Answer: [tex]\left \{ {{y\leq -\frac{5}{3}x + \frac{11}{3}} \atop {y < 8}} \right[/tex]
Step-by-step explanation:
(1, 2) and (4, -3)
First, find the slope (m): [tex]\frac{y_{2}-y_{1}}{x_{2}-x_{1}}[/tex]
m = [tex]\frac{-3-2}{4-1}[/tex]
= [tex]-\frac{5}{3}[/tex]
Next, choose ONE of the points and input the point and slope into the Point-Slope formula: y - y₁ = m(x - x₁)
y - 2 = [tex]-\frac{5}{3}[/tex](x - 1)
y - 2 = [tex]-\frac{5}{3}x[/tex] + [tex]\frac{5}{3}[/tex]
y = [tex]-\frac{5}{3}x[/tex] + [tex]\frac{11}{3}[/tex]
Then, determine which inequality symbol will result in (-2, 8) being False (since it is not a solution):
y ___ [tex]-\frac{5}{3}x[/tex] + [tex]\frac{11}{3}[/tex]
8 ___ [tex]-\frac{5}{3}(-2)[/tex] + [tex]\frac{11}{3}[/tex]
8 ___ [tex]\frac{1}{3}[/tex]
8 > [tex]\frac{1}{3}[/tex]
so ≤ makes the statement False
⇒ y ≤ [tex]-\frac{5}{3}x[/tex] + [tex]\frac{11}{3}[/tex]
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If you need a "system" of inequalities, then you need another equation.
There are infinite possibilities. Graph the points to see what works.
y < 8 or x > -2 are two examples.
---> I just realized that the system can be simpler than what I did above:
[tex]\left \{ {{x>-2} \atop {y < 8}} \right[/tex]
