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Title: How to solve linear equations/ systems by substitution? Post by: datasian on Feb 18, 2013 Can you help me answer the following linear equations by substitution. Please tell me what to do step by step so I can actually follow what you are saying. I figure if people tell me how to do a couple of them, I can understand how to do the rest.
20x-30y= -50 x+2y=1 y=2x-1 2x+y=3 Title: How to solve linear equations/ systems by substitution? Post by: julesarth on Feb 18, 2013 Content hidden
Title: How to solve linear equations/ systems by substitution? Post by: nursetia on Feb 18, 2013 Oh! domo; The same old stuff they throw at everybody. Wait until you get 3 equations with 3 unknowns but you can handle it Our object is to make one of the values on the lower equation equal to its amount on the top equation , only opposite in signs so we can cancel out that value. So: We multiply the botton by a negative 20 and get -20X-40Y=negative 20; subtracting this from the top, we get -70 Y=-70 Now we multiply both side by the negative sign to get 70Y= 70 and Y=1. We test:
X + 2x1= 1; x + 1-2=-1. We test( 20 x -1 ) - (30 x 1) =-50 ; -20 -( + 30) =-50 and -20 & -30 =_50. Now you do the rest. Good Luck. You can do it. |