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January 6th, 2017, 07:29 PM  #1 
Member Joined: Jan 2017 From: US Posts: 86 Thanks: 5  Elimination Method
Hello, I am currently trying to learn how to use the elimination method to solve a system of equations in algebra. I normally don't just ask for the answers to my homework that really is not what I am trying to do here, if someone could just point me in the right direction that would be amazing. I was given this system of equations: 2x+5y=5 3x2y=17 And I have to show all of the steps to solve with the elimination method. I understand the addition method, if that will help. I understand the goal here is to make the coefficients of one variable the same; I just would appreciate it if someone could point me in the right direction to solving these equations! (The final answer will need to be in an ordered pair). Thanks in advance! 
January 6th, 2017, 07:39 PM  #2 
Math Team Joined: Jul 2011 From: Texas Posts: 2,549 Thanks: 1259 
Multiply every term in the first equation by 2 Multiply every term in the second equation by 5 Adding the two resulting equations term for term will "eliminate" the yterms ... solve for x, then use the value you find for x in either of the two original equations to solve for y. 
January 6th, 2017, 08:35 PM  #3 
Member Joined: Jan 2017 From: US Posts: 86 Thanks: 5 
Thank you! So, if I did that correctly, I think this is the answer: 2x(2) + 5y(2)= 5 (2) 4x+10y=10 3x(5)  2y(5) = 17 (5) 15x10y=95 4x+10y=10 + 15x10y=85 ________________ 19x =95 95/19 = 5 x = 5 2(5)+5y=5 3(5)2y=17 After I solved for y, I substituted 1 into the equation as y: 2(5)+5(1)=5 3(5)2(1)=17 Since these are both true, the ordered pair would be (5,1), right? 
January 6th, 2017, 09:23 PM  #4 
Senior Member Joined: Feb 2016 From: Australia Posts: 1,170 Thanks: 384 Math Focus: Yet to find out. 
Answer looks correct to me .

January 7th, 2017, 03:07 AM  #5 
Member Joined: Sep 2016 From: India Posts: 88 Thanks: 30 
$2x+5y=5......(1)$ $3x2y=17.......(2)$ Multiplying equation(1) by $2$ and equation(2) by $5$ we get $4x+10y=10......(3)$ $15x10y=85.......(4)$ Subtracting equation(4) from equation(3) we get $19x=95$ $x=5$ Substitute vaule of $x$ in equation(1) $10+5y=5$ $5y =5$ $y=1$ Hence, $x=5,\;\;\;\;\;\;\;y=1$ 
January 7th, 2017, 08:36 AM  #6 
Math Team Joined: Jul 2011 From: Texas Posts: 2,549 Thanks: 1259  

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