation \( x = 2y - 3 \):
\[
x = 2\left(\frac{14}{5}\right) - 3
\]
\[
x = \frac{28}{5} - 3 = \frac{28}{5} - \frac{15}{5}
\]
\[
x = \frac{13}{5}
\]

### Final solution:
The solution to the system of equations is:
\[
x = \frac{13}{5}, \quad y = \frac{14}{5}
\]

Would you like further details or have any questions?

Here are 5 related questions:
1. How do you solve systems of equations using substitution?
2. Can you solve this system using the elimination method?
3. What are some other methods for solving systems of linear equations?
4. How do systems of equations apply to real-world problems?
5. How can we verify if the solution to the system is correct?

**Tip:** Always check your solution by substituting the values of \(x\) and \(y\) back into the original equations.

Learn how to solve the system of equations 2x + y = 8 and x - 2y = -3 using the substitution method. This step-by-step guide covers key algebraic concepts and provides a clear solution for students in Grades 9-12.

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