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:
- \( 3x + 4y = 12 \)
- \( \frac{5}{2} - 5y = 12 \)
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:
\((x, y) = \left( \frac{98}{15}, -\frac{19}{10} \right)\)