![\begin{cases} 4x+3y=-8\\\\ -8x-6y=16 \end{cases}~\hspace{10em} \begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cbegin%7Bcases%7D%204x%2B3y%3D-8%5C%5C%5C%5C%20-8x-6y%3D16%20%5Cend%7Bcases%7D~%5Chspace%7B10em%7D%20%5Cbegin%7Barray%7D%7B%7Cc%7Cll%7D%20%5Ccline%7B1-1%7D%20slope-intercept~form%5C%5C%20%5Ccline%7B1-1%7D%20%5C%5C%20y%3D%5Cunderset%7By-intercept%7D%7B%5Cstackrel%7Bslope%5Cqquad%20%7D%7B%5Cstackrel%7B%5Cdownarrow%20%7D%7Bm%7Dx%2B%5Cunderset%7B%5Cuparrow%20%7D%7Bb%7D%7D%7D%20%5C%5C%5C%5C%20%5Ccline%7B1-1%7D%20%5Cend%7Barray%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
![4x+3y=-8\implies 3y=-4x-8\implies y=\cfrac{-4x-8}{3}\implies y=\stackrel{\stackrel{m}{\downarrow }}{-\cfrac{4}{3}} x-\cfrac{8}{3} \\\\[-0.35em] ~\dotfill\\\\ -8x-6y=16\implies -6y=8x+16\implies y=\cfrac{8x+16}{-6} \\\\\\ y=\cfrac{8}{-6}x+\cfrac{16}{-6}\implies y=\stackrel{\stackrel{m}{\downarrow }}{-\cfrac{4}{3}} x-\cfrac{8}{3}](https://tex.z-dn.net/?f=4x%2B3y%3D-8%5Cimplies%203y%3D-4x-8%5Cimplies%20y%3D%5Ccfrac%7B-4x-8%7D%7B3%7D%5Cimplies%20y%3D%5Cstackrel%7B%5Cstackrel%7Bm%7D%7B%5Cdownarrow%20%7D%7D%7B-%5Ccfrac%7B4%7D%7B3%7D%7D%20x-%5Ccfrac%7B8%7D%7B3%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20-8x-6y%3D16%5Cimplies%20-6y%3D8x%2B16%5Cimplies%20y%3D%5Ccfrac%7B8x%2B16%7D%7B-6%7D%20%5C%5C%5C%5C%5C%5C%20y%3D%5Ccfrac%7B8%7D%7B-6%7Dx%2B%5Ccfrac%7B16%7D%7B-6%7D%5Cimplies%20y%3D%5Cstackrel%7B%5Cstackrel%7Bm%7D%7B%5Cdownarrow%20%7D%7D%7B-%5Ccfrac%7B4%7D%7B3%7D%7D%20x-%5Ccfrac%7B8%7D%7B3%7D)
one simple way to tell if both equations do ever meet or have a solution is by checking their slope, notice in this case the slopes are the same for both, meaning the lines are parallel lines, however, notice both equations are really the same, namely the 2nd equation is really the 1st one in disguise.
since both equations are equal, their graph will be of one line pancaked on top of the other, and the solutions is where they meet, hell, they meet everywhere since one is on top of the other, so infinitely many solutions.
#1 is 2 #2 is 6 i dont know 3
Answer:
The value for the original mean = 32
Step-by-step explanation:
Here, we want to calculate the original value of the mean.
Let the number of samples be n
Mathematically;
mean = Total value/n
Now, we added 8 to each of values; total value added = 8 * n = 8n
Now, for the new mean of 40; we have
(Total value + 8n)/n = 40
Total value + 8n = 40n
Total value = 40n -8n
Total value = 32n
kindly recall from the beginning of the solution;
mean = Total value/n
mean = 32n/n
mean = 32
So the original value of the mean is 32
I’d just do 25 x 4 and it’s simple as that. Not unless I’m missing something
