Question 257610
WIth these absolute values we must start with 4 inequalities. here are the four:
(i) {{{(4+x) + (5+y) <= 100}}}
(ii) {{{(4+x) - (5+y) <=100}}}
(iii){{{-(4+x) + (5+y) <=100}}}
(iv) {{{-(4+x) - (5+y) <=100}}}
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we can solve each of these for y as
(i) becomes (v) {{{y <= 91-x}}}
(ii) becomes (vi) {{{y >= x-101}}}
(iii) becomes (vii) {{{y<= x + 99}}}
(iv) becomes (viii) {{{y>= -x -109}}}
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Graphing them creates a quadrilateral. We now graph these and look for the minimum y crossing points.  This occurs when (vi) crosses over (viii). We get
{{{y >= x-101}}}
{{{y>= -x -109}}}
adding down, the x's drop out and we get
{{{2y >= -210}}}
and then
{{{y >= -105}}}
So, the minimum value of y is -105.