Matrices and linear systems of Equations

I have two questions:

1. Three tanks can hold 500 kgs of water. The largest tank can hold 150 kgs more than the smallest tank. If each tank’s capacity were increased such that the smallest now holds twice as much, the next thrice and the largest four times as much, the total capacity would be 1,650 kgs. How much can each tank hold presently?

2. Three men have ages which add up to 95 years. The eldest is 15 years older than the youngest. In 10 years, the youngest will be 5 years younger than the next man. How old is each now?

**For the above questions we need to formulate the mathematical equations derived and solve the resulting equation by Cramer’s rule**

Re: Matrices and linear systems of Equations

Quote:

Originally Posted by

**georgiazm** I have two questions:

1. Three tanks can hold 500 kgs of water.

Okay, call the quantity of water each tank can hold, in kgs, x, y, and z, from smallest to largest. x+ y+ z= 500.

Quote:

The largest tank can hold 150 kgs more than the smallest tank.

So z= x+ 150.

Quote:

If each tank’s capacity were increased such that the smallest now holds twice as much

2x instead of x

3y instead of y

Quote:

and the largest four times as much

4z instead of 4

Quote:

, the total capacity would be 1,650 kgs. How much can each tank hold presently?

2x+ 3y+ 4z= 1650. That, together with x+ y+ z= 500, and z= x+ 150 gives you three equations to solve for x, y, and z.

Quote:

2. Three men have ages which add up to 95 years. The eldest is 15 years older than the youngest. In 10 years, the youngest will be 5 years younger than the next man. How old is each now?

Let their ages be x, y, and z, in years, from youngest to oldest. Now YOU set up the equations.

Quote:

**For the above questions we need to formulate the mathematical equations derived and solve the resulting equation by Cramer’s rule**

Re: Matrices and linear systems of Equations

Thanks for the reply basically those are the equations i got but the problem is when i start to solve them i seem to be getting Zeros.

d= 2 3 4

1 1 1

-1 0 1

dx=1650 3 4

500 1 1

150 0 1

solving for dx=1650(1-0) -3(500-150)+ 4(0-150)=1650-1050-600=0

solving for d= 2(1-0)-3(1+1) +4(0+1)=2-6+4=0

since d is working out to be zero doesn't that imply that dx/d=0, dy/d=0 dz/d=0

Is there someplace where I am going wrong or is it just the question.