Solve the following system. x^2 + y^2= 25 2x + y = 10 The solution set is...?

1 answer:

iogann1982 [59]4 years ago

4 0

$x^2 + y^2= 25\\
2x+y=10\\\\
x^2 + y^2= 25\\
y=10-2x\\\\
x^2+(10-2x)^2=25\\
x^2+100-40x+4x^2=25\\
5x^2-40x+75=0\\
x^2-8x+15=0\\
x^2-3x-5x+15=0\\
x(x-3)-5(x-3)=0\\
(x-5)(x-3)=0\\
x=5 \vee x=3\\\\
y=10-2\cdot5 \vee y=10-2\cdot3\\
y=0 \vee y=4\\\\
\boxed{(x,y)=\{(3,4),(5,0)\}}

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Nina [5.8K]

Answer:

Option F

Step-by-step explanation:

F). x + y = 3 -------(1)

x - 3y = -2 --------(2)

Equation (1) minus equation (2)

(x + y) - (x - 3y) = 3 - (-2)

4y = 5

y = 1.25

Hence, y is positive.

G). x + y = 3 --------(1)

x + y = -2 --------(2)

Both the equations represent parallel lines.

There are no solutions of the given equations.

H). x - 3y = -2 -------(1)

x + y = -2 -------(2)

Equation (2) minus equation (1)

(x + y) - (x - 3y) = -2 - (-2)

4y = 0

y = 0

Since, 0 is neither positive nor negative, y will be neither positive nor negative.

J). x + y = 3 -------(1)

x - y = 3 --------(2)

Equation (1) - Equation (2)

(x + y) - (x - y) = 3 - 3

2y = 0

Hence, y is neither negative nor positive.

Therefore, Option F is the answer.

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Answer:

Step-by-step explanation:

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