The graphs of the equations y - 2x = 3 and y = x^2 intersect at two points in the xy-plane… — SAT

The graphs of the equations $y - 2x = 3$ and $y = x^2$ intersect at two points in the $xy$ -plane. If $(x, y)$ represents an intersection point with a positive $x$ -coordinate, what is the value of $y$ ?

A
1
B
3
C
6
9Answer

Answer

Substituting

y = x^2

into the first equation yields

x^2 - 2x = 3

. Setting the quadratic equation to zero gives

x^2 - 2x - 3 = 0

, which factors as

(x - 3)(x + 1) = 0

. This gives the solutions

x = 3

and

x = -1

. Because the question specifies a positive

x

-coordinate,

x

must be 3. Substituting

x = 3

back into

y = x^2

yields

y = 9

. Thus, the correct answer is 9.

Step-by-Step Solution

Substitute

y = x^2

into the equation

y - 2x = 3

x^2 - 2x = 3

This eliminates the variable

y

and creates a single quadratic equation in terms of

x

to find the

x

-coordinates of the intersection points.

Rewrite the quadratic equation in standard form.

x^2 - 2x - 3 = 0

Subtracting 3 from both sides sets the quadratic equation to 0, which is necessary for factoring.

Factor the quadratic equation to find its solutions.

(x - 3)(x + 1) = 0

, which gives

x = 3

x = -1

Factoring allows us to find the roots, which represent the

x

-coordinates of the intersection points.

Apply the constraint that the

x

-coordinate must be positive.

x = 3

The question specifies that

x > 0

, so we discard

x = -1

Substitute

x = 3

back into

y = x^2

to find the corresponding value of

y

y = 3^2 = 9

This determines the

y

-coordinate of the intersection point with the positive

x

-coordinate.

Key Concept

Solving a system of a linear equation and a quadratic equation by substitution.

Question

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