9.2 Systems of linear equations: three variables  (Page 3/8)

Looking at the coefficients of $x,$ we can see that we can eliminate $x$ by adding equation (1) to equation (2).

Next, we multiply equation (1) by $\mathrm{-5}$ and add it to equation (3).

Then, we multiply equation (4) by 2 and add it to equation (5).

The final equation $0=2$ is a contradiction, so we conclude that the system of equations in inconsistent and, therefore, has no solution.

First, we can multiply equation (1) by $\mathrm{-2}$ and add it to equation (2).

We do not need to proceed any further. The result we get is an identity, $0=0,$ which tells us that this system has an infinite number of solutions. There are other ways to begin to solve this system, such as multiplying equation (3) by $\mathrm{-2},$ and adding it to equation (1). We then perform the same steps as above and find the same result, $0=0.$

When a system is dependent, we can find general expressions for the solutions. Adding equations (1) and (3), we have

We then solve the resulting equation for $z.$

We back-substitute the expression for $z$ into one of the equations and solve for $y.$

So the general solution is $\left(x,\frac{5}{2}x,\frac{3}{2}x\right).$ In this solution, $x$ can be any real number. The values of $y$ and $z$ are dependent on the value selected for $x.$