Answer:
I'm sorry is there a chart/graph, etc. to give us more information?
Without the added information there is no way for me to answer.
If you repost this question with a picture I would be more than happy to help.
![\begin{array}{rrrrr} 10x&-&18y&=&2\\ -5x&+&9y&=&-1 \end{array}~\hfill \implies ~\hfill \stackrel{\textit{second equation }\times 2}{ \begin{array}{rrrrr} 10x&-&18y&=&2\\ 2(-5x&+&9y&)=&2(-1) \end{array}} \\\\[-0.35em] ~\dotfill\\\\ \begin{array}{rrrrr} 10x&-&18y&=&2\\ -10x&+&18y&=&-2\\\cline{1-5} 0&+&0&=&0 \end{array}\qquad \impliedby \textit{another way of saying \underline{infinite solutions}}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Brrrrr%7D%2010x%26-%2618y%26%3D%262%5C%5C%20-5x%26%2B%269y%26%3D%26-1%20%5Cend%7Barray%7D~%5Chfill%20%5Cimplies%20~%5Chfill%20%5Cstackrel%7B%5Ctextit%7Bsecond%20equation%20%7D%5Ctimes%202%7D%7B%20%5Cbegin%7Barray%7D%7Brrrrr%7D%2010x%26-%2618y%26%3D%262%5C%5C%202%28-5x%26%2B%269y%26%29%3D%262%28-1%29%20%5Cend%7Barray%7D%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cbegin%7Barray%7D%7Brrrrr%7D%2010x%26-%2618y%26%3D%262%5C%5C%20-10x%26%2B%2618y%26%3D%26-2%5C%5C%5Ccline%7B1-5%7D%200%26%2B%260%26%3D%260%20%5Cend%7Barray%7D%5Cqquad%20%5Cimpliedby%20%5Ctextit%7Banother%20way%20of%20saying%20%5Cunderline%7Binfinite%20solutions%7D%7D)
if we were to solve both equations for "y", we'd get
notice, the 1st equation is really the 2nd in disguise, since both lines are just pancaked on top of each other, every point in the lines is a solution or an intersection, and since both go to infinity, well, there you have it.
Plug in -1 for x
h(-1) = 3(-1) + 4
h(-1) = -3 + 4
Solution: h(-1) = 1
Well, first, you need to know what you want to 'do' to it.
You want to find the description of what 'x' has to be in order to
make that inequality a true statement. Here's one way to do it:
4/5 x > 14
Multiply each side by 5 : 4 x > 70
Divide each side by 4 : x > 17.5 .
That's it. As long as 'x' is more than 17.5 ,
the original inequality is true.
