Elimination and substitution

**Jubbly** · February 24th 2009, 02:18 PM

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?

**Reckoner** · February 24th 2009, 02:34 PM

Originally Posted by Jubbly

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?

How did you get $(2,4)?$ If you try checking your work by substituting, you get

$x=2\Rightarrow y=2$

for the first, and

$x=2\Rightarrow y=5$

for the second. Your point is not on either of the curves, so it certainly could not be in the solution set.

**Jubbly** · February 24th 2009, 03:13 PM

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?

**Reckoner** · February 24th 2009, 03:27 PM

Originally Posted by Jubbly

divided .5x into 4 and got 2

4 divided by 0.5 is 8, not 2:

$\frac4{1/2}=4\cdot\frac21=4\cdot2=8\text.$

The correct solution is $(8,\,8)\text.$

**Jubbly** · February 24th 2009, 04:45 PM

Oh, sorry about that.

If you don't mind can you show me the elimination and substitution equation you used for this. Thanks!

**Reckoner** · February 24th 2009, 05:03 PM

Certainly. We have

$\left\{\begin{array}{rcl} y&=&x\\ y&=&\frac12x+4 \end{array}\right.$

Substitution:

Substituting $x=y$ into the second equation produces

$y=\frac12y+4,$

and solving for $y$ gives

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

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

Elimination:

Let's first rearrange the equations a little,

$\left\{\begin{array}{rcl} y-x&=&0\\ 2y-x&=&8 \end{array}\right..$

Subtract the first equation from the second,

$\left\{\begin{array}{rcl} y-x&=&0\\ y&=&8 \end{array}\right.$

and subtract the second equation from the first:

$\left\{\begin{array}{rcl} -x&=&-8\\ y&=&8 \end{array}\right.\Rightarrow\left\{\begin{array}{ rcl} x&=&8\\ y&=&8 \end{array}\right.$

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