We are given the system of linear equations:

\[
\begin{aligned}
1. \quad 2x - 4y &= 10 \\
2. \quad x + 5y &= 40
\end{aligned}
\]

Let's solve this system using the substitution or elimination method. I'll use substitution to solve it step-by-step.

### Step 1: Solve one equation for one variable
From the second equation:
\[
x + 5y = 40
\]
We can solve for \( x \):
\[
x = 40 - 5y
\]

### Step 2: Substitute into the first equation
Now, substitute \( x = 40 - 5y \) into the first equation:
\[
2(40 - 5y) - 4y = 10
\]
Simplifying:
\[
80 - 10y - 4y = 10
\]
Combine like terms:
\[
80 - 14y = 10
\]
Subtract 80 from both sides:
\[
-14y = -70
\]
Divide by -14:
\[
y = 5
\]

### Step 3: Substitute \( y = 5 \) back into the equation for \( x \)
Now substitute \( y = 5 \) into the equation \( x = 40 - 5y \):
\[
x = 40 - 5(5) = 40 - 25 = 15
\]

### Final Solution:
The solution to the system is:
\[
x = 15, \quad y = 5
\]

Would you like further details or have any questions?

### Related Questions:
1. How can you solve the system using the elimination method?
2. What is the graphical interpretation of this system of equations?
3. How would the solution change if the second equation was \(x + 5y = 50\)?
4. Can you set up a similar system with different coefficients?
5. What are the steps for solving a system with three variables?

### Tip:
Always double-check your substitution in the original equations to ensure the solution satisfies both equations.

Learn how to solve the system of equations 2x - 4y = 10 and x + 5y = 40 using the substitution method. This example includes detailed step-by-step solutions, covering algebraic concepts like linear equations and systems of equations. Suitable for students in Grades 8-10.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation