\begin{aligned} y &= x^2 - 5 \\ y &= 4x \end{aligned} If (x, y) is a solution to the syste… — SAT

\begin{aligned} y &= x^2 - 5 \\ y &= 4x \end{aligned}

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

Answer: 20

Answer

Substituting

y = 4x

into the equation

y = x^2 - 5

yields

4x = x^2 - 5

. Rearranging this equation into standard form gives

x^2 - 4x - 5 = 0

. Factoring the quadratic expression gives

(x - 5)(x + 1) = 0

, which means

x = 5

x = -1

. Substituting these values back into

y = 4x

gives the coordinates of the two solutions:

(5, 20)

and

(-1, -4)

. Since the question specifies that

y > 0

, the correct value is the positive

y

-value, which is 20.

Step-by-Step Solution

Substitute the expression for

y

from the second equation into the first equation.

4x = x^2 - 5

This eliminates

y

and creates a single equation in terms of

x

Subtract

4x

from both sides to write the quadratic equation in standard form.

x^2 - 4x - 5 = 0

Standard quadratic form

ax^2 + bx + c = 0

is required to solve by factoring.

Factor the quadratic equation.

(x - 5)(x + 1) = 0

Factoring helps find the values of

x

that satisfy the equation.

Solve for

x

by setting each factor to zero.

x = 5

x = -1

Applying the zero product property yields the possible

x

-coordinates of the solutions.

Calculate the corresponding

y

-values using the equation

y = 4x

For

x = 5

y = 20

. For

x = -1

y = -4

This determines the coordinates of the intersection points, which are

(5, 20)

and

(-1, -4)

Apply the constraint

y > 0

to identify the correct value of

y

y = 20

Since

-4

is not greater than zero, the only valid solution is

(5, 20)

, giving

y = 20

Key Concept

Solving a system of linear and quadratic equations using substitution.

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