What Value of a Makes Two Linear Equations Parallel?

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Discussion Overview

The discussion revolves around determining the value of the parameter \( a \) that makes the two linear equations \( 2x + ay = 1 \) and \( ax + (a+4)y = 2 \) parallel. The scope includes mathematical reasoning and exploration of conditions for parallel lines.

Discussion Character

  • Mathematical reasoning
  • Exploratory

Main Points Raised

  • One participant suggests that the answer is \( a = -2 \) and seeks a detailed explanation.
  • Another participant explains that to determine if two lines are parallel, their slopes must be equal, providing the slope calculations for both equations.
  • A third participant introduces an alternative condition for parallel lines based on the coefficients of the equations, stating that the ratios of the coefficients must be equal.

Areas of Agreement / Disagreement

Participants present different methods for determining parallelism, but there is no consensus on the value of \( a \) or the correctness of the proposed methods.

Contextual Notes

Some assumptions about the values of \( a \) (e.g., \( a \neq 0 \) and \( a \neq -4 \)) are mentioned, but their implications are not fully explored or resolved.

ThomsonKevin
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Hello everyone, I was wondering if someone could solve this and explain it in detail. The answer is supposed to be a= -2. Let's cut to the chase:

If the two lines 2x + ay = 1 and ax + (a+4)y = 2 are parallel ,what is a?
 
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ThomsonKevin said:
Hello everyone, I was wondering if someone could solve this and explain it in detail. The answer is supposed to be a= -2. Let's cut to the chase:

If the two lines 2x + ay = 1 and ax + (a+4)y = 2 are parallel ,what is a?

Hello! (Wave)We can determine if two straight lines are parallel by checking their slopes.

Suppose that the equation for the first line is $y_1 = m_1x + b_1$ and the equation for the second line $y_2 = m_2x + b_2$.

For two lines to be parallel their slopes have to be same. That is, $m_1 = m_2$.

$$2x+ay=1 \Rightarrow ay=-2x+1 \overset{a \neq 0}{\Rightarrow} y=-\frac{2}{a}x+\frac{1}{a}$$

$$ax+(a+4)y=2 \Rightarrow (a+4)y=-ax+2 \overset{a \neq -4}{\Rightarrow} y=\frac{-a}{a+4}x+\frac{2}{a+4}$$So, it has to hold:

$$-\frac{2}{a}=\frac{-a}{a+4} \Rightarrow -2a-8=-a^2 \Rightarrow 2a+8=a^2 \Rightarrow a^2-2a-8=0$$Solve the equation $a^2-2a-8=0$ and you will find the possible values of $a$.
 
You can also use the fact that the lines $A_1x+B_1y+C_1=0$ and $A_2x+B_2y+C_2=0$ are parallel iff $A_1/A_2=B_1/B_2\ne C_1/C_2$. Here if $A_2=0$, then $A_1/A_2=B_1/B_2$ means, by definition, that $A_1=0$, and similarly for $B_1$, $B_2$.
 
Yes, exactly what I needed, thank you both of you!
 

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