Lutasin ang operatorname{IRR}=22000/(1+r)^2-18000

$$\operatorname { IRR } = \frac { 22000 } { ( 1 + r ) ^ { 2 } } - 18000$$

I-solve ang I (complex solution)

$$\left\{\begin{matrix}I=-\frac{2000\left(9r^{2}+18r-2\right)}{\left(R\left(r+1\right)\right)^{2}}\text{, }&r\neq -1\text{ and }R\neq 0\\I\in \mathrm{C}\text{, }&\left(r=-\frac{\sqrt{11}}{3}-1\text{ or }r=\frac{\sqrt{11}}{3}-1\right)\text{ and }R=0\end{matrix}\right.$$

Mga Hakbang sa Pagso-solve ng Linear Equation

I-multiply ang magkabilang dulo ng equation gamit ang \left(r+1\right)^{2}.

$$IRR\left(r+1\right)^{2}=22000+\left(r+1\right)^{2}\left(-18000\right)$$

I-multiply ang R at R para makuha ang R^{2}.

$$IR^{2}\left(r+1\right)^{2}=22000+\left(r+1\right)^{2}\left(-18000\right)$$

Gamitin ang binomial theorem na \left(a+b\right)^{2}=a^{2}+2ab+b^{2} para palawakin ang \left(r+1\right)^{2}.

$$IR^{2}\left(r^{2}+2r+1\right)=22000+\left(r+1\right)^{2}\left(-18000\right)$$

Gamitin ang distributive property para i-multiply ang IR^{2} gamit ang r^{2}+2r+1.

$$IR^{2}r^{2}+2IR^{2}r+IR^{2}=22000+\left(r+1\right)^{2}\left(-18000\right)$$

Gamitin ang binomial theorem na \left(a+b\right)^{2}=a^{2}+2ab+b^{2} para palawakin ang \left(r+1\right)^{2}.

$$IR^{2}r^{2}+2IR^{2}r+IR^{2}=22000+\left(r^{2}+2r+1\right)\left(-18000\right)$$

Gamitin ang distributive property para i-multiply ang r^{2}+2r+1 gamit ang -18000.

$$IR^{2}r^{2}+2IR^{2}r+IR^{2}=22000-18000r^{2}-36000r-18000$$

I-subtract ang 18000 mula sa 22000 para makuha ang 4000.

$$IR^{2}r^{2}+2IR^{2}r+IR^{2}=4000-18000r^{2}-36000r$$

Pagsamahin ang lahat ng term na naglalaman ng I.

$$\left(R^{2}r^{2}+2R^{2}r+R^{2}\right)I=4000-18000r^{2}-36000r$$

Ang equation ay nasa standard form.

$$\left(R^{2}r^{2}+2rR^{2}+R^{2}\right)I=4000-36000r-18000r^{2}$$

I-divide ang magkabilang dulo ng equation gamit ang R^{2}r^{2}+2rR^{2}+R^{2}.

$$\frac{\left(R^{2}r^{2}+2rR^{2}+R^{2}\right)I}{R^{2}r^{2}+2rR^{2}+R^{2}}=\frac{4000-36000r-18000r^{2}}{R^{2}r^{2}+2rR^{2}+R^{2}}$$

Kapag na-divide gamit ang R^{2}r^{2}+2rR^{2}+R^{2}, ma-a-undo ang multiplication gamit ang R^{2}r^{2}+2rR^{2}+R^{2}.

$$I=\frac{4000-36000r-18000r^{2}}{R^{2}r^{2}+2rR^{2}+R^{2}}$$

I-divide ang 4000-36000r-18000r^{2} gamit ang R^{2}r^{2}+2rR^{2}+R^{2}.

$$I=\frac{2000\left(2-18r-9r^{2}\right)}{R^{2}\left(r+1\right)^{2}}$$

I-solve ang I

$$\left\{\begin{matrix}I=-\frac{2000\left(9r^{2}+18r-2\right)}{\left(R\left(r+1\right)\right)^{2}}\text{, }&r\neq -1\text{ and }R\neq 0\\I\in \mathrm{R}\text{, }&\left(r=-\frac{\sqrt{11}}{3}-1\text{ or }r=\frac{\sqrt{11}}{3}-1\right)\text{ and }R=0\end{matrix}\right.$$

Mga Hakbang sa Pagso-solve ng Linear Equation

I-multiply ang magkabilang dulo ng equation gamit ang \left(r+1\right)^{2}.

$$IRR\left(r+1\right)^{2}=22000+\left(r+1\right)^{2}\left(-18000\right)$$

I-multiply ang R at R para makuha ang R^{2}.

$$IR^{2}\left(r+1\right)^{2}=22000+\left(r+1\right)^{2}\left(-18000\right)$$

Gamitin ang binomial theorem na \left(a+b\right)^{2}=a^{2}+2ab+b^{2} para palawakin ang \left(r+1\right)^{2}.

$$IR^{2}\left(r^{2}+2r+1\right)=22000+\left(r+1\right)^{2}\left(-18000\right)$$

Gamitin ang distributive property para i-multiply ang IR^{2} gamit ang r^{2}+2r+1.

$$IR^{2}r^{2}+2IR^{2}r+IR^{2}=22000+\left(r+1\right)^{2}\left(-18000\right)$$

Gamitin ang binomial theorem na \left(a+b\right)^{2}=a^{2}+2ab+b^{2} para palawakin ang \left(r+1\right)^{2}.

$$IR^{2}r^{2}+2IR^{2}r+IR^{2}=22000+\left(r^{2}+2r+1\right)\left(-18000\right)$$

Gamitin ang distributive property para i-multiply ang r^{2}+2r+1 gamit ang -18000.

$$IR^{2}r^{2}+2IR^{2}r+IR^{2}=22000-18000r^{2}-36000r-18000$$

I-subtract ang 18000 mula sa 22000 para makuha ang 4000.

$$IR^{2}r^{2}+2IR^{2}r+IR^{2}=4000-18000r^{2}-36000r$$

Pagsamahin ang lahat ng term na naglalaman ng I.

$$\left(R^{2}r^{2}+2R^{2}r+R^{2}\right)I=4000-18000r^{2}-36000r$$

Ang equation ay nasa standard form.

$$\left(R^{2}r^{2}+2rR^{2}+R^{2}\right)I=4000-36000r-18000r^{2}$$

I-divide ang magkabilang dulo ng equation gamit ang R^{2}r^{2}+2rR^{2}+R^{2}.

$$\frac{\left(R^{2}r^{2}+2rR^{2}+R^{2}\right)I}{R^{2}r^{2}+2rR^{2}+R^{2}}=\frac{4000-36000r-18000r^{2}}{R^{2}r^{2}+2rR^{2}+R^{2}}$$

Kapag na-divide gamit ang R^{2}r^{2}+2rR^{2}+R^{2}, ma-a-undo ang multiplication gamit ang R^{2}r^{2}+2rR^{2}+R^{2}.

$$I=\frac{4000-36000r-18000r^{2}}{R^{2}r^{2}+2rR^{2}+R^{2}}$$

I-divide ang 4000-18000r^{2}-36000r gamit ang R^{2}r^{2}+2rR^{2}+R^{2}.

$$I=\frac{2000\left(2-18r-9r^{2}\right)}{\left(R\left(r+1\right)\right)^{2}}$$

I-solve ang R (complex solution)

$$\left\{\begin{matrix}R=-\frac{20iI^{-\frac{1}{2}}\sqrt{45r^{2}+90r-10}}{r+1}\text{; }R=\frac{20iI^{-\frac{1}{2}}\sqrt{45r^{2}+90r-10}}{r+1}\text{, }&r\neq -1\text{ and }I\neq 0\\R\in \mathrm{C}\text{, }&\left(r=-\frac{\sqrt{11}}{3}-1\text{ or }r=\frac{\sqrt{11}}{3}-1\right)\text{ and }I=0\end{matrix}\right.$$

I-solve ang R

$$\left\{\begin{matrix}R=\frac{20\sqrt{-\frac{5\left(9r^{2}+18r-2\right)}{I}}}{|r+1|}\text{; }R=-\frac{20\sqrt{-\frac{5\left(9r^{2}+18r-2\right)}{I}}}{|r+1|}\text{, }&\left(I<0\text{ or }r\leq \frac{\sqrt{11}}{3}-1\right)\text{ and }\left(I<0\text{ or }r\geq -\frac{\sqrt{11}}{3}-1\right)\text{ and }r\neq -1\text{ and }\left(I>0\text{ or }r\geq \frac{\sqrt{11}}{3}-1\text{ or }r\leq -\frac{\sqrt{11}}{3}-1\right)\text{ and }I\neq 0\\R\in \mathrm{R}\text{, }&\left(r=-\frac{\sqrt{11}}{3}-1\text{ or }r=\frac{\sqrt{11}}{3}-1\right)\text{ and }I=0\end{matrix}\right.$$