To solve the system of equations given:
- ( 8x - 20y = 12 ) (Equation 1)
- ( 2x - 14y = 8 ) (Equation 2)
Let's solve this system using the substitution or elimination method. Here, we will use the elimination method.
First, we'll simplify both equations if possible.
For Equation 2, we can divide everything by 2:
[
x - 7y = 4 \quad \text{(Equation 2 simplified)}
]
Now, we have:
- ( 8x - 20y = 12 )
- ( x - 7y = 4 )
Next, we can express (x) in terms of (y) from Equation 2:
[
x = 7y + 4
]
Now we substitute this expression for (x) in Equation 1:
[
8(7y + 4) - 20y = 12
]
Distributing 8:
[
56y + 32 - 20y = 12
]
Combining like terms:
[
36y + 32 = 12
]
Now, isolate (y):
[
36y = 12 - 32
]
[
36y = -20
]
[
y = -\frac{20}{36} = -\frac{5}{9}
]
Now that we have (y), we can substitute it back into the equation for (x):
[
x = 7\left(-\frac{5}{9}\right) + 4
]
[
x = -\frac{35}{9} + \frac{36}{9}
]
[
x = \frac{1}{9}
]
Thus, the solution to the system of equations is:
[
x = \frac{1}{9}, \quad y = -\frac{5}{9}
]
To summarize, the solution is:
[
(x, y) = \left(\frac{1}{9}, -\frac{5}{9}\right)
]
