Absolute Value Equations 3

harpazo

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Let's go back to high school together as we explore the world of absolute value equations. Ready to have fun with math?

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TheJason

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\(\displaystyle | x + 20| = \dfrac{3}{4} \)

\(\displaystyle | x + 20| = 0.75 \)

\(\displaystyle (x + 20) =0.75 \)

\(\displaystyle x + 20 = 0.75 \)

\(\displaystyle x + 20 - 20 = 0.75 - 20\)

\(\displaystyle x = -19.25\)

Check

\(\displaystyle |(-19.25) + 20| = 0.75\)

\(\displaystyle |0.75| = 0.75\)

Negative side:

\(\displaystyle | x + 20| = 0.75 \)

\(\displaystyle -(x + 20) = 0.75 \)

\(\displaystyle -x - 20 = 0.75 \)

\(\displaystyle -x - 20 + 20 = 0.75 + 20 \)

\(\displaystyle -x = 20.75\)

\(\displaystyle \dfrac{x}{-1} = \dfrac{20.75}{-1}\)

\(\displaystyle x = -20.75\)

Check

\(\displaystyle |(-20.75) + 20| = 0.75\)

\(\displaystyle |-0.75| = 0.75\)

Two solutions of \(\displaystyle -19.25, -20.75\)
 
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harpazo

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\(\displaystyle | x + 20| = \dfrac{3}{4} \)

\(\displaystyle | x + 20| = 0.75 \)

\(\displaystyle (x + 20) =0.75 \)

\(\displaystyle x + 20 = 0.75 \)

\(\displaystyle x + 20 - 20 = 0.75 - 20\)

\(\displaystyle x = -19.25\)

Check

\(\displaystyle |(-19.25) + 20| = 0.75\)

\(\displaystyle |0.75| = 0.75\)

Negative side:

\(\displaystyle | x + 20| = 0.75 \)

\(\displaystyle -(x + 20) = 0.75 \)

\(\displaystyle -x - 20 = 0.75 \)

\(\displaystyle -x - 20 + 20 = 0.75 + 20 \)

\(\displaystyle -x = 20.75\)

\(\displaystyle \dfrac{x}{-1} = \dfrac{20.75}{-1}\)

\(\displaystyle x = -20.75\)

Check

\(\displaystyle |(-20.75) + 20| = 0.75\)

\(\displaystyle |-0.75| = 0.75\)

Two solutions of \(\displaystyle -19.25, -20.75\)
Do we always get two solutions?
 

TheJason

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MarkFL

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If we are taking the absolute value of a linear function, and we set this absolute value equal to a positive number, then we will get two solutions:

\(\displaystyle |ax+b|=c\) where \(0<c\) and \(a\ne0\)

\(\displaystyle ax+b=\pm c\)

\(\displaystyle x=\frac{-b\pm c}{a}\)

If \(c=0\), then we get one solution, and if \(c<0\) we get no solution.
 
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harpazo

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If we are taking the absolute value of a linear function, and we set this absolute value equal to a positive number, then we will get two solutions:

\(\displaystyle |ax+b|=c\) where \(0<c\) and \(a\ne0\)

\(\displaystyle ax+b=\pm c\)

\(\displaystyle x=\frac{-b\pm c}{a}\)

If \(c=0\), then we get one solution, and if \(c<0\) we get no solution.
Can you share an example when c = 0 and when c < 0?