For a real solution (x, y) to the system of equations below, which of the following is a p… — SAT

For a real solution $(x, y)$ to the system of equations below, which of the following is a possible value of $x + y$ ?

\begin{aligned} x^2 + 3y^2 &= 21 \\ x^2 - 2y &= 5 \end{aligned}

A
-5
B
-3
C
1
5Answer

Answer

The correct answer is 5, which is the sum of the coordinates of the real solution (3, 2) of the system.

Substituting

x^2 = 2y + 5

into the first equation yields the quadratic equation

3y^2 + 2y - 16 = 0

. Solving this quadratic equation gives

y = 2

y = -8/3

. Since

x^2

must be non-negative for real solutions,

y = -8/3

yields no real solutions because

x^2 = 2(-8/3) + 5 = -1/3 < 0

. For

y = 2

, we have

x^2 = 2(2) + 5 = 9

, which gives

x = 3

x = -3

. The real solutions to the system are

(3, 2)

and

(-3, 2)

. The possible values for

x + y

are

3 + 2 = 5

and

-3 + 2 = -1

. Therefore, 5 is a possible value of

x + y

Step-by-Step Solution

Rearrange the second equation to express

x^2

in terms of

y

x^2 = 2y + 5

This allows for direct substitution into the first equation to eliminate the

x

variable.

Substitute the expression for

x^2

into the first equation and simplify the resulting quadratic equation in terms of

y

3y^2 + 2y - 16 = 0

Substituting

2y + 5

for

x^2

x^2 + 3y^2 = 21

gives

(2y + 5) + 3y^2 = 21

, which simplifies to

3y^2 + 2y - 16 = 0

Factor or use the quadratic formula to solve for

y

y = 2

y = -8/3

The quadratic equation factors as

(3y + 8)(y - 2) = 0

, giving the roots

y = 2

and

y = -8/3

Find the corresponding values of

x

for each

y

and determine the real solutions.

For

y = 2

x^2 = 2(2) + 5 = 9 \implies x = \pm 3

. For

y = -8/3

x^2 = 2(-8/3) + 5 = -1/3

, which has no real solutions.

A real number squared cannot be negative, so only

y = 2

yields real solutions:

(3, 2)

and

(-3, 2)

Calculate the possible values of

x + y

for the real solutions.

For

(3, 2)

x + y = 3 + 2 = 5

. For

(-3, 2)

x + y = -3 + 2 = -1

Evaluating the target expression for both valid real solutions gives 5 and -1. Since 5 is the only value present in the options, it is the correct answer.

Key Concept

Solving systems of nonlinear equations algebraically using substitution and analyzing real solutions.

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