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 Algebra Pre-Algebra and Basic Algebra Math Forum

 January 6th, 2017, 06:29 PM #1 Senior Member   Joined: Jan 2017 From: US Posts: 120 Thanks: 6 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 3x-2y=-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, 06:39 PM #2 Math Team   Joined: Jul 2011 From: Texas Posts: 3,045 Thanks: 1627 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 y-terms ... solve for x, then use the value you find for x in either of the two original equations to solve for y. Thanks from Indigo28 January 6th, 2017, 07:35 PM #3 Senior Member   Joined: Jan 2017 From: US Posts: 120 Thanks: 6 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) 15x-10y=-95 4x+10y=-10 + 15x-10y=-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, 08:23 PM #4 Senior Member   Joined: Feb 2016 From: Australia Posts: 1,844 Thanks: 658 Math Focus: Yet to find out. Answer looks correct to me . January 7th, 2017, 02:07 AM #5 Member   Joined: Sep 2016 From: India Posts: 88 Thanks: 30 $2x+5y=-5......(1)$ $3x-2y=-17.......(2)$ Multiplying equation(1) by $2$ and equation(2) by $5$ we get $4x+10y=-10......(3)$ $15x-10y=-85.......(4)$ 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, $x=-5,\;\;\;\;\;\;\;y=1$ Thanks from Indigo28 January 7th, 2017, 07:36 AM   #6
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Quote:
 Originally Posted by deesuwalka $2x+5y=-5......(1)$ $3x-2y=-17.......(2)$ Multiplying equation(1) by $2$ and equation(2) by $5$ we get $4x+10y=-10......(3)$ $15x-10y=-85.......(4)$ Subtracting equation(4) from equation(3) we get $19x=-95$
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