using substitution method solve 4x+y=2 and 5x-5y=-5

To solve the system of equations using the substitution method, we have:

\[
4x + y = 2 \tag{1}
\]
\[
5x - 5y = -5 \tag{2}
\]

### Step 1: Solve for \( y \) in terms of \( x \) using Equation (1)
From Equation (1), we isolate \( y \):
\[
y = 2 - 4x
\]

### Step 2: Substitute \( y \) in Equation (2)
Now, substitute \( y = 2 - 4x \) into Equation (2):
\[
5x - 5(2 - 4x) = -5
\]

Expand and simplify:
\[
5x - 10 + 20x = -5
\]
\[
25x - 10 = -5
\]

Add 10 to both sides:
\[
25x = 5
\]

Divide by 25:
\[
x = \frac{5}{25} = \frac{1}{5}
\]

### Step 3: Substitute \( x = \frac{1}{5} \) back into Equation (1) to find \( y \)
Using Equation (1):
\[
4\left(\frac{1}{5}\right) + y = 2
\]
\[
\frac{4}{5} + y = 2
\]

Subtract \( \frac{4}{5} \) from both sides:
\[
y = 2 - \frac{4}{5} = \frac{10}{5} - \frac{4}{5} = \frac{6}{5}
\]

### Solution
The solution to the system is:
\[
x = \frac{1}{5}, \quad y = \frac{6}{5}
\]

Would you like additional details on this solution, or have any other questions?

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**Further Exploration Questions:**
1. How would the solution change if we used the elimination method?
2. Can you graph these equations and verify the solution visually?
3. What does the solution tell us about the intersection point of the two lines?
4. How would this method apply to a system with three variables?
5. What happens if the system of equations has no solution?

**Tip:** Always check your solution by plugging values back into the original equations to verify they hold true.

Using substitution method solve 4x+y=2 and 5x-5y=-5

This page explains how to solve the system of linear equations 4x + y = 2 and 5x - 5y = -5 using the substitution method. A step-by-step approach simplifies the problem for students in Grades 8-10, helping them grasp core algebraic concepts and techniques.

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