If X+Y+Z=35000, X+38% of Z=11900, Y+62% of Z=23100, find Z

Simplificatation, evaluation, linear equations, linear graphs, linear inequalities, basic word problems, etc.
Madhuri hadwale
Posts: 1
Joined: Wed Mar 02, 2011 10:19 am
Contact:

If X+Y+Z=35000, X+38% of Z=11900, Y+62% of Z=23100, find Z

Postby Madhuri hadwale » Wed Mar 02, 2011 10:34 am

Hello,

I having problem in arithmatic Equation can anybody help me.

EX
If
X+Y+Z=35000
X+38% of Z=11900
Y + 62% of Z =23100

Than Z=??????

User avatar
maggiemagnet
Posts: 319
Joined: Mon Dec 08, 2008 12:32 am
Contact:

Re: If X+Y+Z=35000, X+38% of Z=11900, Y+62% of Z=23100, find

Postby maggiemagnet » Wed Mar 02, 2011 12:13 pm

Madhuri hadwale wrote:If
X+Y+Z=35000
X+38% of Z=11900
Y + 62% of Z =23100

Than Z=??????

Turn the "38%" into a decimal: 0.38. Turn the "of" into "times". So you have "X + 0.38Z = 11900".

Now do the same thing with the third equation. And then solve the second equation for "X=" and the third equation for "Y=". Plug these in for "X" and "Y" in the first equation.

You can solve this for "Z=". :thumb:


Return to “Beginning Algebra”