Q28. - Algebra 2

Question

28: Solve each system of equations.

$\begin{array}{c}e+f=4\\ 2d+4e-f=-3\\ 3e=-3\end{array}$

Step-by-Step Solution

Answer

The solution is $d=3,e=-1,f=5$.

Step-1 – Using the elimination to find the solution.

Given equations are-:

$\begin{array}{c}e+f=\mathrm{4......}\left(1\right)\\ 2d+4e-f=-\mathrm{3......}\left(2\right)\\ 3e=-\mathrm{3......}\left(3\right)\end{array}$

From equation$\left(3\right)$, we get

$\begin{array}{c}3e=-3\\ e=-1\end{array}$

Step-2 – Putting the value of e in equation ( 1 ) , we get

$\begin{array}{c}e+f=4\\ -1+f=4\\ f=4+1\\ f=5\end{array}$

Step-3 – Substitute the value of e and f in equation with all three variables.

$\begin{array}{c}2d+4e-f=-3\\ 2d+4(-1)-5=-3\\ 2d-4-5=-3\\ 2d=-3+9\\ 2d=6\\ d=3\end{array}$