The answer to 1 is common multiple. And the answer to 2 is common denominator
a)
well, she put 4000, and she earned in interest 960, so her accumulated amount is just their sum, 4960.
b)
now, it doesn't say, so we're assuming is <u>simple interest</u>, as opposed to compound interest.

c)
let's make the rate 1% greater then, and check
![\bf ~~~~~~ \textit{Simple Interest Earned} \\\\ I = Prt\qquad \begin{cases} I=\textit{interest earned}\dotfill\\ P=\textit{original amount deposited}\dotfill & \$4000\\ r=rate\to \stackrel{8+1}{9\%}\to \frac{9}{100}\dotfill&0.09\\ t=years\dotfill &3 \end{cases} \\\\\\ I=(4000)(0.09)(3)\implies I=1080 \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{at 9\%}}{1080}-\stackrel{\textit{at 8\%}}{960}\implies \stackrel{\textit{that much more}}{120}](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~%20%5Ctextit%7BSimple%20Interest%20Earned%7D%20%5C%5C%5C%5C%20I%20%3D%20Prt%5Cqquad%20%5Cbegin%7Bcases%7D%20I%3D%5Ctextit%7Binterest%20earned%7D%5Cdotfill%5C%5C%20P%3D%5Ctextit%7Boriginal%20amount%20deposited%7D%5Cdotfill%20%26%20%5C%244000%5C%5C%20r%3Drate%5Cto%20%5Cstackrel%7B8%2B1%7D%7B9%5C%25%7D%5Cto%20%5Cfrac%7B9%7D%7B100%7D%5Cdotfill%260.09%5C%5C%20t%3Dyears%5Cdotfill%20%263%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5C%5C%20I%3D%284000%29%280.09%29%283%29%5Cimplies%20I%3D1080%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Bat%209%5C%25%7D%7D%7B1080%7D-%5Cstackrel%7B%5Ctextit%7Bat%208%5C%25%7D%7D%7B960%7D%5Cimplies%20%5Cstackrel%7B%5Ctextit%7Bthat%20much%20more%7D%7D%7B120%7D)
1) y= -5
2) m = -1
3) k = 3
4) y = -5
Step-by-step explanation:
1) 
Solving:

2) ![8(-1+m)+3 = 2 [m-5 1/2]\\](https://tex.z-dn.net/?f=8%28-1%2Bm%29%2B3%20%3D%202%20%5Bm-5%201%2F2%5D%5C%5C)
Solving:
3)
(Note: Considering 8(12-k)=3(k+21) instead of 8(12-k)=3(k=k+21))
Solving:

4) 
Solving:

Keywords: Solve for variables.
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