\begin{aligned} y &= x^2 - 7 \\ y &= 2x + 1 \end{aligned} If (x, y) is a solution to the s… — SAT

\begin{aligned} y &= x^2 - 7 \\ y &= 2x + 1 \end{aligned}

If $(x, y)$ is a solution to the system of equations above and $y > 0$ , what is the value of $y$ ?

A
4
B
5
C
-3
9Answer

Answer

Substituting the expression for

y

from the second equation into the first equation yields

2x + 1 = x^2 - 7

. Subtracting

2x

and

1

from both sides results in the standard form quadratic equation

x^2 - 2x - 8 = 0

. Factoring this equation gives

(x - 4)(x + 2) = 0

, which yields the solutions

x = 4

and

x = -2

. Substituting these

x

-values back into the linear equation

y = 2x + 1

gives the corresponding

y

-values:

y = 2(4) + 1 = 9

and

y = 2(-2) + 1 = -3

. Since the system specifies the constraint

y > 0

, the correct value of

y

must be

9

Step-by-Step Solution

Substitute the expression for

y

from the linear equation into the quadratic equation.

2x + 1 = x^2 - 7

This eliminates the variable

y

and leaves a single quadratic equation in terms of

x

Rearrange the quadratic equation into standard form

ax^2 + bx + c = 0

x^2 - 2x - 8 = 0

Moving all terms to one side allows us to solve the quadratic equation by factoring.

Factor the quadratic equation.

(x - 4)(x + 2) = 0

, which gives

x = 4

x = -2

Finding the roots of the quadratic equation provides the possible

x

-coordinates of the solution points.

Substitute the

x

-values back into the linear equation

y = 2x + 1

to find the corresponding

y

-values.

For

x = 4

y = 2(4) + 1 = 9

. For

x = -2

y = 2(-2) + 1 = -3

This gives the complete coordinate pairs

(4, 9)

and

(-2, -3)

for the system's solutions.

Apply the constraint

y > 0

to identify the correct solution.

Since

9 > 0

and

-3 < 0

, the correct value of

y

9

Only the solution

(4, 9)

satisfies the condition that

y

must be positive.

Key Concept

Solving linear-quadratic systems of equations by substitution and evaluating solutions under given constraints.

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