Determine whether the system of linear equations has one and only one solution, infinitely many solutions, or no solution.
4x − y = −3
12x − 3y = 9
a. one and only one solution
b. infinitely many solutions
c. no solution
Find the solution, if one exists. (If there are infinitely many solutions, express x and y in terms of the parameter t. If there is no solution, enter NO SOLUTION.)
(x, y) =
multiplying the 1st eqn by 3 gives ... 12x − 3y = -9
this is the 2nd eqn , with a different constant term
this means the lines are parallel
... they never cross, so there is no solution
no soultion
To determine whether the system of linear equations has one and only one solution, infinitely many solutions, or no solution, we can solve the system using any method (such as substitution or elimination) and analyze the resulting equations.
Let's solve this system of linear equations using the method of elimination:
1) Multiply the first equation by 3, and multiply the second equation by -1 to make the coefficients of y the same.
Eq1: 12x - 3y = -9
Eq2: -12x + 3y = -9
2) Add the two equations together:
(12x - 3y) + (-12x + 3y) = -9 + (-9)
0 = -18
The resulting equation is contradictory (0 = -18), which means there are no values of x and y that satisfy both equations simultaneously. Therefore, the system of linear equations has NO SOLUTION.
To determine whether the system of linear equations has one and only one solution, infinitely many solutions, or no solution, we can use the method of solving the system of equations.
First, let's write the system of equations in the form of Ax = b, where A is the coefficient matrix, x is the variable matrix, and b is the constant matrix:
A = [[4, -1], [12, -3]]
x = [[x], [y]]
b = [[-3], [9]]
Now, we can find the determinant of matrix A. If the determinant is non-zero, the system has one and only one solution. If the determinant is zero and the constant matrix b is not zero, the system has no solution. If the determinant is zero and the constant matrix b is also zero, the system has infinite solutions.
To calculate the determinant of A, we can use the formula:
det(A) = ad - bc
In our case,
det(A) = (4)(-3) - (-1)(12) = -12 + 12 = 0
The determinant of A is zero.
Now, let's check the constant matrix b.
If b = [[-3], [9]] is not zero, then the system of equations has no solution.
Since b is not zero, we can conclude that the system of linear equations has no solution.
Therefore, the answer is: c. no solution.
Now, let's find the solution using the parameter t.
Since the system has no solution, there is no specific value for x and y that satisfy both equations. We cannot express x and y in terms of a parameter t.
Therefore, the answer to the second part is: NO SOLUTION.