Question

Hello, I need help with #50, #51, and #52 please.Thank You!

Answer 1

Okay, here we have this:

Considering the provided equation system, we are going to solve it, so we obtain the following:

`\begin{bmatrix}x+4y=10 \\ 3x-4y=6\end{bmatrix}`

We will start by clearing for x for the first equation:

`\begin{gathered} x+4y=10 \\ x+4y-4y=10-4y \\ x=10-4y \end{gathered}`

Now, we are going to replace "x" with this in the second equation:

`\begin{gathered} \begin{bmatrix}3\mleft(10-4y\mright)-4y=6\end{bmatrix} \\ \begin{bmatrix}30-12y-4y=6\end{bmatrix} \\ \begin{bmatrix}30-16y=6\end{bmatrix} \\ \begin{bmatrix}30-16y=6\end{bmatrix} \\ \begin{bmatrix}-16y=-24\end{bmatrix} \\ \begin{bmatrix}y=(-24)/(-16)\end{bmatrix} \\ \begin{bmatrix}y=(3)/(2)\end{bmatrix} \\ \begin{bmatrix}y=1.5\end{bmatrix} \end{gathered}`

And, finally we are going to replace "y" with 1.5 in the equation that we found for x:

`\begin{gathered} x=10-4\cdot1.5 \\ x=10-6 \\ x=4 \end{gathered}`

Finally we obtain that solution of the system is (4, 1.5).