QUESTION IMAGE
Question
which of these systems of equations have a solution of (1, -1)? select all that apply.
$y = -x$
$4x - 4y = 8$
$y = -2x + 1$
$y = 5x - 6$
$x = 1$
$y - 1 = 0$
$x - y = 0$
$x + 6y = -5$
$4x + 2y = 2$
$3x - 5y = 8$
$2x + y = -1$
$y = -3x - 2$
To determine which systems of equations have a solution of \((1, -1)\), we substitute \(x = 1\) and \(y = -1\) into each system and check if both equations are satisfied.
System 1: \(y = -x\) and \(4x - 4y = 8\)
- For \(y = -x\): Substitute \(x = 1\), \(y = -1\).
\( -1 = -1\) (True).
- For \(4x - 4y = 8\): Substitute \(x = 1\), \(y = -1\).
\(4(1) - 4(-1) = 4 + 4 = 8\) (True).
Thus, \((1, -1)\) is a solution.
System 2: \(y = -2x + 1\) and \(y = 5x - 6\)
- For \(y = -2x + 1\): Substitute \(x = 1\), \(y = -1\).
\( -1 = -2(1) + 1 = -2 + 1 = -1\) (True).
- For \(y = 5x - 6\): Substitute \(x = 1\), \(y = -1\).
\( -1 = 5(1) - 6 = 5 - 6 = -1\) (True).
Thus, \((1, -1)\) is a solution.
System 3: \(x = 1\) and \(y - 1 = 0\)
- For \(x = 1\): \(x = 1\) (True).
- For \(y - 1 = 0\): Substitute \(y = -1\).
\( -1 - 1 = -2
eq 0\) (False).
Thus, \((1, -1)\) is not a solution.
System 4: \(x - y = 0\) and \(x + 6y = -5\)
- For \(x - y = 0\): Substitute \(x = 1\), \(y = -1\).
\(1 - (-1) = 2
eq 0\) (False).
Thus, \((1, -1)\) is not a solution.
System 5: \(4x + 2y = 2\) and \(3x - 5y = 8\)
- For \(4x + 2y = 2\): Substitute \(x = 1\), \(y = -1\).
\(4(1) + 2(-1) = 4 - 2 = 2\) (True).
- For \(3x - 5y = 8\): Substitute \(x = 1\), \(y = -1\).
\(3(1) - 5(-1) = 3 + 5 = 8\) (True).
Thus, \((1, -1)\) is a solution.
System 6: \(2x + y = -1\) and \(y = -3x - 2\)
- For \(2x + y = -1\): Substitute \(x = 1\), \(y = -1\).
\(2(1) + (-1) = 2 - 1 = 1
eq -1\) (False).
Thus, \((1, -1)\) is not a solution.
Final Answer:
The systems with solution \((1, -1)\) are:
- \(y = -x\) and \(4x - 4y = 8\)
- \(y = -2x + 1\) and \(y = 5x - 6\)
- \(4x + 2y = 2\) and \(3x - 5y = 8\)
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To determine which systems of equations have a solution of \((1, -1)\), we substitute \(x = 1\) and \(y = -1\) into each system and check if both equations are satisfied.
System 1: \(y = -x\) and \(4x - 4y = 8\)
- For \(y = -x\): Substitute \(x = 1\), \(y = -1\).
\( -1 = -1\) (True).
- For \(4x - 4y = 8\): Substitute \(x = 1\), \(y = -1\).
\(4(1) - 4(-1) = 4 + 4 = 8\) (True).
Thus, \((1, -1)\) is a solution.
System 2: \(y = -2x + 1\) and \(y = 5x - 6\)
- For \(y = -2x + 1\): Substitute \(x = 1\), \(y = -1\).
\( -1 = -2(1) + 1 = -2 + 1 = -1\) (True).
- For \(y = 5x - 6\): Substitute \(x = 1\), \(y = -1\).
\( -1 = 5(1) - 6 = 5 - 6 = -1\) (True).
Thus, \((1, -1)\) is a solution.
System 3: \(x = 1\) and \(y - 1 = 0\)
- For \(x = 1\): \(x = 1\) (True).
- For \(y - 1 = 0\): Substitute \(y = -1\).
\( -1 - 1 = -2
eq 0\) (False).
Thus, \((1, -1)\) is not a solution.
System 4: \(x - y = 0\) and \(x + 6y = -5\)
- For \(x - y = 0\): Substitute \(x = 1\), \(y = -1\).
\(1 - (-1) = 2
eq 0\) (False).
Thus, \((1, -1)\) is not a solution.
System 5: \(4x + 2y = 2\) and \(3x - 5y = 8\)
- For \(4x + 2y = 2\): Substitute \(x = 1\), \(y = -1\).
\(4(1) + 2(-1) = 4 - 2 = 2\) (True).
- For \(3x - 5y = 8\): Substitute \(x = 1\), \(y = -1\).
\(3(1) - 5(-1) = 3 + 5 = 8\) (True).
Thus, \((1, -1)\) is a solution.
System 6: \(2x + y = -1\) and \(y = -3x - 2\)
- For \(2x + y = -1\): Substitute \(x = 1\), \(y = -1\).
\(2(1) + (-1) = 2 - 1 = 1
eq -1\) (False).
Thus, \((1, -1)\) is not a solution.
Final Answer:
The systems with solution \((1, -1)\) are:
- \(y = -x\) and \(4x - 4y = 8\)
- \(y = -2x + 1\) and \(y = 5x - 6\)
- \(4x + 2y = 2\) and \(3x - 5y = 8\)