Question

I need help trying to write a system of linear equations for the graph below

Answer 1

We need to find the equation in slope-intersect form

`y=mx+b`

of the given lines.

For the horizontal line, we can see that it passes through points

`\begin{gathered} (x_1,\text{y}_1)=(0,7) \\ (x_1,y_2)=(4,6) \end{gathered}`

the its slope (m) is given by

`m=(y_2-y_1)/(x_2-x_1)=(6-7)/(4-0)=(-1)/(4)=-(1)/(4)`

Then, the line equation has the form

`y=-(1)/(4)x+b`

where b is the y-intercept. From the picture, we can see that the line crosses the y-axis at y=7, therefore, b=7. Then, the line equation for the horizontal line is

`y=-(1)/(4)x+7`

Similarly, we can apply the same procedure for the other line. We can see that it passes through points

`\begin{gathered} (x_1,\text{y}_1)=(0,-2) \\ (x_2,\text{y}_2)=(4,6) \end{gathered}`

Then, the slope (m) of this line is given by

`m=(y_2-y_1)/(x_2-x_1)=(6-(-2))/(4-0)=(6+2)/(4)=(8)/(4)=2`

Then, the line equation has the form

`y=2x+b`

Since this line crosses y-axis at y=-2 then b=-2. Hence, the equation is

`y=2x-2`

In summary, the system of linear equations is:

`\begin{gathered} y=-(1)/(4)x+7 \\ y=2x-2 \end{gathered}`