Question

section 3.5 homework

1. solve the system by elimination.

$x^2 - 7x + 11 = y - 1$
$-x + y = -4$
a. $\square$ $(0, 7)$
b. $\square$ $(2, 3)$
c. $\square$ $(3, 1)$
d. $\square$ $(4, 0)$
e. $\square$ $(5, 3)$
f. $\square$ no real solution

Explanation:

Step1: Rewrite the system of equations

We have the system:

$$
LATEXBLOCK0
$$

From the second equation, we can express \(y\) in terms of \(x\): \(y=x - 4\).

Step2: Substitute \(y=x - 4\) into the first equation

Substitute \(y\) into \(x^{2}-7x + 11=y - 1\):

$$
LATEXBLOCK1
$$

Step3: Solve the quadratic equation

The quadratic equation \(x^{2}-8x + 16 = 0\) is a perfect square trinomial, which can be written as \((x - 4)^{2}=0\).
So, \(x=4\) (with multiplicity 2).

Step4: Find the value of \(y\)

Substitute \(x = 4\) into \(y=x - 4\), we get \(y=4-4 = 0\).

So the solution of the system is \((4,0)\).

Answer:

d. (4, 0)