To solve the system of the two linear equations:
- \(-3x - 4y = -23\) (Equation 1)
- \(3x - 5y = 5\) (Equation 2)
We can use either the substitution method or the elimination method. In this case, we will use the elimination method.
First, we can multiply Equation 2 by 1 to already make it easier to eliminate \(x\):
- \(-3x - 4y = -23\)
- \(3x - 5y = 5\)
Next, we'll add both equations together to eliminate \(x\):
\[ (-3x - 4y) + (3x - 5y) = -23 + 5 \]
The \(3x\) and \(-3x\) will cancel each other out:
\[ -4y - 5y = -23 + 5 \]
This simplifies to:
\[ -9y = -18 \]
Now, we solve for \(y\):
\[ y = \frac{-18}{-9} = 2 \]
Now that we have \(y = 2\), we can substitute this value back into Equation 2 to find \(x\):
\[ 3x - 5(2) = 5 \]
This simplifies to:
\[ 3x - 10 = 5 \]
Now, we’ll add \(10\) to both sides:
\[ 3x = 15 \]
Dividing by \(3\) gives:
\[ x = 5 \]
So the solution to the system of equations is:
\[ (x, y) = (5, 2) \]
Therefore, the answer is:
\((5, 2)\)
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