Question

1. identify each system of equations that has no solution. *

1 point

i. \\( 4x + 4y = -8 \\)
\\( 2x + 2y = -4 \\)
ii. \\( x - y = -1 \\)
\\( x - y = 1 \\)
iii. \\( 2x + y = -4 \\)
\\( 4x - 2y = 8 \\)
iv. \\( x + y = 8 \\)
\\( x + y = -1 \\)
v. \\( 15x - 5y = -20 \\)
\\( 3x - y = -4 \\)
vi. \\( 7x - 7y = -7 \\)
\\( 2x - 2y = -18 \\)

\\( \square \\) i.
\\( \square \\) ii.
\\( \square \\) iii.
\\( \square \\) iv.
\\( \square \\) v.
\\( \square \\) vi.

Explanation:

To determine which systems have no solution, we analyze each system by checking if they are inconsistent (parallel lines, same slope but different y - intercepts) or dependent (same line, infinitely many solutions). A system \(a_1x + b_1y=c_1\) and \(a_2x + b_2y = c_2\) has no solution if \(\frac{a_1}{a_2}=\frac{b_1}{b_2}
eq\frac{c_1}{c_2}\).

System I:

Equations: \(4x + 4y=-8\) and \(2x + 2y=-4\)
Simplify the first equation by dividing by 4: \(x + y=-2\)
Simplify the second equation by dividing by 2: \(x + y=-2\)
Since both equations are the same, this system has infinitely many solutions, not no solution.

System II:

Equations: \(x - y=-1\) and \(x - y = 1\)
Here, \(a_1 = 1,b_1=-1,c_1=-1\) and \(a_2 = 1,b_2=-1,c_2 = 1\)
\(\frac{a_1}{a_2}=\frac{1}{1}=1\), \(\frac{b_1}{b_2}=\frac{-1}{-1}=1\), \(\frac{c_1}{c_2}=\frac{-1}{1}=-1\)
Since \(\frac{a_1}{a_2}=\frac{b_1}{b_2}
eq\frac{c_1}{c_2}\), this system has no solution.

System III:

Equations: \(2x + y=-4\) and \(4x-2y = 8\)
The slopes of the lines: For \(2x + y=-4\), \(y=-2x - 4\), slope \(m_1=-2\)
For \(4x-2y = 8\), we can rewrite it as \(y = 2x-4\), slope \(m_2 = 2\)
Since the slopes are different, the lines intersect, so the system has a unique solution.

System IV:

Equations: \(x + y=8\) and \(x + y=-1\)
Here, \(a_1 = 1,b_1 = 1,c_1=8\) and \(a_2 = 1,b_2 = 1,c_2=-1\)
\(\frac{a_1}{a_2}=\frac{1}{1}=1\), \(\frac{b_1}{b_2}=\frac{1}{1}=1\), \(\frac{c_1}{c_2}=\frac{8}{-1}=-8\)
Since \(\frac{a_1}{a_2}=\frac{b_1}{b_2}
eq\frac{c_1}{c_2}\), this system has no solution.

System V:

Equations: \(15x-5y=-20\) and \(3x - y=-4\)
Simplify the first equation by dividing by 5: \(3x - y=-4\)
The second equation is \(3x - y=-4\)
Both equations are the same, so the system has infinitely many solutions.

System VI:

Equations: \(7x-7y=-7\) and \(2x-2y=-18\)
Simplify the first equation by dividing by 7: \(x - y=-1\) or \(y=x + 1\)
Simplify the second equation by dividing by 2: \(x - y=-9\) or \(y=x + 9\)
Here, \(a_1 = 1,b_1=-1,c_1=-1\) and \(a_2 = 1,b_2=-1,c_2=-9\)
\(\frac{a_1}{a_2}=\frac{1}{1}=1\), \(\frac{b_1}{b_2}=\frac{-1}{-1}=1\), \(\frac{c_1}{c_2}=\frac{-1}{-9}=\frac{1}{9}\)
Since \(\frac{a_1}{a_2}=\frac{b_1}{b_2}
eq\frac{c_1}{c_2}\), this system has no solution.

Answer:

II. \(x - y=-1\); \(x - y = 1\), IV. \(x + y=8\); \(x + y=-1\), VI. \(7x-7y=-7\); \(2x - 2y=-18\)