Question 1170631
let a = number of butterscotch breads
let b = number of chocolate breads
let c = number of coconut breads.


your solution is:
a = 500
b = 750
c = 900


your original equations are:


2a + 3b + 4c = 6850
a + b + c = 2150
a + 1.5b + 1.5c = 2975


replace the variables with their respective values to get:


2*500+ 3*750 + 4*900 = 6850
500 + 750 + 900 = 2150
500 + 1.5*750 + 1.5*900 = 2975


this confirms the solution is correct.


i used a gaussian elimination calculator such as the one found at <a href = "https://onlinemschool.com/math/assistance/equation/gaus/" target = "_blank">https://onlinemschool.com/math/assistance/equation/gaus/</a>


if you were to do this manually, you would probably do the following, or something that, the goal being to get your matrix into a form of:


<pre>
                    xa            0           0          t1
                    0             yb          0          t2
                    0             0           zc         t3
</pre>


you could stop there, or you could go a step further and get your matrix into a form of:


<pre>
                    1a            0           0          t1
                    0             1b          0          t2
                    0             0           1c         t3
</pre>


both forms will get you the answer.


in the first form, however, you will  need to go one extra step, such as:
a = t1/x
b = t2/y
c = t3/z
while, in the second form, that has already been done and you can just read off the answer as:
a = t1
b = t2
c = t3


the t's in the first form would, naturally, not be the same values as the t's in the second form, since the division was already done in the second form.
the a,b,c in the first form, however, would be the same as the a,b,c in the second form, once the final divisions were performed.


i took a stab at doing this manually, just to see if i could get the same answer.
i was successful.
my effort is shown below.
it's a little sloppy, but you should be able to get the idea from it as to how the procedure works.
the are different procedures that will work.
the one below is just one tactic that was used.
the goal is to get the matrix into one of the forms i shows you above.
i just happened to get to the second form, which was actually quite good for a first effort.


<img src = "http://theo.x10hosting.com/2020/112801.jpg" >
<img src = "http://theo.x10hosting.com/2020/112802.jpg" >


in step 1, i just copied down the matrix.
that's the 3 equations with only the coefficients and the constants showing.


in step 2, i replaced row 3 with the calculation of row 1 minus 2 * row 3 from step 1.
that's what R1 - 2R3 means on row 3 of the matrix in step 1.


in step 3, i replaced row 1 with the calculation of row 1 minus row 2 from step 2.
that's what R1 - R2 means on row 1 of the matrix in step 2.


in step 4, i screwed it up and just crossed it out and repeated step 4 below that.
i replaced row 2 with the calculation of row 2 minus row 1 from step 3.


in step 5, i replaced row 2 with the calculation of row 2 + 2 * row 3 from step 4.


in step 6, i replaced row 2 with the calculation of row 2 * -1 from step 5.


in step 7, i replaced row 1 with the calculation of row 1 - 2 * row 2 from step 6.


in step 8, i replaced row 1 with the calculation of row 1 - 3 * row 2 from step 7.


you are free to use the mechanized calculator to see if you get the same results.
in doing that, you can also see the steps that the calculator used to get to the final answer.
they are not necessarily the steps that i took manually.
there are various ways to get the same answer.
some are more efficient than others.