We will have a system of equations, and we are going to solve it by using substitution.
Let's make the equations.
x+y+z=2100
z+y=x
Since y and z are equal, I think it is easier for us to have them as one variable for now. Let the value of z and y be a.
These are the equations now.
x+2a=2100
2a=x
Since the second equation is giving the value of x, let it replace the value of x also in the first equation.
2a+2a=2100
4a=2100
2100÷4= 525
so, x and y equal 525. Let's put this in our original equation.
x+525+525=2100
x+ 1,050= 2,100
x= 1,050
So, x=1,050, y= 525 =z.
Answer:
the 3rd one is yes the rest are no
Step-by-step explanation:
The question is incomplete. The complete question is as follows:
Solve for X. Assume X is a 2x2 matrix and I denotes the 2x2 identity matrix. Do not use decimal numbers in your answer. If there are fractions, leave them unevaluated.
· X·
=<em>I</em>.
First, we have to identify the matrix <em>I. </em>As it was said, the matrix is the identiy matrix, which means
<em>I</em> =
So,
· X·
= ![\left[\begin{array}{ccc}1&0\\0&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%260%5C%5C0%261%5Cend%7Barray%7D%5Cright%5D)
Isolating the X, we have
X·
=
- ![\left[\begin{array}{ccc}1&0\\0&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%260%5C%5C0%261%5Cend%7Barray%7D%5Cright%5D)
Resolving:
X·
= ![\left[\begin{array}{ccc}2-1&8-0\\-6-0&-9-1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2-1%268-0%5C%5C-6-0%26-9-1%5Cend%7Barray%7D%5Cright%5D)
X·
=![\left[\begin{array}{ccc}1&8\\-6&-10\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%268%5C%5C-6%26-10%5Cend%7Barray%7D%5Cright%5D)
Now, we have a problem similar to A.X=B. To solve it and because we don't divide matrices, we do X=A⁻¹·B. In this case,
X=
⁻¹·![\left[\begin{array}{ccc}1&8\\-6&-10\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%268%5C%5C-6%26-10%5Cend%7Barray%7D%5Cright%5D)
Now, a matrix with index -1 is called Inverse Matrix and is calculated as: A . A⁻¹ = I.
So,
·
=![\left[\begin{array}{ccc}1&0\\0&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%260%5C%5C0%261%5Cend%7Barray%7D%5Cright%5D)
9a - 3b = 1
7a - 6b = 0
9c - 3d = 0
7c - 6d = 1
Resolving these equations, we have a=
; b=
; c=
and d=
. Substituting:
X=
·![\left[\begin{array}{ccc}1&8\\-6&-10\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%268%5C%5C-6%26-10%5Cend%7Barray%7D%5Cright%5D)
Multiplying the matrices, we have
X=![\left[\begin{array}{ccc}\frac{8}{11} &\frac{26}{11} \\\frac{39}{11}&\frac{198}{11} \end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%5Cfrac%7B8%7D%7B11%7D%20%26%5Cfrac%7B26%7D%7B11%7D%20%5C%5C%5Cfrac%7B39%7D%7B11%7D%26%5Cfrac%7B198%7D%7B11%7D%20%20%5Cend%7Barray%7D%5Cright%5D)
Simple...
you have:
1.) Find the slope of the line using (-2,1) and (3,6)
Using


m=1 (slope)
2.)Tell whether the equation, 8y+3=5x+3, is a direct variation.
y=

(None)
3.)What is the slope and y-intercept of the graph of the equation y=2x+3?
Slope =2
y-intercept=3
C.
Thus, your answer.
9t-6-6t=6 -- add 6 to both sides
9t-6t=12 -- combine like terms
3t=12 -- divide by 3
t=4