Answer: [tex]\begin{bmatrix}\mathrm{Solution:}\:&\:x>15\:\\ \:\mathrm{Interval\:Notation:}&\:\left(15,\:\infty \:\right)\end{bmatrix}[/tex]
Step-by-step explanation:
[tex]-3>-\frac{x}{5}[/tex]
[tex]-\frac{x}{5}<-3[/tex]
[tex]5\left(-\frac{x}{5}\right)<5\left(-3\right)[/tex]
[tex]-x<-15[/tex]
[tex]\left(-x\right)\left(-1\right)>\left(-15\right)\left(-1\right)[/tex]
[tex]x>15[/tex]
Hey there! :)
To find if the solution is a part of the Inequality, we can do by using two methods.
Solve The Inequality
We are given the Inequality:
[tex] \displaystyle \large{ - 3 > - \frac{x}{5} }[/tex]
Our goal is to isolate x-term. Since there's 5 as a denominator, what we have to do is to get rid of it by multiplying whole Inequality by 5.
[tex] \displaystyle \large{ - 3 \cdot 5 > - \frac{x}{5} \cdot 5 } \\ \displaystyle \large{ - 15> - x}[/tex]
Since -x is equivalent to -1x.
[tex] \displaystyle \large{ - 15> - 1x}[/tex]
Divide both sides by -1. Do not forget to swap the sign from > to < if we are dividing both sides by negative numbers.
[tex] \displaystyle \large{ \frac{ - 15}{ - 1} > \frac{ - 1x}{ - 1} } \\ \displaystyle \large{ 15 < x}[/tex]
Switch 15 < x to x > 15.
[tex] \displaystyle \large \boxed{x > 15}[/tex]
Since x > 15, that means if we substitute x = 10, we'd get 10 > 15 which is false.
Hence, x = 10 is not a part of the Inequality.
Substitution and Compare
From the Inequality, substitute x = 10 in.
[tex] \displaystyle \large{ - 3 > - \frac{10}{5} }[/tex]
Simplify:
[tex] \displaystyle \large{ - 3 > - 2}[/tex]
As we know that -3 is less than -2 and not greater. The Inequality is false.
Therefore, x = 10 is not a part of the solution.
Answer
