To solve the system of equations:
- \( 6x - 5y = 27 \) (Equation 1)
- \( 3x + 5y = 36 \) (Equation 2)
We can use the elimination or substitution method. Here, I'll use the elimination method.
First, let's eliminate \( y \) by adding both equations. To do that, we will first align the coefficients of \( y \).
From Equation 1, we can multiply it by \( 1 \) (no change): \[ 6x - 5y = 27 \]
From Equation 2, if we multiply it by \( 1 \) (no change as well): \[ 3x + 5y = 36 \]
Now we can add both equations: \[ (6x - 5y) + (3x + 5y) = 27 + 36 \] This simplifies to: \[ 9x = 63 \] Dividing by \( 9 \): \[ x = 7 \]
Now that we have \( x \), we can substitute it back into one of the original equations to find \( y \). We'll use Equation 2: \[ 3(7) + 5y = 36 \] This simplifies to: \[ 21 + 5y = 36 \] Now, isolate \( 5y \): \[ 5y = 36 - 21 \] \[ 5y = 15 \] Dividing by \( 5 \): \[ y = 3 \]
So the solution to the system of equations is: \[ x = 7, \quad y = 3 \]
In ordered pair form, the solution is: \[ (7, 3) \]
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