1/2 * x - 3/4 * y = 1/2 1/5 * x - 3/10 * y = 1/5

You have a system of two linear equations:

1. \(\frac{1}{2}x - \frac{3}{4}y = \frac{1}{2}\)
2. \(\frac{1}{5}x - \frac{3}{10}y = \frac{1}{5}\)

Let's solve this system step by step.

### Step 1: Eliminate fractions

To eliminate fractions, we can multiply both sides of each equation by the least common denominator (LCD) of the denominators in that equation.

#### First equation:
\[
\frac{1}{2}x - \frac{3}{4}y = \frac{1}{2}
\]
The LCD of 2 and 4 is 4. Multiply through by 4:
\[
4 \times \left(\frac{1}{2}x\right) - 4 \times \left(\frac{3}{4}y\right) = 4 \times \frac{1}{2}
\]
\[
2x - 3y = 2
\]

#### Second equation:
\[
\frac{1}{5}x - \frac{3}{10}y = \frac{1}{5}
\]
The LCD of 5 and 10 is 10. Multiply through by 10:
\[
10 \times \left(\frac{1}{5}x\right) - 10 \times \left(\frac{3}{10}y\right) = 10 \times \frac{1}{5}
\]
\[
2x - 3y = 2
\]

### Step 2: Compare the equations

Both equations simplify to:
\[
2x - 3y = 2
\]

This means the two equations are identical, so there is no unique solution. Instead, this system represents **infinitely many solutions**. The two equations describe the same line, and any point on this line will satisfy the system.

### Final Answer:
The system has infinitely many solutions.

Would you like details on how to express the solutions, or do you have any other questions?

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Here are 5 related questions to expand on this:
1. How do you solve a system of equations with different slopes?
2. What is the graphical interpretation of this solution?
3. How can we express the solution of this system parametrically?
4. What happens when two equations are inconsistent?
5. How can substitution be used to solve linear systems?

**Tip:** Always simplify equations by eliminating fractions early to make the calculations easier.

1/2 * x - 3/4 * y = 1/2, 1/5 * x - 3/10 * y = 1/5

Explore how to solve a system of linear equations involving fractions: 1/2 * x - 3/4 * y = 1/2 and 1/5 * x - 3/10 * y = 1/5. Learn to eliminate fractions and identify infinitely many solutions due to identical equations. Suitable for students in Grades 8-10.

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