Question 1062395
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(x+1)^2 - (y-1)^2 = 20
x^2 - (y+2)^2 = 24
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<pre>
(x+1)^2 - (y-1)^2 = 20,    (1)
X^2 - (y+2)^2     = 24.    (2)

Open parentheses:

x^2 + 2x + 1 - y^2 + 2y - 1 = 20,    (3)     
x^2          - y^2 - 4y - 4 = 24.    (4)

Distract (4) from (3) (both sides). You will get

2x + 1 + 6y + 3 = -4,   or

2x + 6y = -8,   or

x + 3y = -4.                          (5)

Express x = -3y - 4 from (5) and substitute it into (2), replacing x. You will get

(-3y - 4)^2 - (y+2)^2 = 24,   or

9y^2 + 24y + 16 - y^2 - 4y - 4 = 24,   or

8y^2 + 20y - 12 = 0,   or

2y^2 + 5y - 3 = 0.

{{{y[1,2]}}} = {{{(-5 +- sqrt(25-4*2*(-3)))/(2*2)}}} = {{{(-5 +- 7)/4}}}.

The roots are {{{y[1]}}} = {{{1/2}}},  {{{y[2]}}} = -3.

For each root {{{y[1,2]}}} find the corresponding value of x.

Can you complete it on your own?
</pre>

If you want to see more similar solved problems, look into the lesson

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/Systems-of-equations/Solving-the-system-of-algebraic-equations-of-degree-2.lesson>Solving systems of algebraic equations of degree 2</A> 

in this site.


Also, you have this free of charge online textbook in ALGEBRA-I in this site

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson>ALGEBRA-I - YOUR ONLINE TEXTBOOK</A>.


The referred lesson is the part of this online textbook under the topic "<U>Systems of equations that are not linear</U>".