Question
Solving the system yz +3 and z+y=8 by graphing shows that there is no solution to the system. Is this statement true or false? Explain.
(1 point)
The statement is false, because the lines are parallel.
The statement is false, because the lines have an intersection point.
The statement is true, because the lines have an intersection point.
The statement is true, because the lines are parallel
(1 point)
The statement is false, because the lines are parallel.
The statement is false, because the lines have an intersection point.
The statement is true, because the lines have an intersection point.
The statement is true, because the lines are parallel
Answers
GPT-4o mini
To analyze the system of equations given:
1. \( yz + 3 = 0 \) or rearranging gives \( yz = -3 \)
2. \( z + y = 8 \) which can be rearranged to \( z = 8 - y \)
Let's rewrite the first equation in terms of \( y \) and \( z \):
From \( yz = -3 \), we get \( z = -\frac{3}{y} \) (for \( y \neq 0 \)).
Now we have two expressions for \( z \):
1. \( z = -\frac{3}{y} \)
2. \( z = 8 - y \)
Setting them equal to find intersection points:
\[
-\frac{3}{y} = 8 - y
\]
Multiplying through by \( y \) (assuming \( y \neq 0 \)):
\[
-3 = (8 - y)y
\]
\[
-3 = 8y - y^2
\]
Rearranging gives us a quadratic equation:
\[
y^2 - 8y - 3 = 0
\]
To determine whether there are solutions for \( y \), we can calculate the discriminant:
\[
D = b^2 - 4ac = (-8)^2 - 4(1)(-3) = 64 + 12 = 76
\]
Since the discriminant is positive (\( D > 0 \)), the quadratic equation has two real solutions for \( y \). For each value of \( y \), we can find a corresponding value of \( z \).
Thus, the system does indeed have solutions, meaning the statement "there is no solution to the system" is false.
Therefore, the correct choice is:
**The statement is false, because the lines have an intersection point.**
1. \( yz + 3 = 0 \) or rearranging gives \( yz = -3 \)
2. \( z + y = 8 \) which can be rearranged to \( z = 8 - y \)
Let's rewrite the first equation in terms of \( y \) and \( z \):
From \( yz = -3 \), we get \( z = -\frac{3}{y} \) (for \( y \neq 0 \)).
Now we have two expressions for \( z \):
1. \( z = -\frac{3}{y} \)
2. \( z = 8 - y \)
Setting them equal to find intersection points:
\[
-\frac{3}{y} = 8 - y
\]
Multiplying through by \( y \) (assuming \( y \neq 0 \)):
\[
-3 = (8 - y)y
\]
\[
-3 = 8y - y^2
\]
Rearranging gives us a quadratic equation:
\[
y^2 - 8y - 3 = 0
\]
To determine whether there are solutions for \( y \), we can calculate the discriminant:
\[
D = b^2 - 4ac = (-8)^2 - 4(1)(-3) = 64 + 12 = 76
\]
Since the discriminant is positive (\( D > 0 \)), the quadratic equation has two real solutions for \( y \). For each value of \( y \), we can find a corresponding value of \( z \).
Thus, the system does indeed have solutions, meaning the statement "there is no solution to the system" is false.
Therefore, the correct choice is:
**The statement is false, because the lines have an intersection point.**