Question

is (-7, 10) a solution to the inequality $y > \frac{-1}{7}x + 9$?
yes
no
submit

Explanation:

Step1: Substitute x and y values

Given the point \((-7, 10)\), so \(x = -7\) and \(y = 10\). Substitute into the inequality \(y > \frac{-1}{7}x + 9\).
Left side: \(y = 10\)
Right side: \(\frac{-1}{7}(-7)+9\)

Step2: Calculate the right side

First, calculate \(\frac{-1}{7}(-7)\) which is \(1\) (since a negative times a negative is positive and \(\frac{7}{7}=1\)). Then add \(9\): \(1 + 9 = 10\). So the right side is \(10\).

Step3: Check the inequality

The inequality is \(y > \frac{-1}{7}x + 9\), which becomes \(10 > 10\)? But \(10\) is not greater than \(10\), it is equal. However, wait, let's re - check the calculation. Wait, \(\frac{-1}{7}\times(-7)=\frac{7}{7}=1\), then \(1 + 9 = 10\). So \(y = 10\) and the right - hand side is also \(10\). But the inequality is \(y>\frac{-1}{7}x + 9\), so \(10>10\) is false? Wait, no, wait, maybe I made a mistake. Wait, \(\frac{-1}{7}x\) when \(x=-7\) is \(\frac{-1}{7}\times(-7)=1\), then \(1 + 9 = 10\). So \(y = 10\) and the right - hand side is \(10\). But the inequality is \(y>\frac{-1}{7}x + 9\), so \(10>10\) is false? Wait, no, wait the original inequality is \(y>\frac{-1}{7}x + 9\). So we have \(10>10\)? No, that's not true. Wait, but maybe I miscalculated. Wait, \(\frac{-1}{7}\times(-7)=1\), \(1 + 9 = 10\). So \(y = 10\) and the right side is \(10\). So \(10\) is not greater than \(10\), so the answer should be no? Wait, no, wait, maybe I messed up the sign. Wait, the inequality is \(y>\frac{-1}{7}x + 9\). Let's substitute again: \(x=-7\), so \(\frac{-1}{7}\times(-7)=1\), then \(1 + 9 = 10\). So \(y = 10\), so \(10>10\) is false. So the answer is no? Wait, but let's check again. Wait, \(\frac{-1}{7}x\) when \(x = - 7\): \(\frac{-1}{7}\times(-7)=1\), then \(1+9 = 10\). So \(y = 10\) and the right - hand side is \(10\). So \(10>10\) is false, so \((-7,10)\) is not a solution. Wait, but maybe the problem has a typo? Or maybe I made a mistake. Wait, no, the calculation seems correct. So the answer is no? Wait, but let's see, maybe I misread the inequality. The inequality is \(y>\frac{-1}{7}x + 9\). So substituting \(x=-7\), \(y = 10\): \(10>\frac{-1}{7}\times(-7)+9\), which is \(10>1 + 9\), \(10>10\), which is false. So the answer is no.

Answer:

no