system of equations ?????

G

Guest

Guest
ax-by=c
bx+ay=d

a,b,c,d represent constants
Solve the system and give the conditions that the system has a unique solution.
 
Hello, bryan-duong!

(1) ax - by = c
(2) bx + ay = d

a,b,c,d represent constants
Solve the system and give the conditions that the system has a unique solution.
I assume you know the Elimination method.

Multiply (1) by a: .a<sup>2</sup>x - aby . = .ac
Multiply (2) by b: .b<sup>2</sup>x + aby .= .bd

Add: .a<sup>2</sup>x + b<sup>2</sup>x .= .ac + bd . . . . (This eliminates the y-terms)

Factor: .(a<sup>2</sup> + b<sup>2</sup>)x .= .ac + bd

. . . . . . . . . . .ac + bd
Then: . x .= .-----------
. . . . . . . . . . .a<sup>2</sup> + b<sup>2</sup>


Now use the same procedure to eliminate the x-terms

. . . . . . . . . . . . . . .ad - bc
. . and get: . y .= .----------
. . . . . . . . . . . . . . .a<sup>2</sup> + b<sup>2</sup>

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

The system will not have a unique solution if the denominators equal zero.
. . This means: .a<sup>2</sup> + b<sup>2</sup> = 0

Clearly this is impossible. **

The system will <u>always</u> have a unique solution.

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Well, okay . . . it is possible if a = 0 and b = 0
. . but then we'd have: .0 = c .and .0 = d
. . and we wouldn't have a system, would we?
 
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