1. $\displaystyle Hi\ Vuong,$

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

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

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

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

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

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

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

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

2. Now how about the problem

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

3. Originally Posted by Vuong

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 $\displaystyle 9x$ in terms of $\displaystyle y$, and substitute into the first equation to find $\displaystyle x$. Then substitute the value of $\displaystyle x$ you found into any of the two equations to find $\displaystyle y$.

4. $\displaystyle Hi\ Vuong,$

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

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

$\displaystyle Parallel\ lines\ never\ intersect.$

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

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

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

$\displaystyle 9x-15y+30=0$

$\displaystyle 9x-15y-30=0$

$\displaystyle or,$

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

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

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

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

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