Friend,
let the number be y
3*y=9
y=9/3
y=3
So the number is 3
2006-11-07 05:13:23
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answer #1
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answered by JACKREX 2
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richard is x and ruel is y. from the difficulty all of us comprehend that x+y=60. additionally, (x-9)=2(y-9). from the 1st equation all of us comprehend that x=60-y, so, ((60-y)-9)=2(y-9). via simplifying this we are able to get the value of y: ((60-y)-9)=2(y-9) (51-y)=2(y-9) 51-y=2y-18 51=2y-18+y 51=3y-18 51+18=3y sixty 9=3y y=13 x=60-y x=60-13 x=40 seven As for the 2d difficulty, the 1st digit is x and the 2d is y. so, x+y=9, and...it does not make sense while u suggested that the huge variety is 12 situations the ten's digit, so i can't somewhat remedy that until eventually u describe this in extra info. The 0.33 difficulty, the three numbers are x,y, and z. What all of us comprehend is that: x+y+z=fifty seven x=y-2 y=z-2 x, y, and z are unusual integers. from the 2d equation all of us comprehend that y=x+2, to that end, x+2=z-2. via including 2 on the two aspects, we see that x+4=z. so we've the value of y and z in terms of x. via inputting those values into the 1st equation, we get x+(x+2)+(x+4)=fifty seven. via simplifying this, we get 3x+6=fifty seven. via subtracting 6 on the two area, we come to the effect 3x=51. dividing the two aspects via 3 supplies us the value of x, x=17. Now that all of us comprehend x=17, y=x+2, so y=17+2=19. y=z-2, so 19=z-2, z=21 ok, so i'm getting the 2d question. the respond is 36.
2016-10-21 10:28:56
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answer #5
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answered by Anonymous
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