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3 years ago

Solve and Check using the removal method:

Mathematics

1 answer:

Luba_88 [7]3 years ago

8 0

9514 1404 393

Answer:

(x, y) = (2, 3)

Step-by-step explanation:

"Removal method" is a new one for me. We presume it is another name for the "elimination" or "addition" method.

Multiply the second equation by 5 and add the first.

(6x -5y) +5(-4x +y) = (-3) +5(-5)

-14x = -28 . . . . . simplify

x = 2 . . . . . . . . . divide by -14

-4(2) +y = -5 . . . substitute for x in the second equation

y = 3 . . . . . . . . . add 8

The solution is (x, y) = (2, 3).

<em>Check</em>

We found y using the second equation, so we know the values of x and y work in that one. We only need to check the values in the first equation.

6(2) -5(3) = -3

12 -15 = -3 . . . . . true; answer checks OK

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miss Akunina [59]

<h2>Hello!</h2>

The answer is:

Center: (1,-2)

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Where,

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The given equation is:

$x^2+y^2-2x+4y-11=0$

So, solving we have:

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$(x^2-2x+(\frac{2}{2})^{2} )+(y^2+4y+(\frac{4}{2})^{2})=11+(\frac{2}{2})^{2} +(\frac{4}{2})^{2}\\\\(x^2-2x+1)+(y^2+4y+4)=11+1+4\\\\(x^2-1)+(y^2+2)=16$

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Now, we have that:

$h=1\\k=-2\\r=\sqrt{16}=4$

So,

Center: (1,-2)

Radius: 4 units.

Have a nice day!

Note: I have attached a picture for better understanding.

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