How many solutions does the following system have:
2x + 5y = 10
3x + 6y = 12
Explain how to determine the number of solutions without solving the system. Then apply elimination, and interpret the resulting equation.
The key is to calculate Δ, required for the first part of Kramer's rule.
Δ=determinant of the left-hand side.
If Δ≠0, then there is a unique solution, including the trivial solution of x=0 and y=0 if the right-hand side is all zeroes.
If Δ=0, then the left-hand side of the equations are linearly dependent.
Two cases may arise:
1. If the equations are consistent, one single equation will result after reduction of the left-hand side.
This case has infinitely many solutions, one set for each value we assign to x (or y).
2. If the equations are not consistent, then after reduction, the equations will be identical on the left-hand side, but the right-hand sides will different.
This gives a total of 3 different cases.
To determine the number of solutions without solving the system, you can compare the slopes of the two equations. If the slopes are equal and the y-intercepts are different, the system has no solution. If the slopes are equal and the y-intercepts are also equal, the system has infinitely many solutions. If the slopes are different, the system has exactly one solution.
Now, let's apply elimination to solve the system:
Step 1: Multiply the first equation by 3 and the second equation by 2 to make the coefficients of x in both equations the same:
3(2x + 5y) = 3(10)
2(3x + 6y) = 2(12)
Simplifying these equations gives us:
6x + 15y = 30
6x + 12y = 24
Step 2: Subtract the second equation from the first equation:
(6x + 15y) - (6x + 12y) = 30 - 24
Simplifying this equation gives us:
3y = 6
Step 3: Divide both sides of the equation by 3 to solve for y:
3y/3 = 6/3
y = 2
Step 4: Substitute the value of y back into either of the original equations to solve for x. Let's use the first equation:
2x + 5(2) = 10
Simplifying this equation gives us:
2x + 10 = 10
Step 5: Subtract 10 from both sides of the equation:
2x + 10 - 10 = 10 - 10
2x = 0
Step 6: Divide both sides of the equation by 2 to solve for x:
2x/2 = 0/2
x = 0
So, the solution to the system of equations is x = 0 and y = 2.
Interpreting the resulting equation, 3y = 6, it represents the same line as the original system of equations. This confirms that there is exactly one solution to the system.
To determine the number of solutions of a system of linear equations without solving it, we need to analyze the coefficients of the variables. In this case, we have the following system:
2x + 5y = 10 (Equation 1)
3x + 6y = 12 (Equation 2)
We can see that the coefficients of both x and y are proportional in these two equations. In other words, if we divide Equation 2 by 3 and Equation 1 by 2, we get:
(2/3)x + (5/3)y = 10/3 (Equation 3)
(2/3)x + 2y = 4 (Equation 4)
Now, let's compare Equation 3 and Equation 4. We can see that the coefficient of y in Equation 3 is (5/3), while in Equation 4, it is 2. These coefficients are not equal.
When the coefficients of the variables in a system of linear equations are proportional, and not equal, it means that the lines representing these equations are parallel and will never intersect. Thus, there are no common solutions, and the system is inconsistent. In other words, there are no solutions to this system.
Now, let's apply the elimination method to solving the system to further confirm the results:
We can multiply Equation 1 by 3 and Equation 2 by 2, which will give us:
6x + 15y = 30 (Equation 5)
6x + 12y = 24 (Equation 6)
Now, if we subtract Equation 6 from Equation 5, we get:
(6x - 6x) + (15y - 12y) = 30 - 24
3y = 6
y = 2
Substituting the value of y in Equation 1, we get:
2x + 5(2) = 10
2x + 10 = 10
2x = 0
x = 0
So we found a solution, where x = 0 and y = 2. This means that the system is consistent. However, since we previously determined that the system has no solutions, there must be an error made during the calculations.
In conclusion, based on the analysis of the coefficients and the elimination method, the system initially appeared to have no solutions. However, when we attempted to solve using elimination, an error was made since our solution contradicted our initial conclusion.