Question

What system of equations is represented by the graph? Write in equations in y=Mx+b form

Answer 1

The Solution:

Given the linear graphs in the Question section, we are required to find the system of equations represented in the given graphs (in the form: y=mx+b)

Step 1:

Pick two points on each of the given lines.

For the red line:

`\begin{gathered} (0,2)\text{ and (-6,0)} \\ (x_1=0,y_1=2) \\ (x_2=-6,y_2=0) \end{gathered}`

For the blue line:

`\begin{gathered} (0,6)\text{ and (-3,1)} \\ (x_1=0,y_1=6) \\ (x_2=-3,y_2=1) \end{gathered}`

Step 2:

The formula for the equation of a line when two points are given is:

`(y-y_1)/(x-x_1)=(y_2-y_1)/(x_2-x_1)`

Step 3:

Substituting the appropriate values in the above formula.

For the red line:

`\begin{gathered} \frac{y-2_{}}{x-0_{}}=\frac{0_{}-2_{}}{-6_{}-0_{}} \\ \\ (y-2)/(x)=(-2)/(-6) \\ \\ (y-2)/(x)=(1)/(3) \end{gathered}`

Cross multiplying, we get

`\begin{gathered} y-2=(1)/(3)x \\ \\ \text{ writing the above equation in the form: y=ma+b} \\ \text{ We have that:} \\ y=(1)/(3)x+2 \end{gathered}`

For the blue line:

`\begin{gathered} \frac{y-6_{}}{x-0_{}}=\frac{1_{}-6_{}}{-3_{}-0_{}} \\ \\ (y-6)/(x)=(-5)/(-3) \\ \\ (y-6)/(x)=(5)/(3) \end{gathered}`

Cross multiplying, we get

`\begin{gathered} y-6=(5)/(3)x \\ \\ y=(5)/(3)x+6 \end{gathered}`

Therefore, the system of equations is:

`\begin{gathered} y=(1)/(3)x+2 \\ \\ y=(5)/(3)x+6 \end{gathered}`