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r-ruslan [8.4K]
3 years ago
5

Solve the system of equations. ​ −5y−10x=45 −3y+10x=−5 ​

Mathematics
2 answers:
Fynjy0 [20]3 years ago
4 0
(-2,-5)
The first goal with equations like these is to get one of the variables to go away. You can do that by adding/subtracting one equation from the other. Then you solve for the one variable. Then you plug that answer into one of the original equations and solve.

postnew [5]3 years ago
3 0

Answer:

X=-2

Y=-5

Step-by-step explanation:

-5y-10x=45 Adding the equations

-3y+10x=-5

= -8y=40

-8y=40

-8y/-8=40/-8

<em><u>Y</u></em><em><u>=</u></em><em><u>-5</u></em>

-5y-10x=45

-5(-5)-10x=45......y is equal to-5

25-10×=45

-10x=45-25

-10x=20

-10x/-10=20/-10

<em><u>X</u></em><em><u>=</u></em><em><u>-2</u></em>

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2 years ago
(Y+3)(y^2-3y+9)<br><br> A. Y^3+27<br> B. Y^3-27<br> C. Y^3-6y^2+27<br> D. Y^3+6y^2+27
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Answer:

A) y^3+27

Step-by-step explanation:

There are two ways of solving this problem:

1. Recognizing this as the factored form of the sum of perfect cubes

2. Distribute and add the like terms.

1. In order to distribute we must multiply y by y^2-3y+9, and then 3 by y^2-3y+9:

(y+3)(y^2-3y+9)=y(y^2-3y+9)+3(y^2-3y+9)

y(y^2-3y+9)+3(y^2-3y+9)=y^3-3y^2+9y+3y^2-9y+27

After we add the positive and negative 3y^2 and 9y, they will cancel out and be gone entirely:

y^3-3y^2+9y+3y^2-9y+27=y^3+27

2. You know how you can factor the difference of perfect squares?

As an example:

a^2-b^2=(a+b)(a-b)

Well, not many people know this but you can actually factor both the sum and difference of perfect cubes:

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Because we have these identities, we can easily establish here that we have the sum of perfect cubes, and that (y+3)(y^2-3y+9)= y^3+3^3 = y^3+27

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Answer:

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