To solve this system of linear equations with fractional coefficients:

Given: $\frac{1}{4}x + \frac{2}{5}y = 6$ $\frac{1}{6}x - \frac{1}{3}y = -8$

Step 1: Clear Fractions

To simplify, we can eliminate the fractions by multiplying each equation by the least common multiple (LCM) of the denominators.

1. For the first equation, the denominators are 4 and 5. The LCM is 20, so multiply the entire equation by 20: $20 \cdot \frac{1}{4}x + 20 \cdot \frac{2}{5}y = 20 \cdot 6$ Simplifying: $5x + 8y = 120$

2. For the second equation, the denominators are 6 and 3. The LCM is 6, so multiply the entire equation by 6: $6 \cdot \frac{1}{6}x - 6 \cdot \frac{1}{3}y = 6 \cdot (-8)$ Simplifying: $x - 2y = -48$

Now we have a new system: $5x + 8y = 120$ $x - 2y = -48$

Step 2: Solve for $x$ and $y$

From the second equation, solve for $x$: $x = -48 + 2y$

Substitute $x = -48 + 2y$ into the first equation: $5(-48 + 2y) + 8y = 120$ Expanding: $-240 + 10y + 8y = 120$ Combine terms: $-240 + 18y = 120$ Add 240 to both sides: $18y = 360$ Divide by 18: $y = 20$

Step 3: Substitute $y = 20$ back to find $x$

Using $x = -48 + 2y$: $x = -48 + 2(20)$ $x = -48 + 40$ $x = -8$

Solution

$x = -8, \quad y = 20$

Would you like further details on any of these steps?

Related Questions:

1. How can you verify the solution of a system of equations?
2. What other methods can be used to solve systems of equations with fractions?
3. Why is it helpful to eliminate fractions when solving equations?
4. What is the substitution method for solving systems of equations?
5. How does the elimination method work for systems of linear equations?

Tip: Clearing fractions in equations is a helpful step to simplify calculations and reduce errors, especially in systems of equations.