To solve this system of equations, let's begin by rewriting the equations for clarity:

1. $\frac{1}{2}x + \frac{1}{4}y = -4$
2. $-\frac{2}{3}x + \frac{1}{2}y = 7$

To eliminate the fractions, we can multiply each equation by the least common multiple of the denominators.

Step 1: Clear the Fractions

For Equation 1:

• The denominators are 2 and 4, so we multiply by 4.

$4 \times \left(\frac{1}{2}x + \frac{1}{4}y\right) = 4 \times -4$ $2x + y = -16$

For Equation 2:

• The denominators are 3 and 2, so we multiply by 6.

$6 \times \left(-\frac{2}{3}x + \frac{1}{2}y\right) = 6 \times 7$ $-4x + 3y = 42$

Now, we have the following system of equations:

1. $2x + y = -16$
2. $-4x + 3y = 42$

Step 2: Solve by Elimination

To eliminate $y$, we can multiply the first equation by 3 and add it to the second equation.

1. $3 \cdot (2x + y) = 3 \cdot -16$ $6x + 3y = -48$

Now we have: $6x + 3y = -48$ $-4x + 3y = 42$

Subtract the second equation from the first:

$(6x + 3y) - (-4x + 3y) = -48 - 42$ $10x = -90$ $x = -9$

Step 3: Substitute $x = -9$ into the First Equation

Now that we have $x = -9$, substitute it back into $2x + y = -16$:

$2(-9) + y = -16$ $-18 + y = -16$ $y = 2$

Solution

The solution to the system is: $x = -9, \quad y = 2$

Would you like a further breakdown of any of these steps or have any questions?

Here are 5 related questions to deepen your understanding:

1. What is the process of clearing fractions in systems of equations?
2. How does elimination work, and when is it most effective?
3. Why is it helpful to eliminate one variable before solving?
4. Could this system also be solved by substitution, and how?
5. How would you check if the solution is correct?

Tip: When dealing with fractions in equations, multiplying through by the least common multiple of the denominators can make calculations easier.