Answer:

Step-by-step explanation:
![36-\left[100\div \:560-32\left(8\div \:4-1\right)\right]\\\\100\div \:560-32\left(8\div \:4-1\right)\\\\=100\div \:560-32\cdot \:1\\\\=\frac{100}{560}-32\cdot \:1\\\\=\frac{100}{560}-32\\\\=-\frac{891}{28}\\\\\\=36-\left(-\frac{891}{28}\right)\\\\36-\left(-\frac{891}{28}\right)\\\\=36+\frac{891}{28}\\\\\frac{36\cdot \:28}{28}+\frac{891}{28}\\\\\frac{36\cdot \:28+891}{28}\\\\=\frac{1899}{28}\\\\\\\frac{1899}{28}(1899\div28)=67\frac{23}{28}](https://tex.z-dn.net/?f=36-%5Cleft%5B100%5Cdiv%20%5C%3A560-32%5Cleft%288%5Cdiv%20%5C%3A4-1%5Cright%29%5Cright%5D%5C%5C%5C%5C100%5Cdiv%20%5C%3A560-32%5Cleft%288%5Cdiv%20%5C%3A4-1%5Cright%29%5C%5C%5C%5C%3D100%5Cdiv%20%5C%3A560-32%5Ccdot%20%5C%3A1%5C%5C%5C%5C%3D%5Cfrac%7B100%7D%7B560%7D-32%5Ccdot%20%5C%3A1%5C%5C%5C%5C%3D%5Cfrac%7B100%7D%7B560%7D-32%5C%5C%5C%5C%3D-%5Cfrac%7B891%7D%7B28%7D%5C%5C%5C%5C%5C%5C%3D36-%5Cleft%28-%5Cfrac%7B891%7D%7B28%7D%5Cright%29%5C%5C%5C%5C36-%5Cleft%28-%5Cfrac%7B891%7D%7B28%7D%5Cright%29%5C%5C%5C%5C%3D36%2B%5Cfrac%7B891%7D%7B28%7D%5C%5C%5C%5C%5Cfrac%7B36%5Ccdot%20%5C%3A28%7D%7B28%7D%2B%5Cfrac%7B891%7D%7B28%7D%5C%5C%5C%5C%5Cfrac%7B36%5Ccdot%20%5C%3A28%2B891%7D%7B28%7D%5C%5C%5C%5C%3D%5Cfrac%7B1899%7D%7B28%7D%5C%5C%5C%5C%5C%5C%5Cfrac%7B1899%7D%7B28%7D%281899%5Cdiv28%29%3D67%5Cfrac%7B23%7D%7B28%7D)
can o<u> help me pleaase and nope</u>
No, we do not multiply the first equation by 2 and then add it to the second equation. We have to multiply the first equation by 4.
To illustrate this point, consider system A:

If we multiply the first equation by 2, we get

Now, if we replace the second equation with the sum of the second equation and
, we get

which simplifies to

This is not an equivalent system to System B. We can see that we ended up with a 2y = 4 equation.
In order to end up with a 2x = 10 second equation, we have to multiply the first equation of system A by 4 to get

If we replace the second equation with the sum of the second equation and
, we get

which simplifies to

Otherwise, you are correct. The solution to system B is the solution to system A. Adding an equation to another does not change the system.
Solution:
we are given that
Mary ran 4 miles in 37 minutes.
we have been asked to find
how long would it take her to run 11 miles?
Since Mary ran 4 miles in 37 minutes, it mean time taken to run 1 miles can be given by

Hence we can write , time taken in running 11 miles as
