Ok so assume that if you have
xy=0 then x and y=0
-1/4(x^2-4x+3)>0
multiply both sides by -4 to clear fraction (-4/-4=1)
flip sign
x^2-4x+3<0
factor
(x-3)(x-1)<0
we know that (+) times (-)=(-) and
(-) times (-)=(+) so we don't want them to be both negative, we want different sign
we cannnot have 3 since it would be (0)(2)<0 which is false
we cannot have 1 either since (-2)(0)<0 is also false
lets see if the solution is in betwen 3 and 1 or outside of 3 and 1
ltry 2
2^3-4(2)+3<0
8-8+3<0
3<0
false
therefor it is outside ie
x>3 and x<1
is simply the difference of both amounts, but firstly let's convert the mixed fractions to improper, and subtract.
![\bf \stackrel{mixed}{4\frac{1}{2}}\implies \cfrac{4\cdot 2+1}{2}\implies \stackrel{improper}{\cfrac{9}{2}} \\\\\\ \stackrel{mixed}{6\frac{7}{16}}\implies \cfrac{6\cdot 16+7}{16}\implies \stackrel{improper}{\cfrac{103}{16}} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \stackrel{Jessie}{\cfrac{103}{16}}-\stackrel{Bryce}{\cfrac{9}{2}}\implies \stackrel{\textit{our LCD is 16}}{\cfrac{(1)103-(8)9}{16}}\implies \cfrac{103-72}{16}\implies \cfrac{31}{16}\implies 1\frac{15}{16}](https://tex.z-dn.net/?f=%5Cbf%20%5Cstackrel%7Bmixed%7D%7B4%5Cfrac%7B1%7D%7B2%7D%7D%5Cimplies%20%5Ccfrac%7B4%5Ccdot%202%2B1%7D%7B2%7D%5Cimplies%20%5Cstackrel%7Bimproper%7D%7B%5Ccfrac%7B9%7D%7B2%7D%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Cstackrel%7Bmixed%7D%7B6%5Cfrac%7B7%7D%7B16%7D%7D%5Cimplies%20%5Ccfrac%7B6%5Ccdot%2016%2B7%7D%7B16%7D%5Cimplies%20%5Cstackrel%7Bimproper%7D%7B%5Ccfrac%7B103%7D%7B16%7D%7D%0A%5C%5C%5C%5C%5B-0.35em%5D%0A%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%0A%5Cstackrel%7BJessie%7D%7B%5Ccfrac%7B103%7D%7B16%7D%7D-%5Cstackrel%7BBryce%7D%7B%5Ccfrac%7B9%7D%7B2%7D%7D%5Cimplies%20%5Cstackrel%7B%5Ctextit%7Bour%20LCD%20is%2016%7D%7D%7B%5Ccfrac%7B%281%29103-%288%299%7D%7B16%7D%7D%5Cimplies%20%5Ccfrac%7B103-72%7D%7B16%7D%5Cimplies%20%5Ccfrac%7B31%7D%7B16%7D%5Cimplies%201%5Cfrac%7B15%7D%7B16%7D)
-x+3x=2x
2x=2+6
2x=8
8/2=4
x=4
