Find the value(s) of k for which the pair of equations:
Equation 1: kx + (k+1)y = 8
Equation 2: 4x + 3ky = 4
(a) no solutions
(b) a unique solution
(c) infinitely many solutions
Can you please use the gradient way and show your working? Thanks a lot
I've calculated the gradient and k = 2 and -2/3
When k = 2,
2x+3y=8
4x+6y=4
When k = -2/3
-2x+y=24
4x-2y=4
The equations are all similar except for the constants and I do not know which is no solutions or infinitely many solutions.
Where do I go from here? Also how do i find the unique solutions? Thx a bunch
Suppose the slopes are equal:
-k/(k+1) = -4/(3k)
-3k^2 = -4k-4
3k^2-4k-4 = 0
(3k+2)(k-2) = 0
k = 2 or -2/3
as you calculated. Using you pairs of equations, we see that they can be written
4x+6y=16
4x+6y=4
2x-y = -24
2x-y = 2
In both cases the lines are distinct and parallel, so there is no solution.
Any other value of k will give the lines different slopes, so they must intersect at one point.
So, there is no value of k which will make the two lines the same, so (c) cannot be obtained.
To determine the value(s) of k for which the pair of equations has no solutions, a unique solution, or infinitely many solutions, we will use the gradient method.
First, let's find the gradients of the two equations.
The gradient of Equation 1 is given by the ratio of the coefficients of x and y, which is k/(k+1).
The gradient of Equation 2 is given by the ratio of the coefficients of x and y, which is 4/(3k).
Now, let's consider the three cases:
a) No solutions:
If the gradients of the two equations are equal, but the constant terms are not equal, there will be no solution.
Setting the gradients equal, we have:
k/(k+1) = 4/(3k)
Cross-multiplying, we get:
3k(k+1) = 4(k)
Expanding and rearranging, we have:
3k^2 + 3k = 4k
Simplifying, we get:
3k^2 + 3k - 4k = 0
Combining like terms, we have:
3k^2 - k = 0
Factoring out k, we have:
k(3k - 1) = 0
So, we have two possible solutions:
k = 0 or k = 1/3
b) A unique solution:
If the gradients of the two equations are different, there will be a unique solution.
Setting the gradients unequal, we have:
k/(k+1) ≠ 4/(3k)
Cross-multiplying, we get:
3k(k+1) ≠ 4(k)
Expanding and rearranging, we have:
3k^2 + 3k ≠ 4k
Simplifying and combining like terms, we have:
3k^2 - k ≠ 0
Since this is a quadratic equation, it will always have a solution. Therefore, there is no value of k for which there is a unique solution.
c) Infinitely many solutions:
If the gradients of the two equations are equal and the constant terms are also equal, there will be infinitely many solutions.
Setting the gradients equal and the constant terms equal, we have:
k/(k+1) = 4/(3k) and 8 = 4
Since the equation 8 = 4 is not true, it implies that there are no values of k for which the constant terms are equal. Therefore, there are no values of k for which there are infinitely many solutions.
In conclusion, the value(s) of k for which the pair of equations has no solutions is k = 0 or k = 1/3, and there are no values of k for which there is a unique solution or infinitely many solutions.
To find the value(s) of k for which the pair of equations has no solutions, a unique solution, or infinitely many solutions, we can use the concept of gradients.
First, let's rewrite the given equations in the standard form of a linear equation (y = mx + c) to make it easier to determine the gradients:
Equation 1: kx + (k+1)y = 8
Rearranging, we get:
(k+1)y = -kx + 8
y = (-k/k+1)x + 8/(k+1)
Equation 2: 4x + 3ky = 4
Rearranging, we get:
3ky = -4x + 4
y = (-4/3k)x + 4/3k
Now, we can compare the gradients (coefficients of x) of the two equations.
For a unique solution:
The gradients of the two equations must be different. So, we need to find the values of k for which:
(-k/k+1) ≠ (-4/3k)
To solve this inequality, we can cross-multiply and simplify:
-3k^2 ≠ -4(k+1)
-3k^2 ≠ -4k - 4
3k^2 - 4k - 4 ≠ 0
Now, we can solve this quadratic equation using the quadratic formula:
k = (-b ± √(b^2 - 4ac)) / (2a)
Plugging in the values, we get:
k = (-(-4) ± √((-4)^2 - 4(3)(-4))) / (2(3))
k = (4 ± √(16 + 48)) / 6
k = (4 ± √64) / 6
k = (4 ± 8) / 6
This gives two potential solutions:
k1 = (-4 + 8) / 6 = 2/3
k2 = (-4 - 8) / 6 = -2
Therefore, if k is equal to -2 or 2/3, the pair of equations will have a unique solution.
For no solutions:
The gradients of the two equations must be the same, and the y-intercepts must be different. So, we need to find the values of k for which:
(-k/k+1) = (-4/3k) and 8/(k+1) ≠ 4/3k
Simplifying the equation:
3k^2 - 4k - 4 = 0
Using the quadratic formula, we get:
k = (4 ± √(16 + 48)) / 6
k = (4 ± √64) / 6
k = (4 ± 8) / 6
Again, we have the same potential solutions:
k1 = (-4 + 8) / 6 = 2/3
k2 = (-4 - 8) / 6 = -2
Therefore, if k is equal to -2 or 2/3, the pair of equations will have no solutions.
For infinitely many solutions:
The gradients of the two equations must be the same, and the y-intercepts must also be the same. So, we need to find the values of k for which:
(-k/k+1) = (-4/3k) and 8/(k+1) = 4/3k
Simplifying the equation:
3k^2 - 4k - 4 = 0
Using the quadratic formula, we again get the same potential solutions:
k1 = (-4 + 8) / 6 = 2/3
k2 = (-4 - 8) / 6 = -2
Therefore, if k is equal to -2 or 2/3, the pair of equations will have infinitely many solutions.
To summarize:
(a) The value(s) of k for which the pair of equations has no solutions are -2 and 2/3.
(b) The value(s) of k for which the pair of equations has a unique solution are also -2 and 2/3.
(c) The value(s) of k for which the pair of equations has infinitely many solutions are -2 and 2/3.