You already know that since you need to pay 10$ for the room you would do 75-10 and you have 65$ left. 12 goes into 65 5 times, because 12x5=60. You then have 5 dollars left. So Carrie can invite 5 people to her birthday party.
There may be more than one way in which to answer this question. I will assume that the "equation" is a linear one: f(x) = mx + b.
Then (16/3) = m(1) + b
This is one equation in two unknowns, so it does not have a unique solution. Was there more to this problem than you have shared?
If we assume that the y-intercept (b) is zero, then y = mx, and
16/3 = 1m, so that m = 16/3, and so y = (16/3)x.
I'm partial to solving with generating functions. Let
![T(x)=\displaystyle\sum_{n\ge0}t_nx^n](https://tex.z-dn.net/?f=T%28x%29%3D%5Cdisplaystyle%5Csum_%7Bn%5Cge0%7Dt_nx%5En)
Multiply both sides of the recurrence by
and sum over all
.
![\displaystyle\sum_{n\ge0}2t_{n+2}x^{n+2}=\sum_{n\ge0}3t_{n+1}x^{n+2}+\sum_{n\ge0}2t_nx^{n+2}](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Csum_%7Bn%5Cge0%7D2t_%7Bn%2B2%7Dx%5E%7Bn%2B2%7D%3D%5Csum_%7Bn%5Cge0%7D3t_%7Bn%2B1%7Dx%5E%7Bn%2B2%7D%2B%5Csum_%7Bn%5Cge0%7D2t_nx%5E%7Bn%2B2%7D)
Shift the indices and factor out powers of
as needed so that each series starts at the same index and power of
.
![\displaystyle2\sum_{n\ge2}2t_nx^n=3x\sum_{n\ge1}t_nx^n+2x^2\sum_{n\ge0}t_nx^n](https://tex.z-dn.net/?f=%5Cdisplaystyle2%5Csum_%7Bn%5Cge2%7D2t_nx%5En%3D3x%5Csum_%7Bn%5Cge1%7Dt_nx%5En%2B2x%5E2%5Csum_%7Bn%5Cge0%7Dt_nx%5En)
Now we can write each series in terms of the generating function
. Pull out the first few terms so that each series starts at the same index
.
![2(T(x)-t_0-t_1x)=3x(T(x)-t_0)+2x^2T(x)](https://tex.z-dn.net/?f=2%28T%28x%29-t_0-t_1x%29%3D3x%28T%28x%29-t_0%29%2B2x%5E2T%28x%29)
Solve for
:
![T(x)=\dfrac{2-3x}{2-3x-2x^2}=\dfrac{2-3x}{(2+x)(1-2x)}](https://tex.z-dn.net/?f=T%28x%29%3D%5Cdfrac%7B2-3x%7D%7B2-3x-2x%5E2%7D%3D%5Cdfrac%7B2-3x%7D%7B%282%2Bx%29%281-2x%29%7D)
Splitting into partial fractions gives
![T(x)=\dfrac85\dfrac1{2+x}+\dfrac15\dfrac1{1-2x}](https://tex.z-dn.net/?f=T%28x%29%3D%5Cdfrac85%5Cdfrac1%7B2%2Bx%7D%2B%5Cdfrac15%5Cdfrac1%7B1-2x%7D)
which we can write as geometric series,
![T(x)=\displaystyle\frac8{10}\sum_{n\ge0}\left(-\frac x2\right)^n+\frac15\sum_{n\ge0}(2x)^n](https://tex.z-dn.net/?f=T%28x%29%3D%5Cdisplaystyle%5Cfrac8%7B10%7D%5Csum_%7Bn%5Cge0%7D%5Cleft%28-%5Cfrac%20x2%5Cright%29%5En%2B%5Cfrac15%5Csum_%7Bn%5Cge0%7D%282x%29%5En)
![T(x)=\displaystyle\sum_{n\ge0}\left(\frac45\left(-\frac12\right)^n+\frac{2^n}5\right)x^n](https://tex.z-dn.net/?f=T%28x%29%3D%5Cdisplaystyle%5Csum_%7Bn%5Cge0%7D%5Cleft%28%5Cfrac45%5Cleft%28-%5Cfrac12%5Cright%29%5En%2B%5Cfrac%7B2%5En%7D5%5Cright%29x%5En)
which tells us
![\boxed{t_n=\dfrac45\left(-\dfrac12\right)^n+\dfrac{2^n}5}](https://tex.z-dn.net/?f=%5Cboxed%7Bt_n%3D%5Cdfrac45%5Cleft%28-%5Cdfrac12%5Cright%29%5En%2B%5Cdfrac%7B2%5En%7D5%7D)
# # #
Just to illustrate another method you could consider, you can write the second recurrence in matrix form as
![49y_{n+2}=-16y_n\implies y_{n+2}=-\dfrac{16}{49}y_n\implies\begin{bmatrix}y_{n+2}\\y_{n+1}\end{bmatrix}=\begin{bmatrix}0&-\frac{16}{49}\\1&0\end{bmatrix}\begin{bmatrix}y_{n+1}\\y_n\end{bmatrix}](https://tex.z-dn.net/?f=49y_%7Bn%2B2%7D%3D-16y_n%5Cimplies%20y_%7Bn%2B2%7D%3D-%5Cdfrac%7B16%7D%7B49%7Dy_n%5Cimplies%5Cbegin%7Bbmatrix%7Dy_%7Bn%2B2%7D%5C%5Cy_%7Bn%2B1%7D%5Cend%7Bbmatrix%7D%3D%5Cbegin%7Bbmatrix%7D0%26-%5Cfrac%7B16%7D%7B49%7D%5C%5C1%260%5Cend%7Bbmatrix%7D%5Cbegin%7Bbmatrix%7Dy_%7Bn%2B1%7D%5C%5Cy_n%5Cend%7Bbmatrix%7D)
By substitution, you can show that
![\begin{bmatrix}y_{n+2}\\y_{n+1}\end{bmatrix}=\begin{bmatrix}0&-\frac{16}{49}\\1&0\end{bmatrix}^{n+1}\begin{bmatrix}y_1\\y_0\end{bmatrix}](https://tex.z-dn.net/?f=%5Cbegin%7Bbmatrix%7Dy_%7Bn%2B2%7D%5C%5Cy_%7Bn%2B1%7D%5Cend%7Bbmatrix%7D%3D%5Cbegin%7Bbmatrix%7D0%26-%5Cfrac%7B16%7D%7B49%7D%5C%5C1%260%5Cend%7Bbmatrix%7D%5E%7Bn%2B1%7D%5Cbegin%7Bbmatrix%7Dy_1%5C%5Cy_0%5Cend%7Bbmatrix%7D)
or
![\begin{bmatrix}y_n\\y_{n-1}\end{bmatrix}=\begin{bmatrix}0&-\frac{16}{49}\\1&0\end{bmatrix}^{n-1}\begin{bmatrix}y_1\\y_0\end{bmatrix}](https://tex.z-dn.net/?f=%5Cbegin%7Bbmatrix%7Dy_n%5C%5Cy_%7Bn-1%7D%5Cend%7Bbmatrix%7D%3D%5Cbegin%7Bbmatrix%7D0%26-%5Cfrac%7B16%7D%7B49%7D%5C%5C1%260%5Cend%7Bbmatrix%7D%5E%7Bn-1%7D%5Cbegin%7Bbmatrix%7Dy_1%5C%5Cy_0%5Cend%7Bbmatrix%7D)
Then solving the recurrence is a matter of diagonalizing the coefficient matrix, raising to the power of
, then multiplying by the column vector containing the initial values. The solution itself would be the entry in the first row of the resulting matrix.
Y= -3 . Plug the -2 into x. 8x-3y=-7
8(-2) -3y= -7