anonymous
  • anonymous
A system of equations is given below. 2x + 7y = 1 -3x – 4y = 5 Create an equivalent system of equations by replacing the first equation by multiplying the first equation by an integer other than 1, and adding it to the second equation. Use any method to solve the equivalent system of equations (the new first equation with the original second equation). Prove that the solution for the equivalent system is the same as the solution for the original system of equations.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
So, I already did some of part a, multiplied the first equation by 2, and got 4x+14y=2. What should I do next?
FortyTheRapper
  • FortyTheRapper
Hmm, that wording is confusing to me
anonymous
  • anonymous
ikr

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anonymous
  • anonymous
Plz help!
anonymous
  • anonymous
multiply the first one by 3 and the second one by 21 (all the way across) then when you add them the x terms will go bye bye
anonymous
  • anonymous
ok not "21" multiply the second one by 2
anonymous
  • anonymous
Thanks! But then what do you do? @satellite73
anonymous
  • anonymous
supposed to do 2*r1 + r2 => r1
anonymous
  • anonymous
then solve system comprised of new r1 and original r2.
anonymous
  • anonymous
add
anonymous
  • anonymous
in a column
anonymous
  • anonymous
the first term of the first equation will be \(6x\) and the first term of the second one will be \(-6x\) when you add the two together, there will be no more \(x\)
anonymous
  • anonymous
\[2x + 7y = 1 \\-3x – 4y = 5\]\[6x+21y=3\\ -6x-8y=10\] now add them up
anonymous
  • anonymous
So it would be 13y=-7?
anonymous
  • anonymous
@satellite73
retirEEd
  • retirEEd
After you multiplied the first equation by two and got you forgot to add the two to create a third equation, correct? eq 1) 2x + 7y = 1 times 2 eq 3) 4x+14y=2 add eq 3 to eq 2 eq 2) -3x – 4y = 5 eq 4) x + 10y = 7 then it says to use the FIRST new equation ( eq 3) to solve the equivalent system with the original eq 2. Why did I have to add eq 2 and 3 together? Or do they want me to solve the equivalent system using eq 2 and 4 ? Which may or may not yield the correct answer.
anonymous
  • anonymous
Ohhh I think I understand now. Thanks @retirEEd Do you maybe know how to do part c?
anonymous
  • anonymous
solve new system of eqn 4 and eqn 2 and show the soultion is the same as the solution of the original system
retirEEd
  • retirEEd
Thanks pgpilot326, I'll try it. I got the solution for eqn 1 and eqn 2 and I like your short for equation better!
retirEEd
  • retirEEd
Yes it works! eqn 1 and eqn 2 solve to be x=-3 and y= 1 as does eqn 2 and eqn 4. Much easier to visualize on a graphing calculator. Three lines all intersection at (-3,1), then typing all of this out. eqn 1) 2x + 7y = 1 x = ((1-7y)/2 Plug x into eqn 2 eqn 2) -3x – 4y = 5 -3[(1-7y)/2] - 4y = 5 -3/2 + 21y/2 - 8y/2 = 10/2 13y/2 = 10/2 +3/2 multiply both sides by 2 to clear the fraction 13y = 13 y = 1 plug y =1 into eqn 1 2x + 7(1) = 1 x = (1-7)/2 = -6/2 x= -3 eqn 4) x + 10y = 7 eqn 2) -3x – 4y = 5 multiply eqn 4 by 3 and add to eqn 2 the x term equates to zero and 30y + (-4y) = 21 + 5 26y = 26 y = 1 as before plug y = 1 into eqn 4 x + 10(1) = 7 x = 7 -10 x = -3 as before

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