Consider the system of equations below: \begin{aligned} y &= 3x^2 - 5x - 4 \\ y &= x^2 - 2… — SAT

Consider the system of equations below:

\begin{aligned} y &= 3x^2 - 5x - 4 \\ y &= x^2 - 2x + 5 \end{aligned}

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

Cevap: 8

The value of

y

is 8.

By setting the two equations equal to each other, we obtain

3x^2 - 5x - 4 = x^2 - 2x + 5

. Simplifying this equation by moving all terms to one side yields

2x^2 - 3x - 9 = 0

. Factoring this quadratic equation gives

(2x + 3)(x - 3) = 0

, which has solutions

x = -1.5

and

x = 3

. Since the problem specifies that

x > 0

, we must use

x = 3

. Substituting

x = 3

into the second equation, we find

y = (3)^2 - 2(3) + 5 = 8

. Substituting into the first equation also yields

y = 3(3)^2 - 5(3) - 4 = 8

. Therefore, the value of

y

is 8.

Set the quadratic expressions equal to each other.

3x^2 - 5x - 4 = x^2 - 2x + 5

Since both equations are solved for

y

, their right-hand sides must be equal at any point of intersection.

Rearrange the terms to set the quadratic equation to zero.

2x^2 - 3x - 9 = 0

Putting the equation in standard form

ax^2 + bx + c = 0

allows us to solve it by factoring.

Factor the quadratic expression to find the roots.

(2x + 3)(x - 3) = 0

, which yields

x = -1.5

x = 3

Factoring shows the values of

x

that satisfy the system.

Select the positive root and substitute it back to find

y

y = 8

The problem specifies

x > 0

, so we use

x = 3

. Substituting

x = 3

into

y = x^2 - 2x + 5

gives the corresponding

y

-value.

Solving a system of nonlinear equations by setting the equations equal to each other and solving the resulting quadratic equation.

Soru