Find conditions on k that will make the following system of equations have a unique solution. To enter your answer, first select whether k should be equal or not equal to specific values, then enter a value or a list of values separated by commas.
Then give a formula in terms of k for the solution to the system, when it exists. Be sure to include parentheses where necessary, e.g. to distinguish 1/(2k) from 1/2k.
kx+y = 5
2x+2ky = 3
The system has a unique solution when k = ???
The unique solution is xy = 0
0
Well, well, well, let's clown around with this system of equations!
For the system to have a unique solution, the determinant of the coefficients matrix should not be zero. In other words, k should not take on specific values. So, we can say that k should NOT equal 1 or -1.
Now, let's find the solution when it exists.
From the first equation, we can isolate x, which gives us:
x = (5 - y)/k
Now, we substitute this into the second equation:
2(5 - y)/k + 2ky = 3
Simplifying this, we get:
(10 - 2y)/k + 2ky = 3
Multiplying through by k to get rid of the denominator, we have:
10 - 2y + 2k^2*y = 3k
Rearranging this equation gives:
(2k^2 - 2)*y = 3k - 10
Solving for y, we find:
y = (3k - 10)/(2k^2 - 2)
Now, let's solve for x by substituting this value of y into the first equation:
x = (5 - (3k - 10)/(2k^2 - 2))/k
Simplifying this, we get:
x = (5k - 3k + 10)/(k(2k^2 - 2))
Finally, we can say that the solution to the system, when it exists, is:
x = (5k - 3k + 10)/(k(2k^2 - 2))
y = (3k - 10)/(2k^2 - 2)
And there you go! That's the unique solution when it exists. Now, go forth and solve those equations with a smile! 🤡
To find the conditions on k that will make the system of equations have a unique solution, we can solve the system using elimination or substitution.
Using elimination:
Multiplying the first equation by 2 and the second equation by 1, we get:
2kx + 2y = 10
2x + 2ky = 3
Now, subtracting the second equation from the first equation, we get:
(2kx + 2y) - (2x + 2ky) = 10 - 3
2kx + 2y - 2x - 2ky = 7
(2k - 2)x + (2 - 2k)y = 7
(2k - 2)x - (2k - 2)y = 7
(2k - 2)(x - y) = 7
To have a unique solution, the coefficients of x and y in the last equation must be non-zero. This means that (2k - 2) must not equal zero.
Therefore, the condition is:
2k - 2 ≠0
Simplifying the inequality:
2k - 2 ≠0
2k ≠2
k ≠1
So, the system has a unique solution when k is not equal to 1.
To find the solution when it exists, we can substitute the value of k in either of the equations and solve for the variables x and y.
Let's solve for x using the first equation:
kx + y = 5
If we assume k ≠0, we can isolate x:
kx = 5 - y
x = (5 - y)/k
Now, let's solve for y using the second equation:
2x + 2ky = 3
Substituting the expression for x:
2(5 - y)/k + 2ky = 3
(10 - 2y)/k + 2ky = 3
10 - 2y + 2k²y = 3k
10 + y(2k² - 2) = 3k
y(2k² - 2) = 3k - 10
y = (3k - 10)/(2k² - 2)
Therefore, the solution to the system, when it exists, is:
x = (5 - y)/k
y = (3k - 10)/(2k² - 2)
To determine the conditions on k that will make the system of equations have a unique solution, we can solve the system of equations using the method of elimination.
First, let's multiply the first equation by 2 to eliminate the k variable:
2(kx + y) = 2(5)
2kx + 2y = 10
Now let's compare the resulting equation with the second equation:
2kx + 2y = 10
2x + 2ky = 3
From this comparison, we can see that the two equations have the same coefficients for both x and y. In order for the system to have a unique solution, these equations must represent the same line. This means that the only condition on k for a unique solution is that the ratios of the coefficients for x and y are the same:
2k/(2) = 1/(k)
Simplifying this equation, we get:
k/1 = 1/k
Cross-multiplying, we have:
k^2 = 1
Taking the square root of both sides, we get:
k = ±1
So the conditions on k for the system of equations to have a unique solution are k = 1 or k = -1.
To find the formula for the solution to the system, we can substitute k = 1 (or k = -1) into one of the original equations and solve for the other variable. Let's use k = 1 for this example.
Substituting k = 1 into the first equation, we get:
x + y = 5
Rearranging this equation, we have:
y = 5 - x
Now substituting this expression for y into the second equation, we get:
2x + 2(1)(5 - x) = 3
Simplifying this equation, we have:
2x + 10 - 2x = 3
10 = 3
Since this equation is not true, it means that there is no solution for the system when k = 1. Therefore, the system has no unique solution when k = 1.
Similarly, when k = -1, substituting into the first equation gives:
-x + y = 5
Rearranging this equation, we have:
y = x + 5
Substituting this expression for y into the second equation, we get:
2x + 2(-1)(x + 5) = 3
Simplifying this equation, we have:
2x - 2x - 10 = 3
-10 = 3
Since this equation is also not true, it means that there is no solution for the system when k = -1. Therefore, the system also has no unique solution when k = -1.
In summary, the conditions on k for the system of equations to have a unique solution are k = 1 or k = -1. However, when substituted into the original equations, both values of k result in inconsistent equations, indicating that the system has no unique solution for any value of k.
The slopes must be different to have a unique solution. So, we need
-k ≠-1/k
k^2 ≠1
To find that solution, multiply the 1st equation by 2k and you have
2k^2 x + 2ky = 10k
2x + 2ky = 3
Now subtract to eliminate y, and you have
(2k^2-2)x = 10k-3
x = (10k-3)/(2k^2-2)
Substituting that in, we have
k(10k-3)/(2k^2-2) + y = 5
y = (3k-10)/(2k^2-2)