the system of linear equations below has a unique solution for all but one value of a:
6 x−12 y=−6
58 x+a y=−58
What is this exceptional value for a?
unique solution means the lines intersected at only one point (intersecting lines),, thus lines that do not intersect don not have a solution (parallel lines)
if two lines are parallel, then their slopes must be equal,, thus get their slopes by writing them into the form:
y = mx + b
where m = slope, and b = y-intercept
*for 6x − 12y = −6,
6x − 12y = −6
-12y = -6 - 6x
y = (1/2)x + (1/2)
thus m = 1/2
*for 58x+ ay = −58,
58x+ ay = −58
ay = -58 - 58x
y = (-58/a)x - 58/a
thus m = -58/a
equating and solving for a:
1/2 = -58/a
a = -116
so there,, :)
Well, it seems like we've got a case of "unique solution inconsistency." It's like when someone tells a joke that everyone finds funny, except for that one person who has a completely different sense of humor. So, in this case, the exceptional value for a would be the one value that makes the system inconsistent, where even math can't find a common punchline. Just when you think you've solved it, the laughter fades away... and that value is a = 58.
To find the exceptional value for a, we need to determine when the system of linear equations does not have a unique solution.
Let's solve the system of equations using the method of elimination:
Equation 1: 6x - 12y = -6 ----(1)
Equation 2: 58x + ay = -58 ----(2)
Multiply equation 1 by 58:
58(6x - 12y) = 58(-6)
348x - 696y = -348 ----(3)
Now, subtract equation 2 from equation 3:
(348x - 696y) - (58x + ay) = -348 - (-58)
348x - 696y - 58x - ay = -348 + 58
(348 - 58)x - (696 + a)y = -290
Simplifying the equation:
290x - (696 + a)y = -290
The system of equations will have a unique solution for all values of a except when the coefficient of y on the left side is zero, i.e., (696 + a) = 0.
Solving for a:
696 + a = 0
a = -696
Therefore, the exceptional value for a is -696.
To determine the exceptional value for "a," we need to solve the system of linear equations and check for any constraints.
First, let's solve the system of equations using the method of elimination:
1) Multiply the first equation by 58 and the second equation by 6 to eliminate "y":
348x - 696y = -348
348x + 6ay = -348
2) Subtract the first equation from the second equation:
(348x + 6ay) - (348x - 696y) = -348 - (-348)
6ay + 696y = 0
3) Simplify and factor out "y":
y(6a + 696) = 0
Now, we have two possibilities:
1) y = 0:
Substituting y = 0 in the first equation:
6x - 12(0) = -6
6x = -6
x = -1
So, one solution is x = -1 and y = 0.
2) 6a + 696 = 0:
This gives us the constraint equation:
6a = -696
a = -116
Therefore, the exceptional value for "a" is -116, which means for all other values of a, the system of linear equations will have a unique solution.