How many solutions does the following system have? {2x+3y=1−3x−2y=−1
The system of equations given is:
2x + 3y = 1
-3x - 2y = -1
To determine the number of solutions, we can use various methods such as substitution, elimination, or graphing. Let's use the elimination method:
Multiply the first equation by 3 and the second equation by 2 to make the coefficients of x in both equations opposite:
(3)(2x + 3y) = (3)(1)
(2)(-3x - 2y) = (2)(-1)
Simplifying, we get:
6x + 9y = 3
-6x - 4y = -2
Adding the two equations together, the x terms cancel out:
6x - 6x + 9y - 4y = 3 - 2
5y = 1
Divide both sides by 5:
y = 1/5
Substitute this value back into either of the original equations, let's use the first equation:
2x + 3(1/5) = 1
2x + 3/5 = 1
2x = 1 - 3/5
2x = 2/5
Divide both sides by 2:
x = 1/5
Therefore, the system has a unique solution:
x = 1/5 and y = 1/5.
To determine the number of solutions for the given system of equations, we can use a method called elimination or substitution.
Given the system:
2x + 3y = 1
-3x - 2y = -1
Let's solve it using the elimination method:
Step 1: Multiply the first equation by 3 and the second equation by 2 to make the coefficients of x equal in both equations.
3(2x + 3y) = 3(1)
2(-3x - 2y) = 2(-1)
This simplifies to:
6x + 9y = 3
-6x - 4y = -2
Step 2: Add the two equations together.
(6x + 9y) + (-6x - 4y) = 3 + (-2)
This simplifies to:
(6x - 6x) + (9y - 4y) = 1
The x-terms cancel out, leaving:
5y = 1
Step 3: Solve for y.
y = 1/5
Now, let's substitute this value of y back into one of the original equations to solve for x. Let's use the first equation:
2x + 3(1/5) = 1
Simplifying, we get:
2x + 3/5 = 1
Subtracting 3/5 from both sides:
2x = 1 - 3/5
2x = 5/5 - 3/5
2x = 2/5
Divide both sides by 2:
x = (2/5) / 2
x = 2/10
x = 1/5
Therefore, the solution to the system of equations is x = 1/5 and y = 1/5.
So, the system has a unique solution, and there is only one solution.
To find out the number of solutions for a system of equations, we can use the concept of linear independence.
First, let's rewrite the system of equations:
2x + 3y = 1
-3x - 2y = -1
We can rearrange the second equation to its standard form:
3x + 2y = 1
Now, let's put these equations into matrix form:
| 2 3 | | x | | 1 |
|-3 -2 | * | y | = |-1 |
To determine the number of solutions, we need to find the rank of the augmented matrix [A|B], where A is the coefficient matrix and B is the constant matrix.
First, let's find the row-echelon form (REF) of the system. Using row operations, we can transform the matrix:
| 2 3 | | x | | 1 |
|-3 -2 | * | y | = |-1 |
R2 -> R2 + (3/2) * R1
| 2 3 | | x | | 1 |
| 0 1 | | y | |-2 |
Now, let's find the REF of the augmented matrix:
| 1 0 | | x | |-7 |
| 0 1 | * | y | = |-2 |
The matrix is now in row-echelon form (REF), and we can count the number of linearly independent rows to determine the number of solutions.
Since both rows are linearly independent, we have two pivots (one in each row). Therefore, this system has a unique solution.
In conclusion, the given system of equations has exactly one solution.