When solving a system algebraically how can you tell if there is no solution

You get a contradiction, i.e. a false statement. This is because you are writing down a set of equation, which already supposes that there does exist at least one solution. So, if there are no solutions that supposition is false, so you must get a contradiction.

When solving a system of equations algebraically, you can determine if there is no solution by performing the following steps:

1. Write down the given system of equations.
2. Simplify the equations, if necessary, to put them in a standard form like Ax + By = C.
3. Compare the coefficients of the variables (x and y) in the two equations. If the coefficients are proportional or equal, it means the two lines representing the equations are parallel. Parallel lines never intersect, which means there is no solution.
- For example, if one equation is 2x + 3y = 5 and the other equation is 4x + 6y = 10, the coefficients of x and y are proportional. Therefore, there is no solution.

4. Another way to determine if there is no solution is by calculating the slopes (or gradients) of the two equations. If the slopes are equal, the lines are parallel, and there is no solution.
- For example, if one equation has a slope of 2 and the other equation has a slope of 2, the lines are parallel, and there is no solution.

So, if you find that the coefficients of x and y are proportional or equal, or if the slopes are equal, it implies that there is no solution to the given system of equations.

To determine if a system of equations has no solution, you need to analyze the coefficients and the constants in the equations. Here is the step-by-step process for solving a system of equations algebraically and identifying if there is no solution:

1. Write down the given equations of the system.
For example, let's say we have the following system:
Equation 1: 2x + 3y = 8
Equation 2: 4x - 6y = 10

2. Select one of the variables (x or y) to eliminate by multiplying an equation or both equations by appropriate numbers such that the coefficients of the selected variable becomes the same (or negative) in both equations.
In our example, suppose we decide to eliminate the variable x by multiplying Equation 1 by 2 and Equation 2 by 1:
Equation 1 (multiplied by 2): 4x + 6y = 16
Equation 2 (multiplied by 1): 4x - 6y = 10

3. Add or subtract the two new equations to eliminate the selected variable. This will give you a new equation with only one variable.
In our example, let's subtract Equation 2 from Equation 1:
(Equation 1) - (Equation 2): 4x + 6y - (4x - 6y) = 16 - 10
Simplifying, we get: 12y = 6 or y = 1/2

4. Substitute the value of the solved variable back into one of the original equations to find the value of the remaining variable.
Using Equation 1, we substitute y = 1/2:
2x + 3(1/2) = 8
Simplifying, we get: 2x + 3/2 = 8
Multiply through by 2 to eliminate the fraction: 4x + 3 = 16
Move the constant to the other side: 4x = 16 - 3 or 4x = 13
Divide both sides by 4: x = 13/4

5. Check if the found solution satisfies the other equation of the system. If it does, then the system has a unique solution. However, if it doesn't, then there is no solution.
Using Equation 2, we substitute x = 13/4 and y = 1/2:
4(13/4) - 6(1/2) = 10
Simplifying, we get: 13 - 3 = 10
This equation is not true, so the system has no solution.

In conclusion, if at any point during the process, you find that the remaining equation is not true, or it simplifies to a contradictory statement, then the system has no solution.