Determine whether the system of linear equations has one and only one solution, infinitely many solutions, or no solution.
5x − 6y = 8
10x − 12y = 16
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) =
double the first equation
They are the same line
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To determine whether the system of linear equations has one and only one solution, infinitely many solutions, or no solution, we'll solve the system using the method of elimination.
1. Let's start by multiplying the first equation by 2 to make the coefficients of x the same:
2(5x - 6y) = 2(8)
10x - 12y = 16
2. Now, we can see that the second equation and the modified first equation have the same coefficients of x and y. Combining them will eliminate the variable x:
(10x - 12y) + (10x - 12y) = 16 + 16
20x - 24y = 32
3. Simplifying the equation, we get:
20x - 24y = 32
4. The coefficients of x and y in the two equations are the same. This means that the two equations represent the same line. Thus, there are infinitely many solutions to this system of linear equations.
Therefore, the answer is b. infinitely many solutions.
To find the solution, we can express x and y in terms of a parameter, let's use t:
x = t
y = (8/6) + (4/6)t
So, (x, y) = (t, (4/6)t + (8/6)) where t is any real number.
To determine whether the system of linear equations has one and only one solution, infinitely many solutions, or no solution, we can begin by putting the system of equations into matrix form.
The given system of equations can be written as:
| 5 -6 | | x | = | 8 |
| 10 -12 | * | y | = | 16 |
Now, we need to find the determinant of the coefficient matrix. If the determinant is non-zero, then the system has a unique solution. If the determinant is zero, we need to further analyze the system.
Determinant of the coefficient matrix:
| 5 -6 |
| 10 -12 |
= (5 * -12) - (-6 * 10)
= -60 + 60
= 0
Since the determinant of the coefficient matrix is zero, we need to analyze the system further.
We can observe that the second equation is simply twice the first equation. This means that the two equations are essentially the same, and there are infinite solutions.
To find the solution, we can express x and y in terms of a parameter t.
Let's assume t = y. Then, we have:
x = (8 + 6t) / 5
Therefore, the solution to the system of equations, if it exists, can be expressed as (x, y) = ((8 + 6t) / 5, t), where t can take any real value.
So, the system of linear equations has infinitely many solutions.
The solution is: (x, y) = ((8 + 6t) / 5, t)