# Solving these Simultaneous Equations

• Darkmisc

#### Darkmisc

Homework Statement
(a+b)x + cy = bc
(b+c)y + ax = -ab
Relevant Equations
x - y = a + c

x = c
y = -a
Hi everyone

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

$$(a+b)x + cy = bc$$
$$ax + (b+c)y = -ab$$

$$a(a+b)x + acy = abc$$
$$a(a+b)x + (a+b)(b+c)y = -ab(a+b)$$

Subtracting the both sides we can delete x to get y
$$[(a+b)(b+c)- ac]y = -ab(a+b)-abc$$
$$y = \frac{-ab(a+b)-abc}{(a+b)(b+c)- ac}=...$$

• Darkmisc
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}##.

Last edited:
• Darkmisc
• Darkmisc