Solve the system of equations:

-3x + 5y = 21

6x - y = -15

To solve the system of equations:

\[
-3x + 5y = 21 \tag{1}
\]
\[
6x - y = -15 \tag{2}
\]

We can use either the substitution method or the elimination method. Here, let's use the **elimination method**.

### Step 1: Multiply both sides of Equation (2) by 5
We do this to align the coefficients of \( y \) in both equations.

\[
5(6x - y) = 5(-15)
\]

This gives:
\[
30x - 5y = -75 \tag{3}
\]

### Step 2: Add Equation (1) and Equation (3)
Now, add Equation (1) to Equation (3) to eliminate \( y \).

\[
(-3x + 5y) + (30x - 5y) = 21 + (-75)
\]

Simplifying:
\[
-3x + 30x = 21 - 75
\]
\[
27x = -54
\]

### Step 3: Solve for \( x \)
\[
x = \frac{-54}{27} = -2
\]

### Step 4: Substitute \( x = -2 \) back into Equation (2)
Substitute the value of \( x \) into Equation (2) to solve for \( y \).

\[
6(-2) - y = -15
\]
\[
-12 - y = -15
\]
\[
-y = -15 + 12
\]
\[
-y = -3
\]
\[
y = 3
\]

### Final Solution:
The solution to the system of equations is:
\[
x = -2, \quad y = 3
\]

Let me know if you'd like further details or have any questions!

---

Here are 5 related questions to expand your understanding:
1. What would happen if we used the substitution method instead?
2. How can we verify that the solution is correct by substitution into the original equations?
3. What if both equations had different coefficients for \( y \)? How would that affect elimination?
4. How does the elimination method compare to the substitution method for larger systems?
5. How do we handle systems of equations with three variables?

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

Solve the system of equations: -3x + 5y = 21 and 6x - y = -15

This page explains how to solve the system of linear equations -3x + 5y = 21 and 6x - y = -15 using the elimination method. It covers key algebraic concepts, step-by-step calculations, and the elimination method. Suitable for students in Grades 8-10.

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