Answer:
![x=\frac{1}{2}\\\\y=-6](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B1%7D%7B2%7D%5C%5C%5C%5Cy%3D-6)
Step-by-step explanation:
Given the following system of equations:
![\left \{ {{4x-11y=68} \atop {6x+5y=-27}} \right.](https://tex.z-dn.net/?f=%5Cleft%20%5C%7B%20%7B%7B4x-11y%3D68%7D%20%5Catop%20%7B6x%2B5y%3D-27%7D%7D%20%5Cright.)
In order to solve the system of equations using the Elimination Method, you can follow these steps:
- Multiply the first equation by -6 and the secondd equation by 4.
- Add both equations.
- Solve for "y".
Then:
![\left \{ {{-24x+66y=-408} \atop {24x+20y=-108}} \right\\.........................\\86y=300\\\\y=\frac{-516}{86}\\\\y=-6](https://tex.z-dn.net/?f=%5Cleft%20%5C%7B%20%7B%7B-24x%2B66y%3D-408%7D%20%5Catop%20%7B24x%2B20y%3D-108%7D%7D%20%5Cright%5C%5C.........................%5C%5C86y%3D300%5C%5C%5C%5Cy%3D%5Cfrac%7B-516%7D%7B86%7D%5C%5C%5C%5Cy%3D-6)
- Substitute the value of "y" into one of the original equations and solve for "x":
![6x+5(-6)=-27\\\\6x=-27+30\\\\x=\frac{3}{6}\\\\x=\frac{1}{2}](https://tex.z-dn.net/?f=6x%2B5%28-6%29%3D-27%5C%5C%5C%5C6x%3D-27%2B30%5C%5C%5C%5Cx%3D%5Cfrac%7B3%7D%7B6%7D%5C%5C%5C%5Cx%3D%5Cfrac%7B1%7D%7B2%7D)
f(x)=(x+a)/b
or bf(x)=x+a
let f(x)=y
by=x+a
flip x and y
bx=y+a
or y=bx-a
or f^{-1}(x)=bx-a
also g(x) is inverse of f(x)
bx-a=cx-d
so b=c,a=d
again let g(x)=y
y=cx-d
flip x and y
x=cy-d
cy=x+d
y=(x+d)/c
or g^{-1}(x)=(x+d)/c
also f(x) is inverse of g(x)
so (x+a)/b=(x+d)/c
so a=d,b=c
so in either case a=d,b=c
take b=c=1
a=d=2
f(x)=(x+2)/1=x+2
g(x)=1x-2=x-2
so f(x) and g(x) are two parallel lines f(x) with y- intercept=1 and slope 0
g(x) with y-intercept -2 and slope 0
if we take b=c=2,a=d=3
f(x)=(x+3)/2=x/2+3/2
g(x)=2x-3
here f(x) is of slope 1/2 and y-intercept 3/2
g(x) is of slope 2 and y intercept -3
part 3.
f(f(x))=g((x+a)/b)=c[(x+a)/b]-d=(c/b)(x+a)-d
(fg)(x) = [f(x)][g(x)]
(fg)(x) = (4x - 5)(3) = 12x - 15
Ok, this definition of f is just a bunch of points (4 to be exact).
so if point (a,b) is part of f, then f(a)=b
f(1) is about point (1,0), so f(1)=0.
g(1) = 1 (due to point (1,1))
g(2/3) = 0
f(2) = 3/4
g(-2) = 3
f(π) = -2
just a lookup once you see what is happening