liizzyliizz
  • liizzyliizz
Solve using elimination. 5x + 2y = 7 10x + 4y = 14 ok so I am reteaching myself algebra 2, and I was doing this, would there be no solution? I am not sure If I did this correct.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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jim_thompson5910
  • jim_thompson5910
let z = 5x+2y
jim_thompson5910
  • jim_thompson5910
double both sides 2z = 2(5x+2y) 2z = 10x+4y
jim_thompson5910
  • jim_thompson5910
so if 5x + 2y = 7 10x + 4y = 14 then we can also say z = 7 2z = 14 where z = 5x+2y

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jim_thompson5910
  • jim_thompson5910
but....what happens when we solve 2z = 14 for z?
liizzyliizz
  • liizzyliizz
I am not sure If I am thinking about this in the correct way, but technically wouldn't z would = 7. and we already know that. So I am a bit lost now, with where you're going with this.
jim_thompson5910
  • jim_thompson5910
2z = 14 ---> z = 7 but z = 7 in the first equation
jim_thompson5910
  • jim_thompson5910
so no matter what happens, we're going to get z = 7
jim_thompson5910
  • jim_thompson5910
so because z = 5x+2y, this means that 5x+2y = 7 and this applies to both equations because the two are really the same equation
Mertsj
  • Mertsj
The problem says to solve by elimination so multiply the first equation by -2 and then add the two equations. You will get the true statement 0=0 which tells you that there is an infinite number of solutions.
liizzyliizz
  • liizzyliizz
oh ok, I got the 0=0 but I wasnt sure if that meant infinite or no solutions. which is why I said no solutions. Ahh thank you for that clarification.
Mertsj
  • Mertsj
If you got a nonsense statement such as 0 = 3, that would indicate no solution.
liizzyliizz
  • liizzyliizz
that makes more sense, I feel so silly now for confusing that.

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