Let's remember orders of operations as we go through the craze of this problem.
We need to multiply or divide from left to right.
3w-6w+3w×7×7×6×5÷9+2
3w-6w+21w×7×6×5÷9+2
3w-6w+147w×6×5÷9+2
3w-6w+882w×5÷9+2
3w-6w+4,410w÷9+2
3w-6w+490w+2
-3w+490w+2
487w+2
To check that we simplified the expression right, I like to put in 1 for w and test it in the original expression and then the simplified expression (not needed).
487+2=489
3-6+3×7×7×6×5÷9+2=489
So that expression equals 487w+2.
Answer:
that would be 4/13
Step-by-step explanation:
i think
Solve for X and y using both equations the point you get will be the point they cross eachother
-x-y=1
y=x+3
-x-(x+3)=1
-x-x-3=1
-2x-3=1
+3 both sides
-2x=4
÷-2 both sides
x=-2
solve for y
y=-2+3
y=1
(-2,1)
![\bf ~\hspace{10em}\textit{function transformations} \\\\\\ \begin{array}{llll} f(x)= A( Bx+ C)^2+ D \\\\ f(x)= A\sqrt{ Bx+ C}+ D \\\\ f(x)= A(\mathbb{R})^{ Bx+ C}+ D \end{array}\qquad \qquad \begin{array}{llll} f(x)=\cfrac{1}{A(Bx+C)}+D \\\\\\ f(x)= A sin\left( B x+ C \right)+ D \end{array} \\\\[-0.35em] ~\dotfill\\\\ \bullet \textit{ stretches or shrinks horizontally by } A\cdot B\\\\ \bullet \textit{ flips it upside-down if } A\textit{ is negative}\\ ~~~~~~\textit{reflection over the x-axis}](https://tex.z-dn.net/?f=%5Cbf%20~%5Chspace%7B10em%7D%5Ctextit%7Bfunction%20transformations%7D%20%5C%5C%5C%5C%5C%5C%20%5Cbegin%7Barray%7D%7Bllll%7D%20f%28x%29%3D%20A%28%20Bx%2B%20C%29%5E2%2B%20D%20%5C%5C%5C%5C%20f%28x%29%3D%20A%5Csqrt%7B%20Bx%2B%20C%7D%2B%20D%20%5C%5C%5C%5C%20f%28x%29%3D%20A%28%5Cmathbb%7BR%7D%29%5E%7B%20Bx%2B%20C%7D%2B%20D%20%5Cend%7Barray%7D%5Cqquad%20%5Cqquad%20%5Cbegin%7Barray%7D%7Bllll%7D%20f%28x%29%3D%5Ccfrac%7B1%7D%7BA%28Bx%2BC%29%7D%2BD%20%5C%5C%5C%5C%5C%5C%20f%28x%29%3D%20A%20sin%5Cleft%28%20B%20x%2B%20C%20%5Cright%29%2B%20D%20%5Cend%7Barray%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cbullet%20%5Ctextit%7B%20stretches%20or%20shrinks%20horizontally%20by%20%7D%20A%5Ccdot%20B%5C%5C%5C%5C%20%5Cbullet%20%5Ctextit%7B%20flips%20it%20upside-down%20if%20%7D%20A%5Ctextit%7B%20is%20negative%7D%5C%5C%20~~~~~~%5Ctextit%7Breflection%20over%20the%20x-axis%7D)
with that template in mind, let's see
down by 5 units, D = -5
to the left by 4 units, C = +4
Answer:
The solution of the inequation 
is ![\left(-\infty,-\frac{2}{3}\right]\cup\left[\frac{5}{2},+\infty\right)](https://tex.z-dn.net/?f=%5Cleft%28-%5Cinfty%2C-%5Cfrac%7B2%7D%7B3%7D%5Cright%5D%5Ccup%5Cleft%5B%5Cfrac%7B5%7D%7B2%7D%2C%2B%5Cinfty%5Cright%29)
.
Step-by-step explanation:
First of all, let simplify and factorize the resulting polynomial:
Roots are found by Quadratic Formula:
![r_{1,2} = \frac{\left[-\left(-\frac{11}{6}\right)\pm \sqrt{\left(-\frac{11}{6} \right)^{2}-4\cdot (1)\cdot \left(-\frac{10}{6} \right)} \right]}{2\cdot (1)}](https://tex.z-dn.net/?f=r_%7B1%2C2%7D%20%3D%20%5Cfrac%7B%5Cleft%5B-%5Cleft%28-%5Cfrac%7B11%7D%7B6%7D%5Cright%29%5Cpm%20%5Csqrt%7B%5Cleft%28-%5Cfrac%7B11%7D%7B6%7D%20%5Cright%29%5E%7B2%7D-4%5Ccdot%20%281%29%5Ccdot%20%5Cleft%28-%5Cfrac%7B10%7D%7B6%7D%20%5Cright%29%7D%20%5Cright%5D%7D%7B2%5Ccdot%20%281%29%7D)
and 
Then, the factorized form of the inequation is:
By Real Algebra, there are two condition that fulfill the inequation:
a) 
b) 
The solution of the inequation is .
