Find the ordered pair of numbers (x,y) for which 123x+321y=345 and 321x+123y=543

Printable View

- Mar 1st 2007, 07:59 AMRimaspairs of numbers
Find the ordered pair of numbers (x,y) for which 123x+321y=345 and 321x+123y=543

- Mar 1st 2007, 10:26 AMSoroban
Hello, Rimas!

Quote:

Find the ordered pair of numbers (x,y) for which:

. . 123x + 321y .= .345

. . 321x + 123y .= .543

. . You*do*know how, right?

- Mar 1st 2007, 01:06 PMRimas
no i dont

- Mar 1st 2007, 01:48 PMJhevon
We have:

123x+321y=345 .........................(1)

321x+123y=543 .........................(2)

Divide the first equation by 123 and the second by 321 to obtain:

x + 2.6098y = 2.8049 ..................(3) = (1)/123

x + 0.3832y = 1.6916 ..................(4) = (2)/321

Now subtract equation (4) from equation (3) to get rid of the x, and then solve for y

=> 2.2266y = 1.1133 ....................(3) - (4)

=> y = 1.1133/2.2266 = 1/2

Plug in the value you got for y into any one of the first equations to find x

but 123x + 321y = 345

=> 123x + (1/2)(321) = 345

=> 123x = 345 - 321/2 = 184.5

=> x = 184.5/123 = 3/2

Verify your answers by plugging in the numbers you got into the left hand side of both equations and see that you get the correct right hand side. These answers were correct, so:

The numbers (x,y) are (3/2 , 1/2)