Determine whether the system of linear equations has one and only one solution, infinitely many solutions, or no solution. x − 3y = −1 4x + 3y = 26 one and only one solution infinitely many solutions 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.)

There is only one solution, x = 5 and y = 2.

Step-by-step explanation:

To answer this question, we have to solve this system of equations.

We have that:

x - 3y = -1

4x + 3y = 26

Writing x as a function of y in the first equation, and replacing in the second, we have that:

x = 3y - 1

Replacing in the second

4x + 3y = 26

4(3y - 1) + 3y = 26

12y - 4 + 3y = 26

15y = 30

y = 2

Since we have 15y = 30, y = 2, there is only one solution.

If we had 0y = 0, there would be infinitely many solutions.

If we had 0y = a, a different of zero, there would be no solution.

Solving for x

x = 3y - 1 = 3*2 - 1 = 5

There is only one solution, x = 5 and y = 2.

it has one solution.

Step-by-step explanation:

The given system of equations is expressed as

x − 3y = −1 - - - - - - - - -1

4x + 3y = 26 - - - - - - - - 2

The first step would be to eliminate y by adding equation 1 to equation 2. It becomes

5x = 25

Dividing the left hand side and the right hand side of the equation by 5, it becomes

5x/5 = 25/5

x = 5

Substituting x = 5 into equation 1, it becomes

5 − 3y = −1

3y = 5 +1 = 6

Dividing the left hand side and the right hand side of the equation by 3, it becomes

3y/3 = 6/3

y = 2

answer: the common difference is -0.10