3x3 systems (word Problems)

I can do some of these exercises, but others just leave me completely baffled.

The sum of three integers is 189. The first integer is 28 less than the second. The second integer is 21 less than the sum of the first and third integers. Find the three integers.

1. x + y + z=189
2. x = y - 28
3. y = x + z - 21

I can't get this to work. Have I mixed up my equations?

These equations look fine. What method are you using for solving? What steps have you taken? Where are you stuck?

Please be specific. Thank you.

Eliz.

I tried solving for the specific variables by plugging the second equation into the 1st by adding 7-28 in place of x. Thus giving me y-28+y+z=189

The third equation states that y=x+z-21

I need another equation and how I have things, it is going in a circle. If I add the 3rd equation in for y the x is then to be replaced as well with the second equation. I am not doing something correct and this is where my problem is.

jhawk555 said:

I tried solving for the specific variables by plugging the second equation into...

Click to expand...

It sounds like you are using substitution/elimination as your solution method. For three and greater variables, this can get very messy, but it is possible.

Since the second equation gives you x in terms of y, I would plug this in for x in the first the third equations. This will give you two equations in y and z, namely:

. . . . .(y - 28) + y + z = 189

. . . . .y = (y - 28) + z - 21

Simplifying, you get:

. . . . .2y + z = 217

. . . . .z = 49

You should be able to solve this for z, and then for y. Back-solve for x.

Eliz.

2. x = y - 28
3. y = x + z - 21

Kwickie: substitute 2 in 3:
y = y - 28 + z - 21
z = 49

...carry on

I Got it Finally!

Thanks. Word Problems are my weakness, but I keep trying.