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Two Equations with Two Unknowns


Date: 05/24/99 at 01:00:47
From: Margie Yeager
Subject: Algebra 2

Solve by adding or subtracting:

1)  2x-3y = 50
    7x+8y = -10

2)  5x+6y = 16
    3x-4y = 2

3)  3x+2y = 14
    2x+3y = 1

4)  2x+3y = 10
    5x-4y = 2

5)  10x+3y = 1
    3x-2y = -10
 
6)  5x+2y = 9
    2x+3y = -3


Date: 05/24/99 at 18:03:00
From: Doctor Rick
Subject: Re: Algebra 2

Hi, Margie. I will do the first of your problems to show you the 
technique.

Before you add or subtract, you want to multiply one or both equations 
by constants (that is, multiply both sides of one equation by the same 
number). Your goal is to eliminate one variable when you do the 
addition or subtraction. 

Let's eliminate the y. To do this, we want to have the coefficient of 
y in the two equations be either the same or the opposite. One way to 
do this is to multiply each equation by the coefficient of y in the 
OTHER equation.

Multiply the first equation on both sides by 8:

  16x - 24y = 400

Multiply the second equation on both sides by 3:

  21x + 24y = -30

Now the coefficients of y are opposite, so we can add the two 
equations:

  16x - 24y = 400
  21x + 24y = -30
  ---------------
  37x +  0y = 370

The y term has vanished:

  37x = 370

Divide both sides by 37:

  x = 370/37 = 10

We aren't done yet because we only have a value for x, not for y. 
Choose either of the original equations, substitute x = 10, then solve 
for y:

     2x - 3y = 50
  2(10) - 3y = 50
     20 - 3y = 50
         -3y = 50 - 20 = 30
           y = 30/(-3) = -10

The solution set is { x = 10, y = -10 }.
  
Last but not least, check the answer by substituting these values in 
both equations:

  2x - 3y = 50              7x + 8y = -10
  2(10) - 3(-10) = 50       7(10) + 8(-10) = -10
  20 + 30 = 50              70 - 80 = -10
  50 = 50                   -10 = -10

We're done! Now it's your turn.

- Doctor Rick, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Linear Equations

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