https://file.aiquickdraw.com/mathbot/file/1727283973930p96essjq.jpg

The image shows a system of linear equations:

\[
\begin{aligned}
1. \quad 4x + y &= 2 \\
2. \quad 2x - y &= 0
\end{aligned}
\]

We can solve this system using either substitution or elimination. Let's proceed with the **elimination method** by adding the two equations together to eliminate \( y \).

First, add both equations:
\[
(4x + y) + (2x - y) = 2 + 0
\]
This simplifies to:
\[
6x = 2
\]
Solving for \( x \):
\[
x = \frac{2}{6} = \frac{1}{3}
\]

Now, substitute \( x = \frac{1}{3} \) back into one of the original equations, say the first one:
\[
4\left(\frac{1}{3}\right) + y = 2
\]
Simplifying:
\[
\frac{4}{3} + y = 2
\]
Subtract \( \frac{4}{3} \) from both sides:
\[
y = 2 - \frac{4}{3} = \frac{6}{3} - \frac{4}{3} = \frac{2}{3}
\]

Thus, the solution to the system of equations is:
\[
x = \frac{1}{3}, \quad y = \frac{2}{3}
\]

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

Here are 5 related questions for practice:
1. How would you solve the system if you used the substitution method?
2. Can you write a word problem that matches this system of equations?
3. What is the graphical interpretation of the solution \( x = \frac{1}{3}, y = \frac{2}{3} \)?
4. How would the solution change if the second equation was \( 2x - y = 1 \)?
5. What is the determinant of the coefficient matrix for this system?

**Tip:** When solving systems of equations, both substitution and elimination are useful; choose the one that seems easier based on the equations!

Solve the system of equations: 4x + y = 2 and 2x - y = 0

Learn how to solve the system of linear equations 4x + y = 2 and 2x - y = 0 using the elimination method. This problem covers key algebraic concepts and is suitable for students in Grades 8-10.

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