Assuming the equation is:
5/(x-5) = x/(x-5) - 5x/4
We first multiply by the LCD: 4(x-5)
20 = 4x - 5x(x-5)
20 = 4x - 5x^2 + 25x
5x^2 - 29x + 20 = 0
(5x - 4)(x - 5) = 0
x = 4/5, 5
Substituting x = 5 gives denominators of 0, which is extraneous.
Substituting x = 4/5 gives a valid equation, so this is the only correct solution.
2nd one maybe I’m not to sure srry
Answer:
C. ![\left[\begin{array}{ccc}-24&0&-3\\8&-48&56\\25&-6&10\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-24%260%26-3%5C%5C8%26-48%2656%5C%5C25%26-6%2610%5Cend%7Barray%7D%5Cright%5D)
Step-by-step explanation:
The given matrices are;
and
![C=\left[\begin{array}{ccc}12&0&\frac{3}{2}\\1&-6&7\end{array}\right]](https://tex.z-dn.net/?f=C%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D12%260%26%5Cfrac%7B3%7D%7B2%7D%5C%5C1%26-6%267%5Cend%7Barray%7D%5Cright%5D)
![FC=\left[\begin{array}{cc}-2&0\\0&8\\2&1\end{array}\right]\left[\begin{array}{ccc}12&0&\frac{3}{2}\\1&-6&7\end{array}\right]](https://tex.z-dn.net/?f=FC%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D-2%260%5C%5C0%268%5C%5C2%261%5Cend%7Barray%7D%5Cright%5D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D12%260%26%5Cfrac%7B3%7D%7B2%7D%5C%5C1%26-6%267%5Cend%7Barray%7D%5Cright%5D)
![FC=\left[\begin{array}{ccc}12(-2)+0(1)&0(-2)+-6(0)&\frac{3}{2}(-2)+7(0)\\12(0)+8(1)&0(0)+8(-6)&0(\frac{3}{2})+8(7)\\2(12)+1(1)&2(0)+1(-6)&2(\frac{3}{2})+1(7)\end{array}\right]](https://tex.z-dn.net/?f=FC%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D12%28-2%29%2B0%281%29%260%28-2%29%2B-6%280%29%26%5Cfrac%7B3%7D%7B2%7D%28-2%29%2B7%280%29%5C%5C12%280%29%2B8%281%29%260%280%29%2B8%28-6%29%260%28%5Cfrac%7B3%7D%7B2%7D%29%2B8%287%29%5C%5C2%2812%29%2B1%281%29%262%280%29%2B1%28-6%29%262%28%5Cfrac%7B3%7D%7B2%7D%29%2B1%287%29%5Cend%7Barray%7D%5Cright%5D)
This simplifies to;
![FC=\left[\begin{array}{ccc}-24&0&-3\\8&-48&56\\25&-6&10\end{array}\right]](https://tex.z-dn.net/?f=FC%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-24%260%26-3%5C%5C8%26-48%2656%5C%5C25%26-6%2610%5Cend%7Barray%7D%5Cright%5D)
The correct answer is C