Question 123204
The best way is to plot these
(1) {{{3x + y < 9}}}
(2) {{{x - y < 7}}}
(3) {{{x > 0}}}
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Find {{{y}}} in terms of {{{x}}}
(1) {{{3x + y < 9}}}
{{{y < -3x + 9}}}
(2) {{{x - y < 7}}}
{{{-y < -x + 7}}}
multiply both sides by {{{-1}}} and reverse the < sign
{{{y > x - 7}}}
(3) {{{x >0}}} No {{{y}}} involved here. This is saying that {{{x}}}
can never be {{{0}}} or negative. It's domain is the right half of
the graph
I'll do a plot of
{{{y < -3x + 9}}} and
{{{y > x - 7}}}
 {{{ graph( 600, 600, -2, 30, -15, 15, -3x + 9,x - 7)}}}
The solution is the little triangular area bounded by the {{{y}}} axis
and the lines. Now see which points are in that area
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(2,5) looks to be just outside, and it fails (1) because
{{{y < -3x + 9}}}
{{{5 < (-3)*2 + 9}}}
{{{5 < -6 + 9}}}
{{{5 < 3}}} This is clearly not true
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(0,9) This looks like it might be in the area, but it fails (1) also
{{{y < -3x + 9}}}
{{{9 < (-3)*0 + 9}}}
{{{9 < 9}}} This is clearly not true also
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(3,-2) This fails in the area
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(-1,0) {{{x}}} can't be negative, as I said, and it is
also clearly outside the area
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So, only (3,-2) satisfies the inequalities answer