Solving these Simultaneous Equations

In summary, the authors of the problem suggest using the inverse of the given matrix to solve for the solution.
  • #1
Darkmisc
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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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  • #2
[tex](a+b)x + cy = bc[/tex]
[tex]ax + (b+c)y = -ab[/tex][tex]a(a+b)x + acy = abc[/tex]
[tex]a(a+b)x + (a+b)(b+c)y = -ab(a+b)[/tex]Subtracting the both sides we can delete x to get y
[tex][(a+b)(b+c)- ac]y = -ab(a+b)-abc[/tex]
[tex]y = \frac{-ab(a+b)-abc}{(a+b)(b+c)- ac}=...[/tex]
 
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  • #3
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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  • #4
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1. How do I solve simultaneous equations?

To solve simultaneous equations, you need to use a method called substitution or elimination. In substitution, you solve for one variable in one equation and substitute it into the other equation. In elimination, you add or subtract the equations to eliminate one variable and then solve for the remaining variable.

2. What is the purpose of solving simultaneous equations?

The purpose of solving simultaneous equations is to find the values of the variables that satisfy both equations at the same time. This can be useful in solving real-world problems, such as finding the intersection point of two lines or determining the quantities of two different items that need to be mixed together.

3. Can simultaneous equations have more than two variables?

Yes, simultaneous equations can have any number of variables. However, the number of equations should be equal to the number of variables in order to find a unique solution. If there are more equations than variables, the system is considered overdetermined and may not have a solution.

4. What if I get a negative or zero solution when solving simultaneous equations?

If you get a negative or zero solution when solving simultaneous equations, it means that the equations are inconsistent and do not have a solution. This could happen if the equations represent parallel lines or if there is an error in the equations.

5. Can simultaneous equations be solved using a graphing calculator?

Yes, simultaneous equations can be solved using a graphing calculator. Most graphing calculators have a function for solving systems of equations, where you can input the equations and the calculator will give you the solutions. However, it is still important to understand the underlying concepts and methods for solving simultaneous equations by hand.

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