Q16. - Algebra 2

Question

Solve each system of equations 

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

Step-by-Step Solution

Answer

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

Step 1 – 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.

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

Step 2 – 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.