(K+1)x+2y=6; M=k-2
oops
-(k+1) = 2k - 4
-k-1 = 2k-4
3k=3
k=1
To find the values of x and y that satisfy the given equations, let's substitute the value of M into the first equation.
Given:
(K+1)x + 2y = 6
M = K - 2
We will substitute M into the first equation:
(K+1)x + 2y = 6
(K - 2)x + 2y = 6
Now, we have a system of two equations with two variables. To solve this system, we can either use the substitution method or the elimination method.
Let's use the elimination method to solve the system.
Multiply the second equation by (K + 1):
(K + 1)(K - 2)x + (K + 1)(2y) = 6(K + 1)
Expand the equation:
(K^2 - K - 2)x + (2K + 2)y = 6K + 6
Now, we have two equations:
(K + 1)x + 2y = 6
(K^2 - K - 2)x + (2K + 2)y = 6K + 6
We can eliminate the variable y by multiplying the first equation by (K^2 - K - 2) and the second equation by -2:
(K^2 - K - 2)(K + 1)x + (K^2 - K - 2)(2y) = 6(K^2 - K - 2)
-2(K + 1)x - 2(2K + 2)y = -2(6K + 6)
Expand the equations:
(K^3 - K^2 - 2K - K^2 + K + 2)x + (2K^2 - 2K - 4y) = 6K^2 - 6K - 12
-2Kx - 2y - 4Kx - 4y = -12K - 12
Combine like terms:
(K^3 - 2K^2 - 4K)x - 6y = 6K^2 - 8K - 12
We now have the following equations:
(K^3 - 2K^2 - 4K)x - 6y = 6K^2 - 8K - 12
(K^2 - K - 2)x + 2y = 6
Now, solve this system to find the values of x and y.
Steve, You made another mistake in that second edit.
Once you reach -k-1=2k-4.
You need to subtract 2k from the right side.
That gets you -3k-1=-4.
You add 1 to get rid of that negative one which gets -3k=-3. Solve for k=1
Apparently you mean that M is the slope. Most folks use lower-case letters.
In any case, we have
2y = -(k+1)x + 6
y = -(k+1)/2 x + 3
The slope is -(k+1)/2
-(k+1)/2 = k-2
-(k+1) = 2k - 2
-k-1 = 2k-2
3k = 1
k = 1/3