what is the solution of the following linear equations: x+2y=-1 and 1/2x-3=y

To solve the system of equations:

\[
x + 2y = -1
\]
\[
\frac{1}{2}x - 3 = y
\]

we can use substitution or elimination. Here, I'll proceed with substitution by solving for \( y \) in terms of \( x \) from the second equation and substituting it into the first equation.

### Step 1: Solve for \( y \) in terms of \( x \)

The second equation is:
\[
\frac{1}{2}x - 3 = y
\]
Rewrite it as:
\[
y = \frac{1}{2}x - 3
\]

### Step 2: Substitute \( y = \frac{1}{2}x - 3 \) into the first equation

Substitute this expression for \( y \) into the first equation:
\[
x + 2\left(\frac{1}{2}x - 3\right) = -1
\]

### Step 3: Simplify and solve for \( x \)

Expanding inside the parentheses:
\[
x + x - 6 = -1
\]
Combine like terms:
\[
2x - 6 = -1
\]
Add 6 to both sides:
\[
2x = 5
\]
Divide by 2:
\[
x = \frac{5}{2}
\]

### Step 4: Substitute \( x = \frac{5}{2} \) back into \( y = \frac{1}{2}x - 3 \) to find \( y \)

Substitute \( x = \frac{5}{2} \) into the equation for \( y \):
\[
y = \frac{1}{2}\left(\frac{5}{2}\right) - 3
\]
\[
y = \frac{5}{4} - 3
\]
Convert 3 to a fraction with a denominator of 4:
\[
y = \frac{5}{4} - \frac{12}{4}
\]
\[
y = \frac{-7}{4}
\]

### Solution

The solution to the system is:
\[
x = \frac{5}{2}, \quad y = -\frac{7}{4}
\]

Would you like further details or have any questions?

Here are five related questions to explore further:
1. How can we verify the solution for a system of equations?
2. What are some other methods for solving systems of linear equations?
3. How does the substitution method compare to the elimination method?
4. What would the solution look like on a graph?
5. How does solving for one variable simplify substitution?

**Tip:** In linear equations, always check your solution by substituting the values back into both equations.

What is the solution of the following linear equations: x+2y=-1 and 1/2x-3=y

Discover step-by-step how to solve the system of linear equations x + 2y = -1 and 1/2x - 3 = y using substitution. Learn key algebraic methods including substitution, fraction manipulation, and combining like terms, perfect for Grades 8-10 students.

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