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Hitman42 [59]
2 years ago
11

Solve the system of equations using substitution and Elimination

Mathematics
1 answer:
juin [17]2 years ago
5 0
<h2>Greetings!</h2>

Answer:

y = \frac{-18}{7} and x = \frac{50}{7}

Step-by-step explanation:

To solve simultaneous equations, you need to have the number in front of both x's or y's the same. (signs doesn't matter)

To get -x to -10x we simply  need to multiply the first equation by 10:

-x * 10 = -10x

-9y * 10 = -90y

16 * 10 = 160

-10x - 90y = 160

Now we can add the two equations:

-10x + 10x = 0

-90y + 20y = -70y

160 + 20 = 180

-70y = 180

70y = -180

7y = -18

y = \frac{-18}{7}

Now plug \frac{-18}{7} into the second equation:

10x + 20(\frac{-18}{7}) = 20

10x - \frac{360}{7} = 20

Move the \frac{360}{7} over to the other side, making it a positive:

10x = 20 + \frac{360}{7}

10x = \frac{500}{7}

Divide both sides by 10:

x = \frac{50}{7}

So y = \frac{-18}{7} and x = \frac{50}{7}


<h2>Hope this helps!</h2>
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The statement "The domain of (fg)(x) consists of the numbers x that are in the domains of both f and g" is FALSE.

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THe given statement is "The domain of (fg)(x) consists of the numbers x that are in the domains of both f and g".

Now we assume the g(x)=x+2 and f(x)=\frac{1}{x-6}

So here since g(x) is a polynomial function so it exists for all real x.

f(x)=\frac{1}{x-6}<em>  </em>does not exists when x=6, so the domain of f(x) is given by all real x except 6.

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(fg)(x)=f(g(x))=f(x+2)=\frac{1}{(x+2)-6}=\frac{1}{x-4}

So now (fg)(x) does not exists when x=4, the domain of (fg)(x) consists of all real value of x except 4.

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Both equations have the same solutions.

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Please consider the complete question:

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