Consider the system of equations below: \begin{aligned} x^2 + y^2 &= 9 \\ y &= x - 3 \end{… — ACT

Consider the system of equations below:

\begin{aligned} x^2 + y^2 &= 9 \\ y &= x - 3 \end{aligned}

If $(x, y)$ is a solution to the system such that $x > 0$ , what is the value of $x + y$ ?

The correct answer is

3

. Substituting

y = x - 3

into

x^2 + y^2 = 9

gives

x^2 + (x - 3)^2 = 9

. Expanding this yields

2x^2 - 6x = 0

, which factors as

2x(x - 3) = 0

. Since

x > 0

, the

x

-coordinate must be

3

. Substituting

x = 3

back into the linear equation gives

y = 3 - 3 = 0

. Therefore,

x + y = 3 + 0 = 3

Substitute the linear equation into the quadratic equation.

x^2 + (x - 3)^2 = 9

This reduces the system of two variables to a single quadratic equation in terms of

x

Expand the squared binomial and simplify the equation.

x^2 + (x^2 - 6x + 9) = 9 \implies 2x^2 - 6x = 0

Expanding

(x - 3)^2

yields

x^2 - 6x + 9

. Subtracting

9

from both sides simplifies the equation.

Factor the quadratic equation to solve for

x

2x(x - 3) = 0 \implies x = 0

x = 3

Factoring out the greatest common factor

2x

reveals the roots of the quadratic equation.

Apply the constraint

x > 0

to find the correct

x

-coordinate.

x = 3

The problem specifies that the solution

(x, y)

must have

x > 0

Find the corresponding

y

-coordinate and calculate

x + y

y = 3 - 3 = 0 \implies x + y = 3 + 0 = 3

Substituting

x = 3

into the linear equation gives

y = 0

, and summing them gives the required value.

Solving systems of linear and quadratic equations by substitution

Question