Question

Which of these equations, when combined into a set of equations with \(4 x=2 y-6\), will result in no solutions to the set? (A) \(y=x-4\) (B) \(y=2 x+10\) (C) \(y=4 x-1\) (D) \(y=\frac{1}{4} x-6\)

Short answer

Option B will result in no solutions: \(y = 2x + 10\).

Step-by-step solution

Rewrite the given equation

Start by analyzing the given equation \(4x = 2y - 6\). Rearrange it to express \(y\) in terms of \(x\).Add 6 to both sides: \[4x + 6 = 2y\]Then divide both sides by 2:\[y = 2x + 3\]

Compare with the option equations

The solution for \(y\) from the given equation is \(y = 2x + 3\). Now, compare this equation with each option to see which combination results in a contradiction (i.e., no solutions).

Analyze each option individually

**Option A**: \(y = x - 4\)Equating: \(2x + 3 = x - 4\)Subtract \(x\) from both sides: \(x + 3 = -4\)Subtract 3: \(x = -7\)There is a solution, \(x = -7\).**Option B**: \(y = 2x + 10\)Equating: \(2x + 3 = 2x + 10\)Subtract \(2x\) from both sides: \(3 = 10\)This is a contradiction, so there are no solutions.**Option C**: \(y = 4x - 1\)Equating: \(2x + 3 = 4x - 1\)Subtract \(2x\) from both sides: \(3 = 2x - 1\)Add 1: \(4 = 2x\)Divide by 2: \(x = 2\)There is a solution, \(x = 2\).**Option D**: \(y = \frac{1}{4}x - 6\)Equating: \(2x + 3 = \frac{1}{4}x - 6\)Multiply everything by 4 to clear the fraction:\(8x + 12 = x - 24\)Subtract \(x\): \(7x + 12 = -24\)Subtract 12: \(7x = -36\)Divide by 7: \(x = -\frac{36}{7}\)There is a solution.

Key concepts

No Solution Equations

In mathematics, especially when solving systems of equations, we often encounter situations where an equation has no solutions. Simply put, a no solution equation occurs when there is no value or set of values that satisfy all equations in the system simultaneously. In a system of linear equations, if two lines are parallel but have different y-intercepts, they will never meet, leading to no solutions. This results in a condition known as a contradiction.
A common indicator of no solution in algebra is when simplifying equations results in a false statement, such as "5 = 7." This indicates a contradiction, meaning there are no solutions that satisfy both equations in the system. In our exercise, option B, where the equation became "3 = 10," perfectly exemplifies this scenario.
Recognizing no solution equations is crucial, as it helps in understanding the nature of the system. These equations reflect situations where two or more conditions cannot be met simultaneously, revealing an important aspect of problem-solving in mathematics.

Algebra Problem Solving

Algebra problem solving is a fundamental skill that requires understanding how to manipulate equations to find values for unknown variables. The process usually involves a series of steps: rewriting equations, simplifying, and making strategic decisions on how to proceed to find solutions efficiently.
In solving our initial equation, the first step required rearranging it to express one variable, typically y, in terms of the other, x. For example, converting "4x = 2y - 6" to "y = 2x + 3" was necessary to make it easier to compare and solve with other equations. This is part of finding a system's standard form, which is a common technique in algebra.
Problem solving does not end at finding just any solution. We also check for contradictions or inconsistencies, such as in option B of our problem. Through practice, one can improve the ability to solve algebraic problems effectively. This means knowing when to combine terms, when equations signify no solution, or when to apply deeper algebraic methods, like substitution or elimination.

Equation Contradictions

Equation contradictions occur when simplifying the given mathematical statements results in an impossibility. These contradictions indicate that there is no possible solution that will satisfy all the equations involved in the system.
In our problem, we investigated different equations to see if they could coexist with the equation "y = 2x + 3" without contradiction. For Option B, the attempt to equate led to "3 = 10," which clearly can't be true. This is a textbook example of a contradiction.
Recognizing and understanding contradictions is a pivotal skill in algebra. It helps in validating solutions and ensuring consistency when working with multiple equations. Spotting these contradictions and labeling equations that yield them as no solutions simplifies complex problem sets. This approach not only saves time but also enhances the overall comprehension of algebraic systems.