Solving 3 equation

Question

I have a problem with these three equations. I try to solve them by assumption and no thing

$$2 = \frac{8}{R1[\frac{1}{R1}+\frac{1}{R2}+\frac{1}{R3}]}$$

$$3 = \frac{8}{R2[\frac{1}{R1}+\frac{1}{R2}+\frac{1}{R3}]}$$

$$1 = \frac{8}{R3[\frac{1}{R1}+\frac{1}{R2}+\frac{1}{R3}]}$$

Answer 1

$$2 = \frac{8}{R1[\frac{1}{R1}+\frac{1}{R2}+\frac{1}{R3}]}$$

$$3 = \frac{8}{R2[\frac{1}{R1}+\frac{1}{R2}+\frac{1}{R3}]}$$

$$1 = \frac{8}{R3[\frac{1}{R1}+\frac{1}{R2}+\frac{1}{R3}]}$$

Let $\frac{1}{R1}=x$, $\frac{1}{R2}=y$, $\frac{1}{R3}=z$

So now,

$x+y+z=4x$, that is, $3x-y-z=0 -(1)$

$x+y+z=\frac{8}{3}y$, that is, $3x-5y+3z=0 - (2)$

$x+y+z=8z$, that is, $x+y-7z=0 - (3)$

Now solve the 3 equations to get unique solution and replace $(R1, R2, R3)$ with $(\frac{1}{x},\frac{1}{y},\frac{1}{z})$ to get the required answer.

Also, the trivial solution (x,y,z)=(0,0,0) won't work for this.

Answer 2

Let

$$1/R_1 = x, 1/R_1 = y, 1/R_3 = z$$

$$3x-y-z=0 \tag1 $$

$$3x-5y+3z=0 \tag2 $$

$$x+y-7z=0 \tag3 $$

Three lines through origin.

$$ x= y= z= 0, \, R_1 = R2=R_3= \infty.$$

Answer 3

Adding the three equations,

$$6=\left(\frac1{R_1}+\frac1{R_2}+\frac1{R_3}\right)\frac8{\dfrac1{R_1}+\dfrac1{R_2}+\dfrac1{R_3}}.$$

This will be hard to achieve...

Answer 4

Thanks for all solutions In my problem i want to achieve this equation vo =2v1 +3V2 +V3 using one op amp i tried with three resistor R1 R2 R3 connected to non inverting input with gain = 8 , my analysis lead to the problem that i have posted
with procedure of Maadhav Gupta and adding other resistor R4 connected to ground and non inverting input and have any value i try with 9 problem can be solved . The three equation will be the same but adding 9 to this part [1/R1 +1/R2 +1/R3 ] now x=9 , y =27/2 , z =9/2 and R4 =1/9 as i assumed