Answer:
c. 4 4÷2 =8 hope it helps
-4m + 8m = -16
4m = -16
m = -16 / 4
m = -4
Answer:
![\left[\begin{array}{ccc}x\\y\\\end{array}\right] = \left[\begin{array}{ccc}3\\2\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx%5C%5Cy%5C%5C%5Cend%7Barray%7D%5Cright%5D%20%3D%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D3%5C%5C2%5C%5C%5Cend%7Barray%7D%5Cright%5D)
Solution of the given system of equations
x = 3 and y= 2
Step-by-step explanation:
<u><em>Step(i):</em></u>-
Given system of equations are
4x - y = 10 ...(i)
8x +5y = 34...(ii)
The matrix form
A X = B
![\left[\begin{array}{ccc}4&-1\\8&5\\\end{array}\right] \left[\begin{array}{ccc}x\\y\\\end{array}\right] = \left[\begin{array}{ccc}10\\34\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D4%26-1%5C%5C8%265%5C%5C%5Cend%7Barray%7D%5Cright%5D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx%5C%5Cy%5C%5C%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D10%5C%5C34%5C%5C%5Cend%7Barray%7D%5Cright%5D)
<u><em>Step(ii):-</em></u>
<u><em>By using matrix inversion method</em></u>
<u><em></em></u>
<u><em></em></u>
<u><em></em></u>
|A| = ad-b c = 4(5) - 8(-1) = 20+8 = 28
Given matrix
![A = \left[\begin{array}{ccc}4&-1\\8&5\\\end{array}\right]](https://tex.z-dn.net/?f=A%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D4%26-1%5C%5C8%265%5C%5C%5Cend%7Barray%7D%5Cright%5D)
![AdjA = \left[\begin{array}{ccc}d&-b\\-c&a\\\end{array}\right]](https://tex.z-dn.net/?f=AdjA%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dd%26-b%5C%5C-c%26a%5C%5C%5Cend%7Barray%7D%5Cright%5D)
![A djA = \left[\begin{array}{ccc}5&1\\-8&4\\\end{array}\right]](https://tex.z-dn.net/?f=A%20djA%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D5%261%5C%5C-8%264%5C%5C%5Cend%7Barray%7D%5Cright%5D)
Inverse of the matrix
<u><em></em></u>
![A^{-1} = \frac{1}{|A|} A djA =\frac{1}{28} \left[\begin{array}{ccc}5&1\\-8&4\\\end{array}\right]](https://tex.z-dn.net/?f=A%5E%7B-1%7D%20%3D%20%5Cfrac%7B1%7D%7B%7CA%7C%7D%20A%20djA%20%3D%5Cfrac%7B1%7D%7B28%7D%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D5%261%5C%5C-8%264%5C%5C%5Cend%7Barray%7D%5Cright%5D)
<u><em>Step(iii)</em></u>:-
The solution of the given system of equations by using matrix inversion method
X = A⁻¹ B
![X = A^{-1}B =\frac{1}{28} \left[\begin{array}{ccc}5&1\\-8&4\\\end{array}\right]\left[\begin{array}{ccc}10\\34\\\end{array}\right]](https://tex.z-dn.net/?f=X%20%3D%20A%5E%7B-1%7DB%20%20%3D%5Cfrac%7B1%7D%7B28%7D%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D5%261%5C%5C-8%264%5C%5C%5Cend%7Barray%7D%5Cright%5D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D10%5C%5C34%5C%5C%5Cend%7Barray%7D%5Cright%5D)
![X = A^{-1}B =\frac{1}{28} \left[\begin{array}{ccc}5X10+1X34\\-8X10+4X34\\\end{array}\right]](https://tex.z-dn.net/?f=X%20%3D%20A%5E%7B-1%7DB%20%20%3D%5Cfrac%7B1%7D%7B28%7D%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D5X10%2B1X34%5C%5C-8X10%2B4X34%5C%5C%5Cend%7Barray%7D%5Cright%5D)
![\left[\begin{array}{ccc}x\\y\\\end{array}\right] =\frac{1}{28} \left[\begin{array}{ccc}84\\56\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx%5C%5Cy%5C%5C%5Cend%7Barray%7D%5Cright%5D%20%3D%5Cfrac%7B1%7D%7B28%7D%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D84%5C%5C56%5C%5C%5Cend%7Barray%7D%5Cright%5D)
![\left[\begin{array}{ccc}x\\y\\\end{array}\right] = \left[\begin{array}{ccc}3\\2\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx%5C%5Cy%5C%5C%5Cend%7Barray%7D%5Cright%5D%20%3D%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D3%5C%5C2%5C%5C%5Cend%7Barray%7D%5Cright%5D)
<u><em>Final answer</em></u>:-
Solution of the given system x = 3 and y= 2
The points of intersection are (-2, -2) and (5, 5), so clearly y = x. This reduces the choices to the first and third ones.
Since x-4 is the denominator of all choices, its value at the point (5, 5) will be 1. Hence you need to find the choice that has a numerator of 5 when x=5.
The first choice has a numerator of -5-10 = -15, so that's not it.
The third choice has a numerator of -5+10 = 5, so that's the one you're looking for.
The best choice is the third one, ...
... y = negative x plus ten divided by the quantity x minus four , y = x
It’s excellent and good and ever had and