Question

Which of the following uses the substitution method to solve linear-quadratic system: y = x^2 - 4x + 8, y = 3x - 4

Answer 1

To solve the linear-quadratic system using the substitution method, we will substitute one equation into the other. The given equations are:

1. \( y = x^2 - 4x + 8 \) (quadratic equation)
2. \( y = 3x - 4 \) (linear equation)

We can substitute the expression for \( y \) from the linear equation into the quadratic equation. This gives us:

\[
3x - 4 = x^2 - 4x + 8
\]

Next, we'll rearrange the equation to one side:

\[
0 = x^2 - 4x - 3x + 8 + 4
\]

Combining like terms results in:

\[
0 = x^2 - 7x + 12
\]

Now, we can rewrite this as:

\[
x^2 - 7x + 12 = 0
\]

Next, we can factor the quadratic equation:

\[
(x - 3)(x - 4) = 0
\]

Setting each factor to zero gives us the solutions for \( x \):

1. \( x - 3 = 0 \) ⟹ \( x = 3 \)
2. \( x - 4 = 0 \) ⟹ \( x = 4 \)

Now we can find the corresponding \( y \) values by substituting these \( x \) values back into the linear equation \( y = 3x - 4 \).

For \( x = 3 \):

\[
y = 3(3) - 4 = 9 - 4 = 5
\]

For \( x = 4 \):

\[
y = 3(4) - 4 = 12 - 4 = 8
\]

Thus, the solutions to the system of equations are:

1. \( (3, 5) \)
2. \( (4, 8) \)

So the solution set of the system is \( (3, 5) \) and \( (4, 8) \).