The equation 5 y + 9 − 12 y = − 6 y + 3 − 14 y is classified as a conditional equation, which has a unique solution. The solution set is { -\frac{6}{13} }.
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Combine like terms on both sides of the equation: 5 y + 9 − 12 y = − 6 y + 3 − 14 y becomes − 7 y + 9 = − 20 y + 3 .
Add 20 y to both sides: − 7 y + 20 y + 9 = 3 , which simplifies to 13 y + 9 = 3 .
Subtract 9 from both sides: 13 y = − 6 .
Divide by 13 to find the solution: y = − 13 6 .
Explanation
Understanding the Problem We are given the equation 5 y + 9 − 12 y = − 6 y + 3 − 14 y . Our goal is to determine whether this equation is a conditional equation, a contradiction, or an identity, and then to find its solution set.
Simplifying the Equation First, we simplify both sides of the equation by combining like terms. On the left side, we have 5 y − 12 y + 9 = − 7 y + 9 . On the right side, we have − 6 y − 14 y + 3 = − 20 y + 3 . So the equation becomes − 7 y + 9 = − 20 y + 3 .
Isolating the Variable Next, we want to isolate the variable y . We add 20 y to both sides of the equation to get − 7 y + 20 y + 9 = − 20 y + 20 y + 3 , which simplifies to 13 y + 9 = 3 .
Further Isolation Now, we subtract 9 from both sides of the equation to get 13 y + 9 − 9 = 3 − 9 , which simplifies to 13 y = − 6 .
Solving for y Finally, we divide both sides by 13 to solve for y : 13 13 y = 13 − 6 , which gives us y = − 13 6 .
Determining the Equation Type and Solution Set Since we found a unique solution for y , the equation is a conditional equation. The solution set is { 13 − 6 } .
Examples
In electrical circuits, equations like these can help determine the current flowing through different parts of the circuit. By setting up an equation that represents the relationships between voltage, resistance, and current, we can solve for unknown values and ensure the circuit functions as intended. This is crucial for designing and troubleshooting electronic devices.