Jesset C.

asked • 03/11/20# Algebra Problem! Please Help Me! I Need This Answered Today!

Solve the system:

x^{2 }+ y^{2 }= 37

3x - 9 = y

There are two solutions: (x_{1}, y_{1}) (x_{2}, y_{2}).

Evaluate: x_{1}+y_{1}+x_{2}+y_{2}=

(A) 0

(B) (-21+2/3)

(C) (3+3/5)

(D) (7+3/7)

(E) None of the above

For this I got; **(C) (3+3/5)**. Is this the correct answer?

## 2 Answers By Expert Tutors

Yes, you are correct.

You can also solve this the direct way by substituting y = 3x - 9 into the first equation, the result is a quadratic equation in x which factors cleanly. Then just take the two possible values for x, find the corresponding y's, and add the results.

Hello Jesset!

From this equation

3x - 9 = y

(3x - 9)^{2} = y^{2}

Replacing y^{2} in x^{2 }+ y^{2 }= 37

(3x - 9)^{2} + x^{2 }= 37

9x^{2 }-54x + 81 + x^{2} = 37

10x^{2} -54x +44 = 0^{ }

From this equation, we have, x1 + x2 = 54/10 = 27/5 (the sum of the roots)

3x1 - 9 = y1

3x2 - 9 = y2 (summing these equations)

3*(x1 + x2) -18 = y1 + y2

Replacing (x1 + x2) by 27/5 :

3*(27/5) -18 = y1 + y2

81/5 - 90/5 = y1 + y2

y1 + y2 = -9/5

x1 + x2 + y1 + y2 = (x1 + x2) + (y1 + y2) = 27/5 - 9/5 = 18/5 = 3 + 3/5

**Letter C, you are correct!**

Feel free to send me any doubts related to Algebra, ok? Best regards!

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Dayna T.

03/11/20