Solve (x+y)^2=5 | Microsoft Math Solver

Solve for x

x = - y + 5

x = - (y + 5)

Steps by Finding Square Root

(x + y)^{2} = 5

Take the square root of both sides of the equation.

x + y = 5

x + y = - 5

Subtract

y

from both sides of the equation.

x + y - y = 5 - y

x + y - y = - 5 - y

Subtracting

y

from itself leaves

0

x = 5 - y

x = - 5 - y

Subtract

y

from

5

x = - y + 5

Subtract

y

from

- 5

x = - y - 5

The equation is now solved.

x = - y + 5

x = - y - 5

Solve for y

y = - x + 5

y = - (x + 5)

Steps by Finding Square Root

(x + y)^{2} = 5

Take the square root of both sides of the equation.

y + x = 5

y + x = - 5

Subtract

x

from both sides of the equation.

y + x - x = 5 - x

y + x - x = - 5 - x

Subtracting

x

from itself leaves

0

y = 5 - x

y = - 5 - x

Subtract

x

from

5

y = - x + 5

Subtract

x

from

- 5

y = - x - 5

The equation is now solved.

y = - x + 5

y = - x - 5

Graph

Quiz

Algebra

5 problems similar to:

(x + y)^{2} = 5

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How do you convert

2 x + y^{2} = 5

to polar form?

https://socratic.org/questions/how-do-you-convert-2x-y-2-5-to-polar-form

Please see the explanation for steps leading to the polar form: Explanation: Substitute

(r cos (θ))

for x and

r^{2} sin^{2} (θ)

...

p (x + y)^{2} = 5

and

q (x - y)^{2} = 3

, then the simplified value of

p^{2} (x + y)^{2} + 4 p q x y - q^{2} (x - y)^{2}

is? [closed]

https://math.stackexchange.com/questions/2353419/if-pxy2-5-and-q-x-y2-3-then-the-simplified-value-of-p2xy24pqx

(x + y)^{2} = 5 / p

(x - y)^{2} = 3 / q

⟹ 4 x y = 5 / p - 3 / q

How do you determine whether the graph of

x + y^{2} = 1

is symmetric with respect to the x axis, y axis, the line y=x or y=-x, or none of these?

https://socratic.org/questions/how-do-you-determine-whether-the-graph-of-x-y-2-1-is-symmetric-with-respect-to-t

x + y^{2} = 1

is symmetric wrt the

x -

axis Explanation: Let's look at the graph of

x + y^{2} = 1

in relation to the

x -

...

Find all ordered pairs such that

x + y^{2} = 2

and

y + x^{2} = 2

https://math.stackexchange.com/q/2593833

If we leave

y

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y = 2 - x^{2}

. Then putting it in the first equation, we get

x + (2 - x^{2}) = 2 ⟹ x^{4} - 4 x^{2} + x + 2 = 0

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x = 1

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How do you differentiate

x + y^{2} = 25

https://socratic.org/questions/how-do-you-differentiate-x-y-2-25

- \frac{1}{2 y}

Explanation:

(\frac{d}{d x}) (x + y^{2}) = (\frac{d}{d x}) 25

...

Continuity of a

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when

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when

y < 0

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https://math.stackexchange.com/questions/2297697/continuity-of-a-fx-y-x2-when-y-geq-0-x2-when-y-0

Your function is continuous when

y \neq = 0

since

(x, y) \mapsto - x^{2}

and

(x, y) \mapsto x^{2}

are both continuous everywhere. Now you can look at

lim_{y \to 0^{+}} f (x, y)

and

lim_{y \to 0^{-}} f (x, y) .

...

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x+y=\sqrt{5} x+y=-\sqrt{5}

Take the square root of both sides of the equation.

x+y-y=\sqrt{5}-y x+y-y=-\sqrt{5}-y

Subtract y from both sides of the equation.

x=\sqrt{5}-y x=-\sqrt{5}-y

Subtracting y from itself leaves 0.

x=-y+\sqrt{5}

Subtract y from \sqrt{5}.

x=-y-\sqrt{5}

Subtract y from -\sqrt{5}.

x=-y+\sqrt{5} x=-y-\sqrt{5}

The equation is now solved.

y+x=\sqrt{5} y+x=-\sqrt{5}

Take the square root of both sides of the equation.

y+x-x=\sqrt{5}-x y+x-x=-\sqrt{5}-x

Subtract x from both sides of the equation.

y=\sqrt{5}-x y=-\sqrt{5}-x

Subtracting x from itself leaves 0.

y=-x+\sqrt{5}

Subtract x from \sqrt{5}.

y=-x-\sqrt{5}

Subtract x from -\sqrt{5}.

y=-x+\sqrt{5} y=-x-\sqrt{5}

The equation is now solved.

Examples

Quadratic equation

x^{2} - 4 x - 5 = 0

Trigonometry

4 sin θ cos θ = 2 sin θ

Linear equation

y = 3 x + 4

Arithmetic

699 * 533

Matrix

[2534] [2 - 1 0135]

Simultaneous equation

{8 x + 2 y = 46 7 x + 3 y = 47

Differentiation

\frac{d}{d x} \frac{( 3 x ^{2} - 2 )}{( x - 5 )}

Integration

\int_{0}^{1} x e^{- x^{2}} d x

Limits

x \to - 3 lim \frac{x ^{2} - 9}{x ^{2} + 2 x - 3}

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