Question

Is each ordered pair a solution of the inequality? $$\frac{5}{6} x+\frac{5}{3} y>4 ;(6,-12),(8,-8)$$

Short answer

Neither of the ordered pairs \((6,-12)\) and \((8,-8)\) are solutions to the inequality \(\frac{5}{6}x + \frac{5}{3}y > 4\).

Step-by-step solution

Substitute the first ordered pair

Substitute \(x = 6\) and \(y = -12\) into the inequality \(\frac{5}{6}x + \frac{5}{3}y > 4\). So, we get \(\frac{5}{6} * 6 + \frac{5}{3} * -12 > 4\), which simplifies to \(5 - 20 > 4\). Evaluate to see if true or false.

Evaluate the first outcome

From Step 1, we find that the inequality \(5 - 20 > 4\) is false, because \(-15 > 4\) is not true. Therefore, the ordered pair \( (6,-12)\) is not a solution to the given inequality.

Substitute the second ordered pair

Substitute \(x = 8\) and \(y = -8\) into the inequality \(\frac{5}{6}x + \frac{5}{3}y > 4\). So, we get \(\frac{5}{6} * 8 + \frac{5}{3} * -8 > 4\), which simplifies to \(\frac{40}{6} - \frac{40}{3} > 4\). Evaluate to see if true or false.

Evaluate the second outcome

From Step 3, we find that the inequality \( \frac{40}{6} - \frac{40}{3} > 4\) simplifies to \(6.666 - 13.333 > 4\). However, \(-6.666 > 4\) is also not true. Therefore, the ordered pair \( (8,-8)\) is not a solution to the given inequality either.