Solving Systems

**Archie Meade** · December 29th 2009, 05:30 PM

$Hi\ Vuong,$

$You\ may\ consider\ simple\ substitutions\ also\ as\ follows..$

$3x-2y=10,\ hence\ 3x=2y+10,\ x=\frac{2y+10}{3}$

$This\ x\ may\ now\ be\ used\ to\ solve\ for\ y\ in\ the\ 2nd\ equation$

$5x+3y=15,\ hence\ \frac{10y+50}{3}+3y=15$

$10y+50+9y=45,\ 19y=45-50=-5,\ y=-\frac{5}{19}$

$Now\ use\ this\ y\ to\ find\ x.$

$You\ may\ use\ cancellation\ or\ substitution\ methods.$

$The\ second\ set\ of\ equations\ is\ just\ a\ single\ equation.$

**Vuong** · December 29th 2009, 09:50 PM

Now how about the problem

3x-5y+10=0
-9x+15y=-30

**Bacterius** · December 29th 2009, 10:00 PM

Originally Posted by Vuong

Now how about the problem

3x-5y+10=0
-9x+15y=-30

Can't you solve it after all the different ways we showed you ?

Here is my hint : in the second equation, try to express $9x$ in terms of $y$ , and substitute into the first equation to find $x$ . Then substitute the value of $x$ you found into any of the two equations to find $y$ .

**Archie Meade** · December 30th 2009, 01:12 AM

$Hi\ Vuong,$

$Pairs\ of\ linear\ simultaneous\ equations\ in\ two\ variables\ are\ line\ equations.$

$This\ time\ your\ two\ lines\ are\ parallel.$

$Parallel\ lines\ never\ intersect.$

$The\ solution\ of\ a\ pair\ of\ such\ equations\ is\ the\ point\ of\ intersection.$

$There\ is\ no\ point\ of\ intersection\ for\ these\ two\ lines.$

$Your\ equations\ may\ be\ re-written$

$9x-15y+30=0$

$9x-15y-30=0$

$or,$

$y=\frac{3}{5}x+2$

$y=\frac{3}{5}x-2$

$The\ lines\ have\ the\ same\ slope\ but\ cross\ the\ y-axis\ at\ 2\ and\ -2.$

$Cancellation\ and\ substitution\ yield\ no\ solutions.$

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