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Ostrovityanka [42]
3 years ago
15

Solve for x & y: x^2+y^2=20. 3x+y=2

Mathematics
2 answers:
Norma-Jean [14]3 years ago
6 0
You're solving a system here, easy peasy. Solve the second equation for y to get y = 2 - 3x. Now in the first equation, sub in this y value where you see a y and now the equation is in terms of x only: x^2 + (2-3x)^2 = 20. You have to FOIL out the (2-3x)^2 first to get x^2 + 4 - 12x + 9x^2 = 20. Now combine like terms to get 10x^2 - 12x + 4 = 20. Move the 20 over by subtraction to get
10x^2 - 12x - 16 = 0. Factor this now, taking a 2 out first: 2(5x^2 - 6x -8)=0 and you get factors of 2(x-2)(5x+4)=0. x=2 or x = -4/5. If x = 2, then y = -4. If x = -4/5, then y = 22/5
Lerok [7]3 years ago
5 0

Answer:

1

Step-by-step explanation:


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X=16-4y. systems of equations 3x+4y=8<br> substitution
alina1380 [7]

Answer:

x=-4  y=5

Step-by-step explanation:

Rewrite equations:

x=−4y+16;3x+4y=8

Step: Solvex=−4y+16for x:

x=−4y+16

Step: Substitute−4y+16forxin3x+4y=8:

3x+4y=8

3(−4y+16)+4y=8

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−8y+48+−48=8+−48(Add -48 to both sides)

−8y=−40

−8y

−8

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−40

−8

(Divide both sides by -8)

y=5

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x=−4y+16

x=(−4)(5)+16

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Answer:

x=−4 and y=5

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