Q19. - Algebra 2

Question

Solve each system of equations 

$\begin{array}{l}4a-2b+8c=30\\ a+2b-7c=-12\\ 2a-b+4c=15\end{array}$

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 $2a-b+4c=15$ by $2$and add the new resultant equation to the equation $4a-2b+8c=30$.

Perform the subtraction operation on the new two resultant equations

$\begin{array}{l}\underset{\_}{\begin{array}{l}2a-b+4c=15\\ 4a-2b+8c=30\end{array}}\begin{array}{c}\to \text{multiply by 2}\to \\ \end{array}\underset{\_}{\begin{array}{l}4a-2b+8c=30\\ (-)4a-2b+8c=30\end{array}}\\ 0=0\end{array}$

Step 2 – Conclude the solutions of the given system of equations.

Here, $0=0$. This means the third equation $2a-b+4c=15$is a multiple of first equation $4a-2b+8c=30$.

 So, they are the same plane.

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