Answer:
Step-by-step explanation:
f(x) = 2x - 3
f(-2) = 2(-2) -3 = -4 - 3 = -7
f(-1) = 2(-1) -3 = -2 - 3 = - 5
f(0) = 2(0) -3 = 0 - 3 = -3
f(1) = 2(1) -3 = 2 - 3 = -1
f(2) = 2(2) -3 = 4 - 3 = 1
f(3) = 2(3) -3 = 6 - 3 = 3
Answer:
x = 7 and y = -2
Given:
x = 5y + 17 _____first equation
2x + 3y = 8 ____second equation
Steps:
2x + 3y = 8
<em>plug in x = 5y + 17 into second equation</em>
<em></em>
2(5y + 17)+3y = 8 [ open the brackets ]
10y + 34 + 3y = 8
10y + 3y = 8 - 34 [ collect and group terms ]
13y = -26 [ divide both sides by 13 ]
y = -2 [ final answer ]
solve for x:
x = 5y + 17
x = 5(-2) + 17
x = 7
![\bf n^{th}\textit{ term of an arithmetic sequence} \\\\ a_n=a_1+(n-1)d\qquad \begin{cases} n=n^{th}\ term\\ a_1=\textit{first term's value}\\ d=\textit{common difference} \end{cases} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ a_n=2-5(n-1)\implies a_n=\stackrel{\stackrel{a_1}{\downarrow }}{2}+(n-1)(\stackrel{\stackrel{d}{\downarrow }}{-5})](https://tex.z-dn.net/?f=%5Cbf%20n%5E%7Bth%7D%5Ctextit%7B%20term%20of%20an%20arithmetic%20sequence%7D%20%5C%5C%5C%5C%20a_n%3Da_1%2B%28n-1%29d%5Cqquad%20%5Cbegin%7Bcases%7D%20n%3Dn%5E%7Bth%7D%5C%20term%5C%5C%20a_1%3D%5Ctextit%7Bfirst%20term%27s%20value%7D%5C%5C%20d%3D%5Ctextit%7Bcommon%20difference%7D%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%20a_n%3D2-5%28n-1%29%5Cimplies%20a_n%3D%5Cstackrel%7B%5Cstackrel%7Ba_1%7D%7B%5Cdownarrow%20%7D%7D%7B2%7D%2B%28n-1%29%28%5Cstackrel%7B%5Cstackrel%7Bd%7D%7B%5Cdownarrow%20%7D%7D%7B-5%7D%29)
so, we know the first term is 2, whilst the common difference is -5, therefore, that means, to get the next term, we subtract 5, or we "add -5" to the current term.
just a quick note on notation:
![\bf \stackrel{\stackrel{\textit{current term}}{\downarrow }}{a_n}\qquad \qquad \stackrel{\stackrel{\textit{the term before it}}{\downarrow }}{a_{n-1}} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{current term}}{a_5}\qquad \quad \stackrel{\textit{term before it}}{a_{5-1}\implies a_4}~\hspace{5em}\stackrel{\textit{current term}}{a_{12}}\qquad \quad \stackrel{\textit{term before it}}{a_{12-1}\implies a_{11}}](https://tex.z-dn.net/?f=%5Cbf%20%5Cstackrel%7B%5Cstackrel%7B%5Ctextit%7Bcurrent%20term%7D%7D%7B%5Cdownarrow%20%7D%7D%7Ba_n%7D%5Cqquad%20%5Cqquad%20%5Cstackrel%7B%5Cstackrel%7B%5Ctextit%7Bthe%20term%20before%20it%7D%7D%7B%5Cdownarrow%20%7D%7D%7Ba_%7Bn-1%7D%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Bcurrent%20term%7D%7D%7Ba_5%7D%5Cqquad%20%5Cquad%20%5Cstackrel%7B%5Ctextit%7Bterm%20before%20it%7D%7D%7Ba_%7B5-1%7D%5Cimplies%20a_4%7D~%5Chspace%7B5em%7D%5Cstackrel%7B%5Ctextit%7Bcurrent%20term%7D%7D%7Ba_%7B12%7D%7D%5Cqquad%20%5Cquad%20%5Cstackrel%7B%5Ctextit%7Bterm%20before%20it%7D%7D%7Ba_%7B12-1%7D%5Cimplies%20a_%7B11%7D%7D)
Answer:
the solution to the equation is 8. Any value apart from 8 is not a solution
Step-by-step explanation:
Given the equation
6/3x+2 - 1/x-4 = 0
6/3x+2 = 1/x-4
Cross multiply
6(x-4) = 3x+2
6x - 24 = 3x+2
6x-3x = 2 + 24
3x = 24
x = 24/3
x = 8
Hence the solution to the equation is 8. Any value apart from 8 is not a solution
