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=(3)/(2)x+4,(2) \end{gathered}`

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

`y-y=x+6-((3)/(2)x+4)`

Now we solve the parenthesis using the distributive law:

`y-y=x+6-(3)/(2)x-4`

Now we add like terms;

`0=-(x)/(2)+2`

Now we add x/2 to both sides:

`\begin{gathered} (x)/(2)=-(x)/(2)+(x)/(2)+2 \\ (x)/(2)=2 \end{gathered}`

Now we multiply both sides by 2:

`\begin{gathered} (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)`