# How do you solve the system #(x+y)^2+3(x-y) = 30# and #xy+3(x-y) = 11# ?

#(1+sqrt(6), -1+sqrt(6))#

#(1-sqrt(6), -1-sqrt(6))#

#(5, -2)#

#(2, -5)#

Take note of this:

Using the first provided equation, we obtain:

Thus:

Thus:

Thus

Thus, remedies:

Using the first provided equation, we obtain:

Thus:

Thus:

or

Thus, the solutions:

graph{(xy+3(x-y)-11) = 0 [-7.22, 12.78, -6.36, 3.64]}

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To solve the system of equations:

- Expand the first equation: ( (x+y)^2 + 3(x-y) = 30 ) becomes ( x^2 + 2xy + y^2 + 3x - 3y = 30 ).
- Expand the second equation: ( xy + 3(x-y) = 11 ) becomes ( xy + 3x - 3y = 11 ).
- Rearrange both equations to isolate terms: ( x^2 + 2xy + y^2 + 3x - 3y - 30 = 0 ) and ( xy + 3x - 3y - 11 = 0 ).
- Use substitution or elimination method to solve for one variable in terms of the other.
- Once one variable is solved for, substitute its value back into one of the original equations to solve for the other variable.
- Once both variables are found, verify the solutions by substituting them back into both original equations to ensure they satisfy both equations.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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