floree Posts: 5, Reputation: 1 New Member #1 Mar 26, 2008, 04:29 AM
Length and width of rectangle elimination or substitution
The perimeter of a rectangle is 60 inches (let the length be l and width be w). We then took half the length and also doubled the width and kept the perimeter equal to 60 inches. Write and equation for the perimeter of the first rectangle and combine like terms. Then do the same for the perimeter of the second rectangle. Set up the system of equations to model this problem.

Would the equations be? 60=2l +2w and 60=1/2l+2w (2)

Solve the system using elimination or substitution to the find the length and width of the original retangle.
 rodandy12 Posts: 227, Reputation: 24 Full Member #2 Mar 26, 2008, 04:42 AM
So, next get the second equation in a form similar to the first equation. Do you know how to solve the system of equations using elimination?
 floree Posts: 5, Reputation: 1 New Member #3 Mar 26, 2008, 04:45 AM
No I have difficulty with elimination
 galactus Posts: 2,271, Reputation: 282 Ultra Member #4 Mar 26, 2008, 06:31 AM

$2L+2W=60$

Now, we are halving the length and doubling the width:

$2(\frac{L}{2})+2(2W)=60$

There is the system to solve. Substitution works well if you solve the last one for L and sub into the first.
 rodandy12 Posts: 227, Reputation: 24 Full Member #5 Mar 26, 2008, 11:51 AM
Elimination just means you eliminate a variable. You do this by solving for a variable using one of the equations and substitute that value into the other equation.

Say you have x + 2y = 5 and 3x - 4y = 12. Solving for x in the first equation gives x = 5 - 2y. You now have a value for x. Substitute that value for x into the second equation and you have eliminated x. You also have an equation that only has y and constants in it. You can easily solve for y. Then take the value for y you just got and substitute it into either of the equations to solve for x.

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