I have this problem: y=1x and y=.5x+4

I got (2,4) and was wondering if you guys can show me the steps for both elimination and substitution. Also how would I graph this? just put one dot on the graph? (Wondering)

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- Feb 24th 2009, 02:18 PMJubblyElimination and substitution
I have this problem: y=1x and y=.5x+4

I got (2,4) and was wondering if you guys can show me the steps for both elimination and substitution. Also how would I graph this? just put one dot on the graph? (Wondering) - Feb 24th 2009, 02:34 PMReckoner
How did you get $\displaystyle (2,4)?$ If you try checking your work by substituting, you get

$\displaystyle x=2\Rightarrow y=2$

for the first, and

$\displaystyle x=2\Rightarrow y=5$

for the second. Your point is not on either of the curves, so it certainly could not be in the solution set. - Feb 24th 2009, 03:13 PMJubbly
Well, I did 1x=.5x+4

I subtracted .5x from both sides making 1x and .5x

now its .5x=4

divided .5x into 4 and got 2

so x=2?

Can you show me a step by step of how your doing it and how your substituting? - Feb 24th 2009, 03:27 PMReckoner
- Feb 24th 2009, 04:45 PMJubbly
Oh, sorry about that.

If you don't mind can you show me the elimination and substitution equation you used for this. Thanks! - Feb 24th 2009, 05:03 PMReckoner
Certainly. We have

$\displaystyle \left\{\begin{array}{rcl}

y&=&x\\

y&=&\frac12x+4

\end{array}\right.$

Substitution:

Substituting $\displaystyle x=y$ into the second equation produces

$\displaystyle y=\frac12y+4,$

and solving for $\displaystyle y$ gives

$\displaystyle \frac12y=4\Rightarrow y=8.$

Back-substituting, we get $\displaystyle x=8.$

Elimination:

Let's first rearrange the equations a little,

$\displaystyle \left\{\begin{array}{rcl}

y-x&=&0\\

2y-x&=&8

\end{array}\right..$

Subtract the first equation from the second,

$\displaystyle \left\{\begin{array}{rcl}

y-x&=&0\\

y&=&8

\end{array}\right.$

and subtract the second equation from the first:

$\displaystyle \left\{\begin{array}{rcl}

-x&=&-8\\

y&=&8

\end{array}\right.\Rightarrow\left\{\begin{array}{ rcl}

x&=&8\\

y&=&8

\end{array}\right.$