Determine if the following system of equations has no solutions, infinitely many solutions or exactly one solution.2, x, plus, 3, y, equals, minus, 12x+3y=−14, x, plus, 6, y, equals, 14x+6y=1
Question
Determine if the following system of equations has no solutions, infinitely many solutions or exactly one solution.2, x, plus, 3, y, equals, minus, 12x+3y=−14, x, plus, 6, y, equals, 14x+6y=1
Solution
The system of equations given is:
- 2x + 3y = -14
- x + 6y = 1
To determine if the system has no solutions, infinitely many solutions or exactly one solution, we can use the method of substitution or elimination. Here, I'll use the method of elimination.
First, we can multiply the second equation by 2 to make the coefficients of x in both equations the same:
2*(x + 6y) = 2*1 2x + 12y = 2
Now, we subtract the new second equation from the first one:
(2x + 3y) - (2x + 12y) = -14 - 2 2x + 3y - 2x - 12y = -16 -9y = -16
Solving for y, we get:
y = -16 / -9 y = 16/9
Substitute y = 16/9 into the first equation:
2x + 3*(16/9) = -14 2x + 48/9 = -14 2x = -14 - 48/9 2x = -126/9 - 48/9 2x = -174/9 x = -174/18 x = -87/9
So, the solution to the system of equations is x = -87/9 and y = 16/9. Therefore, the system has exactly one solution.
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