To find systems of equations that have the solution \((-2, 1)\), we can create two equations that both hold true when \(x = -2\) and \(y = 1\). Here are a couple of examples:
Example 1:
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First equation:
\[ 3x + 2y = 0 \] Plugging in \((-2, 1)\): \[ 3(-2) + 2(1) = -6 + 2 = -4 \quad \text{(not true, so let's adjust this equation)} \]To make it true, let’s use:
\[ x + 2y = 0 \] Check: \[ -2 + 2(1) = -2 + 2 = 0 \quad \text{(true)} \] -
Second equation:
\[ y - x = 3 \] Check: \[ 1 - (-2) = 1 + 2 = 3 \quad \text{(true)} \]
Our system of equations can thus be: \[ \begin{cases} x + 2y = 0 \ y - x = 3 \end{cases} \]
Example 2:
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First equation:
\[ 2x + y = -3 \] Check: \[ 2(-2) + 1 = -4 + 1 = -3 \quad \text{(true)} \] -
Second equation:
\[ 4x - 2y = -10 \] Check: \[ 4(-2) - 2(1) = -8 - 2 = -10 \quad \text{(true)} \]
Our second system of equations can be: \[ \begin{cases} 2x + y = -3 \ 4x - 2y = -10 \end{cases} \]
Any system of equations that results in the point \((-2, 1)\) as a solution will work. You can create many such systems by adjusting coefficients and constants in similar linear forms, ensuring that they both maintain the same solution.