Question

Solve each system of equations

$\begin{array}{l}4a+2b-6c=2\\ 6a+3b-9c=3\\ 8a+4b-12c=6\end{array}$

Short answer

The solution set of the given system of equations is infinitely many solutions.

Step-by-step solution

– Use the elimination method to get the system of equations in two variables.

Multiply the equation $4a+2b-6c=2$by $3$and multiply the equation $6a+3b-9c=3$by $2$.

Perform the subtraction operation on the new two resultant equations.

$\begin{array}{l}\underset{\_}{\begin{array}{l}4a+2b-6c=2\\ 6a+3b-9c=3\end{array}}\begin{array}{c}\to \text{multiplyby3}\to \\ \to \text{multiplyby2}\to \end{array}\underset{\_}{\begin{array}{l}12a+6b-18c=6\\ (-)12a+6b-18c=6\end{array}}\\ 0=0\end{array}$

– Conclude the solutions of the given system of equations.

Here, $0=0$. This means the second equation $6a+3b-9c=3$is a multiple of first equation $4a+2b-6c=2$.

So, they are the same plane.

Hence, the solution of the given system of equations is infinitely many solutions.