3x+4y=-2, -3x-5y = 1 solve by substitution method 

To solve the system of equations using the substitution method, we can follow these steps:

Given the system:

1. \( 3x + 4y = -2 \)
2. \( -3x - 5y = 1 \)

### Step 1: Solve one equation for one variable
Let's take the first equation and solve for \( x \) in terms of \( y \).

From \( 3x + 4y = -2 \):
\[
3x = -2 - 4y
\]
\[
x = \frac{-2 - 4y}{3}
\]

### Step 2: Substitute this expression for \( x \) into the second equation
Now substitute \( x = \frac{-2 - 4y}{3} \) into the second equation \( -3x - 5y = 1 \):

\[
-3\left( \frac{-2 - 4y}{3} \right) - 5y = 1
\]

Simplify:
\[
(-3) \times \frac{-2 - 4y}{3} = 2 + 4y
\]
So, the equation becomes:
\[
2 + 4y - 5y = 1
\]
Simplify further:
\[
2 - y = 1
\]
\[
-y = 1 - 2
\]
\[
-y = -1
\]
\[
y = 1
\]

### Step 3: Substitute \( y = 1 \) back into the equation for \( x \)
Now, substitute \( y = 1 \) into the expression for \( x \):
\[
x = \frac{-2 - 4(1)}{3} = \frac{-2 - 4}{3} = \frac{-6}{3} = -2
\]

### Solution:
The solution to the system is:
\[
x = -2, \quad y = 1
\]

Would you like more details or have any questions?

Here are 5 related questions:
1. What if both equations had no solution (inconsistent system)?
2. How would you solve this system using the elimination method?
3. What happens if both equations describe the same line?
4. How can you verify if the solution is correct?
5. Can substitution be used when both equations are in standard form?

**Tip**: Always substitute your values back into the original equations to verify the solution is correct!

3x + 4y = -2, -3x - 5y = 1 solve by substitution method

Learn how to solve the system of linear equations 3x + 4y = -2 and -3x - 5y = 1 using the substitution method. This step-by-step guide covers key algebraic techniques for solving equations in two variables. Ideal for students in Grades 8-10.

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