Answer
$R_{1}=10$ Ω
$R_{2}=4$ Ω
Work Step by Step
To solve this system of equations, we use the graphing method.
$R_{1}+R_{2}=14$
$R_{1}-R_{2}=6$
Taking the first equation, we solve for $R_{2}$.
$R_{1}+R_{2}=14$
$R_{2}=14-R_{1}$
Find three solutions:
For $R_{1}$=2,
$R_{2}=14-R_{1}$
$R_{2}=14-2$
$R_{2}=12$
For $R_{1}$=0,
$R_{2}=14-R_{1}$
$R_{2}=14-0$
$R_{2}=14$
For $R_{1}$=-2,
$R_{2}=14-R_{1}$
$R_{2}=14-(-2)$
$R_{2}=14+2$
$R_{2}=16$
With the three points, $(2,12), (0,14), (-2,16)$, we can graph the straight line that goes through these points.
Taking the second equation, we solve for $R_{2}$.
$R_{1}-R_{2}=6$
$R_{2}=R_{1}-6$
Find three solutions:
For $R_{1}$=2,
$R_{2}=R_{1}-6$
$R_{2}=2-6$
$R_{2}=-4$
For $R_{1}$=0,
$R_{2}=R_{1}-6$
$R_{2}=0-6$
$R_{2}=-6$
For $R_{1}$=-2,
$R_{2}=R_{1}-6$
$R_{2}=-2-6$
$R_{2}=-8$
With the three points, $(2,-4), (0,-6), (-2,-8)$, we can graph the straight line that goes through these points.
The intersection point between these two lines is the answer to the system of equations.
