b) \( \begin{array}{l}x+y=0.5 \\ x+2 y=0.5\end{array} \)
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Bonus Knowledge
To solve the system of equations given, we can first use substitution or elimination. Let's solve for \(y\) in the first equation: \(y = 0.5 - x\). Now substitute this expression for \(y\) into the second equation: \( x + 2(0.5 - x) = 0.5 \) leads to \( x + 1 - 2x = 0.5 \). Rearranging gives \( -x + 1 = 0.5 \), and solving for \(x\) yields \( x = 0.5 \). Now substituting \(x\) back into \(y = 0.5 - x\) will give \(y = 0\). Thus, the solution is \((x, y) = (0.5, 0)\). These equations can be quite useful in fields like economics or engineering, where systems of linear equations model constraints or relationships. For example, they could represent a budget constraint (where \(x\) and \(y\) are different expenses) and a production scenario (showing the relationship between resources). When working with equations, it's easy to make simple mistakes. One common issue is misapplying the substitution, like forgetting to distribute correctly when substituting values. Always double-check your arithmetic and rearrangements, and consider graphing the equations for a visual confirmation of your solution!
