Let's solve the three systems of equations one by one.
1) System of Equations:
- \( x - 3y = -7 \) (Equation 1)
- \( 2x = 6y - 14 \) (Equation 2)
First, let's rearrange Equation 2:
\[ 2x + 14 = 6y \implies y = \frac{2x + 14}{6} = \frac{x + 7}{3} \]
Now, we can substitute \( y \) from this expression into Equation 1:
\[ x - 3\left(\frac{x + 7}{3}\right) = -7 \]
This simplifies to:
\[ x - (x + 7) = -7 \] \[ x - x - 7 = -7 \implies -7 = -7 \]
This is a true statement, meaning the equations represent the same line.
Conclusion: The system has infinite solutions.
2) System of Equations:
- \( 9x + 8y = 15 \) (Equation 1)
- \( 9x + 8y = 30 \) (Equation 2)
Here, we can see that the left-hand sides of both equations are the same, but the right-hand sides are different (\(15\) and \(30\)).
Conclusion: The system has no solutions (the lines are parallel).
3) System of Equations:
- \( 5x - 9y = -43 \) (Equation 1)
- \( 3x + 8y = 68 \) (Equation 2)
We will use the elimination method for this system. We can eliminate \(y\) by manipulating the equations. Let's multiply Equation 1 by \(8\) and Equation 2 by \(9\) to align the coefficients of \(y\):
\[ 8(5x - 9y) = 8(-43) \implies 40x - 72y = -344 \quad (Equation 3) \] \[ 9(3x + 8y) = 9(68) \implies 27x + 72y = 612 \quad (Equation 4) \]
Now, we add Equations 3 and 4 to eliminate \(y\):
\[ (40x - 72y) + (27x + 72y) = -344 + 612 \] This simplifies to:
\[ 67x = 268 \implies x = \frac{268}{67} = 4 \]
Now that we have \(x\), we can substitute it back to find \(y\). We can use either original equation; let’s use Equation 1:
\[ 5(4) - 9y = -43 \] \[ 20 - 9y = -43 \implies -9y = -43 - 20 \] \[ -9y = -63 \implies y = \frac{-63}{-9} = 7 \]
Conclusion: The system has one solution \( (x, y) = (4, 7) \).
Summary of Results:
- System 1: Infinite solutions.
- System 2: No solutions.
- System 3: One solution \((4, 7)\).
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