Question

Write a system of linear inequalities so the points (1, 2) and (4, -3) are solutions of the system, but the point (-2, 8) is not a solution of the system.

Answer 1

Answer: $\left \{ {{y\leq -\frac{5}{3}x + \frac{11}{3}} \atop {y < 8}} \right$

Step-by-step explanation:

(1, 2) and (4, -3)

First, find the slope (m): $\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

m = $\frac{-3-2}{4-1}$

   = $-\frac{5}{3}$

Next, choose ONE of the points and input the point and slope into the Point-Slope formula: y - y₁ = m(x - x₁)

y - 2 = $-\frac{5}{3}$(x - 1)

y - 2 = $-\frac{5}{3}x$ + $\frac{5}{3}$

y      = $-\frac{5}{3}x$ + $\frac{11}{3}$

Then, determine which inequality symbol will result in (-2, 8) being False (since it is not a solution):

y ___ $-\frac{5}{3}x$ + $\frac{11}{3}$

8 ___ $-\frac{5}{3}(-2)$ + $\frac{11}{3}$

8 ___ $\frac{1}{3}$

8 > $\frac{1}{3}$

so ≤ makes the statement False

⇒ y ≤ $-\frac{5}{3}x$ + $\frac{11}{3}$

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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:

$\left \{ {{x>-2} \atop {y < 8}} \right$