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xBabyGirlx
Using elimination, substitution solve c and b C. y = 1\2x - 6 2x + 6y = 19 B. 7x - 3y = 20 5x + 3y = 16
okay solve the second one for y
so for elimination, i suggest multiplying the first equation by negative 4 to get rid of x, than solve for y. for the substitution part, any one will work just fine, just isolate one variable
7x - 3y = 20 5x + 3y = 16 To eliminate you add together the top and bottom equation so you get 12x = 36 x= 3 Then you can work out y 5x3 +3y = 16 15+3y = 16 3y= 1 y= 1/3
watch it though inspy, i think that was a .5x
The first one is wrong. I forgot to multiply the -6 by 6
y=.5x-6 and 2x+6y=19 so multiply the first eq by 4 to get: 4y=2x-24 = 4y-2x=-24 now take that and add the two to get: 10y=5 so y =1/2 then just solve for x using the known y-value for the substitution: remember from your first eq that y=.5x-6, so just sub that into the second eq: 2x+6(.5x-6)=19 and solve
y = 1\2x - 6 2x + 6y = 19 6y = 3x-36 6y= 19-2x 3x-36=19-2x 5x= 55 x= 11 y= -0.5 Thanks Mr Doe
i get it thanks you guys