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The inequality y<x-5 shows that the required y values are hosted by the region under the line x-5.
You need to write the next inequality the same with the first inequality.
You need to isolate the variable y to the left side and the variable x to the right side such that:
`2x + 4y gt 8 =gt x + 2y gt 4`
Subtract x both sides such that: `2y gt 4 - x` .
Dividing by 2 both sides yields:
`y gt (4-x)/2`
The inequality `ygt(4-x)/2 ` shows that the required y values are hosted by the region above the line `(4-x)/2` .
The solution to the simultaneous inequalities is the region found to the right of the point of intersection of the lines `x - 5 and (4-x)/2` .
The exact interval of x values that verify both inequalities may be found equating`x - 5 = (4 - x)/2 =gt 2x - 10 = 4- x =gt 3x = 10+4 =gt x = 14/3 ~~ 4.6` .
The solution to the simultaneous inequalities is `(4.6 ; oo).`
in the first equation you have y<x-5. Well first find your points. In this case you would go with your b and your slope. B or y-int = (0,-5) and since m is your slope and M in this case is 1, your slope is 1/1.
for the next one, you can go about doing it two ways, Either putting it into y>mx+b or you can do the intercept method. I personally like intercept method for this setup. So plug zero into y and solve for x which gives you (4,0)
to find your y-int, plug 0 in for x and solve for y. Which gives you (0,2)
So for the first line at (0,-5) with a slope of 1, your are going to shade below the line. For the second line at (4,0) and (0,2) you are going to shade above the line.
remember when drawing both of these lines they are going to be dotted lines not solid because this is less than and greater than, not less than OR equal too or greater than OR equal too.
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