System of equations with only variables

How do you solve a system of equations with variables on both sides? I have a problem solving this:

$a + 3b = a + 2c$
$2a = b + 2d$
$c + 3d = 3a$
$2c = 3b$

I tried substituting $c$ for $1.5b$ but all I end up with is the two middle equations beeing multiples of eachother and $a = 0.5b + d$
According to the book the answer is
$a = c$ and
$b = d = (2/3)c$

So how can you come to the stage where each variable can be expressed in a single other variable?

You are correct about the dependence of the last three equations. Also note you can subtract $a$ from both sides of the first and get the fourth. You actually only have two equations in four unknowns. As the equations are homogeneous, there is nothing to set the scale. You can multiply any solution by a constant to get another solution, so you will have to express the solution as ratios of variables. You need three equations to have a unique (up to scale) solution. The book answer satisfies the equations, but there are other ways. You have $c=3b/2$ and $a=d+b/2$ as you found. Anything that satisfies these will satisfy the original set. To show the book is wrong, you can just pick $a=2c$, solve the second to find $2c=c/3+d,\ d=5c/3$ and verify this is another solution.

• Yes ok, I thought it was really strange, thanks! Commented Feb 5, 2015 at 16:52

You have four equations and four unknowns. If this system of equations has a solution, you have everything you need to solve it. One thing I learned while taking algebra is to consider the system as a matrix. First set each equation equal to zero.

$$0a+3b-2c-0d=0 \\ 2a-1b+0c-2d=0 \\ -3a+0b+1c+3d = 0 \\ 0a-3b +2c+0d = 0$$

The next part can easily be calculated using a graphing calculator. Make a $4\times 5$ and fill it out with the coefficients of your equation above. It should look like this: $$\begin{array}{columns} 0 & 3 & -2 & 0 & 0\\ 2 & -1 & 0 & -2 & 0\\ -3 & 0 & 1 & 3 & 0\\ 0 & -3 & 2 & 0& 0\\ \end{array}$$

Now you want to reduce the matrix to "Row reduced echelon form" which will ideally convert the matrix into something that looks like

$$\begin{array}{columns} 1 & 0 & 0 & 0 & x_1\\ 0 & 1 & 0 & 0 & x_2\\ 0 & 0 & 1 & 0 & x_3\\ 0 & 0 & 0 & 1 & x_4\\ \end{array}$$

where $a=x_1, \space b = x_2, \space c=x_3$ and $d = x_4$ is your final answer. There should be a button under matrix tools on the calculator that is titled "rref" and you want to calculate $\text{rref}(M)$ where $M$ is the coefficient matrix above. If your matrix cannot be completely reduced into a diagonal of $1's$ with a single column of numbers on the far right, then there is not a unique solution to the system. There might be no solutions, or there could be infinite. At any rate, knowing how to do this on a calculator is extremely useful and can save a lot of time!