To solve the given system of equations, let's first rewrite the equations for clarity:

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

Step 1: Eliminate Fractions by Multiplying Both Sides by a Common Denominator

For the first equation:

$\frac{1}{6}x - \frac{1}{3}y = -6$

The least common multiple (LCM) of 6 and 3 is 6. Multiply everything by 6:

$x - 2y = -36 \quad \text{(Equation 1)}$

For the second equation:

$\frac{1}{4}x + \frac{3}{5}y = 2$

The LCM of 4 and 5 is 20. Multiply everything by 20:

$5x + 12y = 40 \quad \text{(Equation 2)}$

Step 2: Solve the System of Equations Using the Elimination Method

Now, we have the system:

1. $x - 2y = -36$
2. $5x + 12y = 40$

Let's eliminate $x$ by making the coefficients of $x$ the same in both equations. We can multiply the first equation by 5:

$5(x - 2y) = 5(-36)$

$5x - 10y = -180 \quad \text{(Equation 3)}$

Now, we have:

1. $5x - 10y = -180$
2. $5x + 12y = 40$

Subtract Equation 1 from Equation 2 to eliminate $x$:

$(5x + 12y) - (5x - 10y) = 40 - (-180)$

$5x + 12y - 5x + 10y = 220$

$22y = 220$

$y = 10$

Step 3: Substitute $y$ Back to Find $x$

Substitute $y = 10$ back into Equation 1:

$x - 2(10) = -36$

$x - 20 = -36$

$x = -16$

Solution

The solution to the system of equations is:

$x = -16, \quad y = 10$

Would you like more details or have any questions?

Related Questions

1. How do you solve a system of equations using the substitution method?
2. What are the common applications of solving systems of equations?
3. How can you determine if a system of equations has no solution, one solution, or infinitely many solutions?
4. What is the graphical interpretation of solving a system of equations?
5. How does the determinant of a coefficient matrix relate to solving systems of linear equations?

Tip

When dealing with fractions in a system of equations, it's often helpful to clear the fractions by multiplying through by the least common denominator. This simplifies the equations and makes further algebraic manipulation easier.