Solving these Simultaneous Equations

AI Thread Summary
The discussion revolves around solving a system of simultaneous equations involving variables x and y. Initial attempts show that the equations can be manipulated to derive relationships between the variables, leading to the conclusion that bc = -ab and c = -a. A matrix approach is suggested, utilizing the inverse of a 2x2 matrix to solve the equations more efficiently. The determinant condition for matrix invertibility is noted, ensuring the method is applicable. Ultimately, the correct solutions for the variables are identified as x = c and y = -a.
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Homework Statement
(a+b)x + cy = bc
(b+c)y + ax = -ab
Relevant Equations
x - y = a + c

Answer:
x = c
y = -a
Hi everyone

Could someone please help with the above equation?

Here is the working for my attempt

ax + bx + cy = bc
by + cy + ax = -ab

ax + bx + cy = bc
by + cy + ax = -ab

b(x-y) = b(a + c)
x - y = a + c

a^2 + ac + ay + ab + bc + by + cy = bc
a+2 + ac + ay + by + by = -ab

b(a+c) = bc + ab
0 = 0

Is it correct to conclude from this that bc = -ab, and that c = -a?

If so, can I substitute that into
y (a + b + c) = -ab

to get y = -a?

The correct answer is x=c and y =-a, which fits into the above equations

Thanks
 
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(a+b)x + cy = bc
ax + (b+c)y = -aba(a+b)x + acy = abc
a(a+b)x + (a+b)(b+c)y = -ab(a+b)Subtracting the both sides we can delete x to get y
[(a+b)(b+c)- ac]y = -ab(a+b)-abc
y = \frac{-ab(a+b)-abc}{(a+b)(b+c)- ac}=...
 
A somewhat simpler way to solve this system, and possibly the technique the authors of the problem had in mind, is to use the inverse of the given matrix.
The inverse of the 2x2 matrix ##A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}## is ##A^{-1} = \frac 1 {det(A)}\begin{bmatrix} d & -b \\ -c & a\end{bmatrix}##.

The determinant of A is ##det(A) = ad - bc##. As long as ##ad - bc \ne 0##, the matrix is invertible; i.e., the inverse of A exists.

For this problem, ##A = \begin{bmatrix} a + b & -c \\ -a & a + b \end{bmatrix}##

To solve the matrix equation ##A\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} bc \\ -ab \end{bmatrix}##, apply the inverse, ##A^{-1}##, to both sides of the matrix equation above to obtain the solution ##\begin{bmatrix} x \\ y \end{bmatrix}##.
 
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