Xavier gave Yvonne and Zeena the same number of marbles as each already had. Then Yvonne gave Xavier and Zeena the same number of marbles as each already had. Then Zeena gave Xavier and Yvonne the same number of marbles as each already had. At that point, each person had 48 marbles. How many fewer marbles did Xavier have at the end than he had at the start?

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Cậu Bé Ngu Ngơ 2 hour ago (12:51)

Let x, y, z be the number of marbels each person has.

After Xavier shared the marbles, Yvonne had 2y, Zeena had 2z and Xavier had x - y - z.

After Yvonne shared the marbles, Xavier had 2(x - y - z), Zeena had 4z and Yvonne had : 2y - (x - y - z) - 2z = 3y - x - z

After Zeena shared the marbles, Xavier had 4(x - y - z), Yvonne had 2(3y - x - z) and Zeena had : 4z - 2(x - y - z) - (3y - x - z) = 7z - x - y

We have :

⎧⎪⎨⎪⎩4(x−y−z)=482(3y−x−z)=487z−x−y=48⇒⎧⎪⎨⎪⎩x−y−z=12−x+3y−z=24−x−y+7z=48{4(x−y−z)=482(3y−x−z)=487z−x−y=48⇒{x−y−z=12−x+3y−z=24−x−y+7z=48⇒⎧⎪⎨⎪⎩x=78y=42z=24⇒{x=78y=42z=24

The answer is : x−4(x−y−z)=78−4.(78−42−24)=78−4.12=30

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Phan Thanh Tinh  Coordinator 17 hour ago (21:59)

Let x, y, z be the number of marbels each person has.

After Xavier shared the marbles, Yvonne had 2y, Zeena had 2z and Xavier had x - y - z.

After Yvonne shared the marbles, Xavier had 2(x - y - z), Zeena had 4z and Yvonne had : 2y - (x - y - z) - 2z = 3y - x - z

After Zeena shared the marbles, Xavier had 4(x - y - z), Yvonne had 2(3y - x - z) and Zeena had : 4z - 2(x - y - z) - (3y - x - z) = 7z - x - y

We have :

⎧⎪⎨⎪⎩4(x−y−z)=482(3y−x−z)=487z−x−y=48⇒⎧⎪⎨⎪⎩x−y−z=12−x+3y−z=24−x−y+7z=48{4(x−y−z)=482(3y−x−z)=487z−x−y=48⇒{x−y−z=12−x+3y−z=24−x−y+7z=48⇒⎧⎪⎨⎪⎩x=78y=42z=24⇒{x=78y=42z=24

The answer is : x−4(x−y−z)=78−4.(78−42−24)=78−4.12=30