## Solving a system of equations

Quadratic equations and inequalities, variation equations, function notation, systems of equations, etc.
Luke53
Posts: 54
Joined: Sun Mar 13, 2011 9:46 am
Contact:

### Solving a system of equations

How to solve?
x/10 = y/14 = z/35 = 1/(10x + 14y + 35z)
Have no clue how to start solving this system.
Luke.

stapel_eliz
Posts: 1628
Joined: Mon Dec 08, 2008 4:22 pm
Contact:
How to solve?
x/10 = y/14 = z/35 = 1/(10x + 14y + 35z)
One method might be to solve x/10 = y/14 and x/10 = z/35 for y and z in terms of x. Then plug this into x/10 = 1/(10x + 14y + 35z), and solve for the value(s) of x.

Whatever method you use, it's likely gonna be messy, since this isn't linear.

Luke53
Posts: 54
Joined: Sun Mar 13, 2011 9:46 am
Contact:

### Re: Solving a system of equations

One method might be to solve x/10 = y/14 and x/10 = z/35 for y and z in terms of x. Then plug this into x/10 = 1/(10x + 14y + 35z), and solve for the value(s) of x.

Whatever method you use, it's likely gonna be messy, since this isn't linear.

stapel_eliz

This method works out very well, tried it out and works fine, (and not too messy):
y = 14x / 10 ; z = 35x / 10 susbt. this in the eqn. x/10 = 1/(10x + 14y +35 z) gives: x/10 = 1/(10x + 14*(14x/10) + 35*(35x/10)
x/10 = 1/(10x +19.6x + 122.5x) => x/10 = 1/152.1x and so 152.1x² = 10 => 152.1x² - 10 = 0 (quadratic) with solutions: x = 0.25641 or - 0.25641
It's easy now to find the values for y and z. (y = 0.358974 and z = 0.897435)

Thanks .

Luke.

Return to “Intermediate Algebra”