for part c, i get the following:
6000 currently live on campus and 3000 currently live off campus.
since x is the number who live on campus this semester and y is the number who currrently live off campus this summer, then you get:
x = 6000 and y = 3000.
your first equation is the ones who will be living on campus next semester.
of the 6000 that are currently living on campus, 70% will still live on campus next semester.
of the 3000 that are currently living off campus, 10% will be living on campus next semester.
.7x + .1y = u becomes .7*6000 + .1*3000 = 4500 that will be living on campus next semeste.
your second equation is the ones who will be living off campus next semeste.
of the 6000 that are currently living on campus, 30% will live off campus next semester.
of the 3000 that are currently living off campus, 90% will continue to live off campus next semester.
.3x + .9y = v becomes .3*6000 + .9*3000 = 4500 that will be living off campus next semester.
for part a i get the following:
the matrix is:
tableau 1 row 1 equals .7 .1 u
tableau 1 row 2 equals .3 .9 v
normally, u and v would be constants.
since i didn't know their value up front (i learaned it in part c), i left them as u and v.
for part b, i get the following:
start with:
tableau 1 row 1 equals .7 .1 u
tableau 1 row 2 equals .3 .9 v
tableau 2 row 1 equals 9 * tableau 1 row 1 minus tableau 1 row 2.
tableau 2 row 2 equals tableau 1 row 2.
you get:
tableau 2 row 1 equals 6 0 9u - v
tableau 2 row 2 equals .3 .9 v
tableau 3 row 1 equals tableau 2 row 1
tableau 3 row 2 = tableau 2 row 2 minus .05 * tableau 2 row 1.
you get:
tableau 3 row 1 equals 6 0 9u - v
tableau 3 row 2 equals 0 .9 v - .45u + .05v
tableau 4 row 1 equals tableau 3 row 1 divided by 6.
tableau 4 row 2 equals tableau 3 row 2 divided by .9.
you get:
tableau 4 row 1 equals 1 0 (9u - v) / 6
tableau 4 row 2 equals 0 1 (1.05v - .45u) / .9
that completes part b.
from part c, we know that both u and v = 4500 each.
to confirm part b was done correctly, replace u and v with 4500 to get:
tableau 4 row 1 equals 1 0 (9*4500-4500)/6 = 6000
tableau 4 row 2 equals 0 1 (1.05*4500 - .45*4500) / .9 = 3000
since we already know that x was equal to 6000 and y was equal to 3000 because that was given to us, this confirms the matrix equation was modeled correctly.
i believe this satisfies the requirements for this problem.
let me know if you have any questions.
solving part c first helped me in analyzing parts a and b because i had some number to check my answers for part a and b against.
here's a reference on matrix operations you might find helpful.
https://www.studypug.com/algebra-help/the-three-types-of-matrix-row-operations?camp_id=1026250253&grp_id=112411027342&kw=&mt=&source=g&pos=&tgt=dsa-19959388920&loc=9003567&device=c&adid=529422243034&gclid=EAIaIQobChMItIvzlf-59AIVg2pvBB3u6wjTEAMYAiAAEgLzxPD_BwE
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