Solve Systems by Substitution
x+y=-1
3x-2y=2
and
x=y-2
3x-4y=-13

Answers

Answer 1
Answer: Problem one: 
1.) Isolate one variable of one equation
x+y-x=-1-x
y=-1-x
2.) Plug in equation for isolated value into other original equation
3x-2(-1-x)=2
3.) Distribute
3x+2+2x=2
4.) Add like terms
5x+2=2
5.) Subtract two from both side
5x=0
6.) Divide both sides by 5
x=0
7.) Plug in numerical value for x into one equation (original or not)
0+y=1
8.) Subtract zero from both sides
y=1
For the first problem, the answer is: x=0, y = 1

For the second problem, follow very similar steps:
3(y-2)-4y=-13
3y-6-4y=-13
-y-6=-13
-y=-7
y=7

x=y-2
x=7-2
x=5
That's your answer for the second problem: x=5, y=7

Hope this helps!
Answer 2
Answer: 1x + 1y = -1 ⇒ 3x - 3y = -3
3x -  2y = 2  ⇒ 3x - 2y = 2
                                -y = -5
                                  y = 5
                       3x - 2(5) = 2
                         3x - 10 = 2
                              +10  +10
                                3x = 12
                                 3      3
                                 x = 4
                          (x , y) = (4, 5)

1x - 1y = -2   ⇒ 3x - 3y = -6
3x - 4y = -13 ⇒ 3x - 4y = -13
                                   y = 7
                        3x - 4(7) = -13
                         3x - 28 = -13
                              +28    +28
                                3x = 15
                                 3      3
                                  x = 5
                            (x, y) = (5, 7)

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Alice solved this division problem.8/4287= 535 7/8

Which expression could be used to check her work?

Answers

8.7 is the answer hope this helps 
im not sure,...but she could do 4287 divided by 8 then u could do 7 divided by 8 then multiply...IDK

PLZ HELP FASTsimplify

1+[2x-3 (1.8divided by0.2)]

X is not multiply

A: 2x-26
B:2x+28
C:3x-27
D:-7x+1

Answers

The answer is 2x-26
1+2x-27  9*3= 27
1-27+2x   You would combine like terms here
2x-26
Work is attached... :D

If you have doubts...check out the questions I've answered... :P

Plz help #30 plz explain
\sqrt{200 {x}^2 {y}^2 {z}^2

Answers

For number 30, its easier if its broken down so were going to make it √200*x^2*y^2*z^2. We can't get a square root of 200 so we will break it into 100*2 to get √2*100*x^2*y^2*z^2. Now let's look at what we can take the square roots of. 2 doesn't have a nice sq. rt. so we will leave it there. 100 has the rt. of 10 so we can make 10√2*x^2*y^2*z^2. X^2 has the root x and we can follow this with the other variables to get 10xyz√2. There is your final simplified answer.

Roots have a property which allows for separation of terms based on wether they are being multiplies:
100*2*x^2...
10xyz * \/2

What are two inequalities to compare -17 and -22

Answers

The two inequalities are

-17 > -22

- 22 < -17

What is inequalities ?

In mathematics, an inequality is a relation which makes a non-equal comparison between two numbers or other mathematical expressions. It is used most often to compare two numbers on the number line by their size.

Comparing the two numbers, if plotted on a graph, we can see that -17 is much nearer to the point zero and -22 is farther from point zero. Therefore, we can say that -17 is greater than -22 or -22 is less than -17.

This can be represented with the help of two inequalities, greater than and less than.

first, -17 > -22   (greater than)

second, - 22 < -17 (less than)

Read more about inequalities at:

brainly.com/question/20383699

#SPJ2

the inequalities to compare the numbers is that they are all negative numbers

What is 423.408 rounded to the neatest hundredth

Answers

The hundredth is the second decimal place. You know the rounding rules are where you look at the previous place and if it's 5 or above the number place your rounding to goes up. So the answer is 423.41
423.408 = 423.41
0.08 is nearer to 0.1 than 0.001

What is 12 and 5 tenths in decimal form

Answers

to do that is by the number 12 and  is 12. and the 5 tenth is . 5 then put them together equals 12.5
that would be 12.5... hope i helped:)