Respuesta :

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:

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

For the blue line:

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

Step 2:

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

[tex]\frac{y-y_1}{x-x_1}=\frac{y_2-y_1}{x_2-x_1}[/tex]

Step 3:

Substituting the appropriate values in the above formula.

For the red line:

[tex]\begin{gathered} \frac{y-2_{}}{x-0_{}}=\frac{0_{}-2_{}}{-6_{}-0_{}} \\ \\ \frac{y-2}{x}=\frac{-2}{-6} \\ \\ \frac{y-2}{x}=\frac{1}{3} \end{gathered}[/tex]

Cross multiplying, we get

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

For the blue line:

[tex]\begin{gathered} \frac{y-6_{}}{x-0_{}}=\frac{1_{}-6_{}}{-3_{}-0_{}} \\ \\ \frac{y-6}{x}=\frac{-5}{-3} \\ \\ \frac{y-6}{x}=\frac{5}{3} \end{gathered}[/tex]

Cross multiplying, we get

[tex]\begin{gathered} y-6=\frac{5}{3}x \\ \\ y=\frac{5}{3}x+6 \end{gathered}[/tex]

Therefore, the system of equations is:

[tex]\begin{gathered} y=\frac{1}{3}x+2 \\ \\ y=\frac{5}{3}x+6 \end{gathered}[/tex]

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