Question

Part EThe equations you wrote in parts B and D form a system of equations. Solve the system of equations algebraically. What is the solution? Is it thesame solution that you found graphically? Equation 1: y=x+6 Equation2: y=3/2x+4

Answer 1

We are given the following system of equations:

$\begin{gathered} y=x+6,(1) \\ y=\frac{3}{2}x+4,(2) \end{gathered}$

To determine the solution we will subtract equation (2) from equation (1):

$y-y=x+6-(\frac{3}{2}x+4)$

Now we solve the parenthesis using the distributive law:

$y-y=x+6-\frac{3}{2}x-4$

Now we add like terms;

$0=-\frac{x}{2}+2$

Now we add x/2 to both sides:

$\begin{gathered} \frac{x}{2}=-\frac{x}{2}+\frac{x}{2}+2 \\ \frac{x}{2}=2 \end{gathered}$

Now we multiply both sides by 2:

$\begin{gathered} \frac{2x}{2}=4 \\ x=4 \end{gathered}$

Therefore, x = 4

Now we replace the value of "x" is equation (1):

$\begin{gathered} y=x+6 \\ y=4+6 \\ y=10 \end{gathered}$

Therefore, the solution is:

$(x,y)=(4,10)$