Question

the area bounded by the lines |2x-3|+|3y+4|=5 is
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Answer 1

Correct Question:-

$\sf \huge{the \: area \: \: bounded \: \: by \: \: the \: \: lines \: \: |2x - 3| + |3y + 2| = 6}$

Solution:

This problem is not difficult, but tedious. Here is what needs to be done.Consider four cases2x-3>0, 3y-2>0 then (2x-3)+(3y-2) =6 or y=-(2/3)x+11/3

This problem is not difficult, but tedious. Here is what needs to be done.Consider four cases2x-3>0, 3y-2>0 then (2x-3)+(3y-2) =6 or y=-(2/3)x+11/32x-3>0, 3y-2<0 then (2x-3)-(3y-2) =6 or y= (2/3)x +7/3

This problem is not difficult, but tedious. Here is what needs to be done.Consider four cases2x-3>0, 3y-2>0 then (2x-3)+(3y-2) =6 or y=-(2/3)x+11/32x-3>0, 3y-2<0 then (2x-3)-(3y-2) =6 or y= (2/3)x +7/32x-3<0, 3y-2>0 then -(2x-3)+(3y-2) =6 or y =(2/3)x+11/3

This problem is not difficult, but tedious. Here is what needs to be done.Consider four cases2x-3>0, 3y-2>0 then (2x-3)+(3y-2) =6 or y=-(2/3)x+11/32x-3>0, 3y-2<0 then (2x-3)-(3y-2) =6 or y= (2/3)x +7/32x-3<0, 3y-2>0 then -(2x-3)+(3y-2) =6 or y =(2/3)x+11/32x-3<0, 3y-2<0 then -(2x-3)-(3y-2) =6 or y= -(2/3)x + 1/3

This problem is not difficult, but tedious. Here is what needs to be done.Consider four cases2x-3>0, 3y-2>0 then (2x-3)+(3y-2) =6 or y=-(2/3)x+11/32x-3>0, 3y-2<0 then (2x-3)-(3y-2) =6 or y= (2/3)x +7/32x-3<0, 3y-2>0 then -(2x-3)+(3y-2) =6 or y =(2/3)x+11/32x-3<0, 3y-2<0 then -(2x-3)-(3y-2) =6 or y= -(2/3)x + 1/3Draw the four straight lines and look at the rectangle. It is a parallelogram because the first and the third line have the same slope, so do the second and the fourth line. We also know the coordinates of all four vertices. We can easily determine the area between the top two segments and the x-axis (sum of two trapezoids) and the area between bottom two segments and the x-axis (sum of two other trapezoids) and then deduct the two areas from each other.

MSD❣️

Answer 2

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hope it helps , ans in attachment