Question 914249
First solve {{{-2x+y=24}}} for y. So isolate y.


{{{-2x+y=24}}}


{{{y = 2x+24}}} add 2x to both sides


Now that y is fully isolated, we plug this into the second equation. This is where the term "substitution" comes from. We substitute y for 2x+24


{{{5y =10x+120}}}


{{{5(y) =10x+120}}}


{{{5(2x+24) =10x+120}}} replace EVERY copy of y with 2x+24


Now we have eliminated every copy of y. We can now solve for x.


{{{5(2x+24)=10x+120}}} Start with the given equation.



{{{10x+120=10x+120}}} Distribute.



{{{10x=10x+120-120}}} Subtract {{{120}}} from both sides.



{{{10x-10x=120-120}}} Subtract {{{10x}}} from both sides.



{{{0x=120-120}}} Combine like terms on the left side.



{{{0x=0}}} Combine like terms on the right side.



{{{0=0}}} Simplify.



Since this equation is <font size=4><b>ALWAYS</b></font> true for any x value, this means x can equal any number. So there are an infinite number of solutions.


So there are <font size=5 color="red">infinitely many solutions</font> for this system. 


This means that the system is <font size=5 color="red">dependent</font> (one equation "depends" on the other; I like to think of it as one line leaning on the other)


Graphically, these 2 equations form the same line. One line lies perfectly on top of the other. This results in infinitely many intersections (recall that an intersection visually corresponds to a solution).


Let me know if you need more help or if you need me to explain a step in more detail.
Feel free to email me at <a href="mailto:jim_thompson5910@hotmail.com?Subject=I%20Need%20Algebra%20Help">jim_thompson5910@hotmail.com</a>
or you can visit my website here: <a href="http://www.freewebs.com/jimthompson5910/home.html">http://www.freewebs.com/jimthompson5910/home.html</a>


Thanks,


Jim