Question

3*+4y=5/2*-5y=12

Answer 1

It looks like there are two equations in your expression. To solve them, let's assume you intended to write them in a clearer format. Here are the equations:

1. \( 3x + 4y = 12 \)
2. \( \frac{5}{2} - 5y = 12 \)

Let's solve them step by step.

### Equation 1: \( 3x + 4y = 12 \)

To express \(x\) in terms of \(y\), we can rearrange the first equation:

\[ 3x = 12 - 4y \]
\[ x = \frac{12 - 4y}{3} \]

### Equation 2: \( \frac{5}{2} - 5y = 12 \)

To solve for \(y\), rearrange the equation:

First, isolate \(5y\):

\[ -5y = 12 - \frac{5}{2} \]

To combine the right side, convert \(12\) into halves:

\[ 12 = \frac{24}{2} \]
\[ -5y = \frac{24}{2} - \frac{5}{2} \]
\[ -5y = \frac{24 - 5}{2} \]
\[ -5y = \frac{19}{2} \]

Now divide both sides by \(-5\):

\[ y = -\frac{19}{10} \]

### Now Substitute \(y\) into Equation 1

Now that we have \(y\), substitute it back into the first equation to find \(x\):

\[ x = \frac{12 - 4(-\frac{19}{10})}{3} \]

Calculating:

\[ x = \frac{12 + \frac{76}{10}}{3} \]
Convert \(12\) to tenths:

\[ x = \frac{\frac{120}{10} + \frac{76}{10}}{3} = \frac{\frac{196}{10}}{3} \]
\[ x = \frac{196}{30} = \frac{98}{15} \]

### Final Answer

The solution to the equations is:

\[ x = \frac{98}{15}, \quad y = -\frac{19}{10} \]

You can express this as:

\((x, y) = \left( \frac{98}{15}, -\frac{19}{10} \right)\)