Lutasin ang {l}{text{I}5x+2y+z=19}{piquad2x-3y-2z=10}{text{IIT}10x+y-2z=42}

$$\left. \begin{array} { l } { \text { I } 5 x + 2 y + z = 19 } \\ { \pi \quad 2 x - 3 y - 2 z = 10 } \\ { \text { IIT } 10 x + y - 2 z = 42 } \end{array} \right.$$

I-solve ang x, y, z

$$x = \frac{16}{\pi + 3} \approx 2.605187433$$

$$y = \frac{16 {(\pi - 1)}}{\pi + 3} \approx 5.579250268$$

$$z=\frac{9-13\pi }{\pi +3}\approx -5.184437701$$

Maiikling Hakbang sa Paggamit ng Substitution

Lutasin ang 5x+2y+z=19 para sa z.

$$z=-5x-2y+19$$

I-substitute ang -5x-2y+19 para sa z sa pangalawa at pangatlong equation.

$$\pi \times 2x-3y-2\left(-5x-2y+19\right)=10 10x+y-2\left(-5x-2y+19\right)=42$$

Lutasin ang mga equation na ito para sa y at x nang naaayon.

$$y=-10x-2\pi x+48 x=4-\frac{1}{4}y$$

I-substitute ang -10x-2\pi x+48 para sa y sa equation na x=4-\frac{1}{4}y.

$$x=4-\frac{1}{4}\left(-10x-2\pi x+48\right)$$

Lutasin ang x=4-\frac{1}{4}\left(-10x-2\pi x+48\right) para sa x.

$$x=16\left(\pi +3\right)^{-1}$$

I-substitute ang 16\left(\pi +3\right)^{-1} para sa x sa equation na y=-10x-2\pi x+48.

$$y=-10\times 16\left(\pi +3\right)^{-1}-2\pi \times 16\left(\pi +3\right)^{-1}+48$$

Kalkulahin ang y mula sa y=-10\times 16\left(\pi +3\right)^{-1}-2\pi \times 16\left(\pi +3\right)^{-1}+48.

$$y=-16\left(3+\pi \right)^{-1}+16\left(3+\pi \right)^{-1}\pi$$

I-substitute ang -16\left(3+\pi \right)^{-1}+16\left(3+\pi \right)^{-1}\pi para sa y at ang 16\left(\pi +3\right)^{-1} para sa x sa equation na z=-5x-2y+19.

$$z=-5\times 16\left(\pi +3\right)^{-1}-2\left(-16\left(3+\pi \right)^{-1}+16\left(3+\pi \right)^{-1}\pi \right)+19$$

Kalkulahin ang z mula sa z=-5\times 16\left(\pi +3\right)^{-1}-2\left(-16\left(3+\pi \right)^{-1}+16\left(3+\pi \right)^{-1}\pi \right)+19.

$$z=9\left(3+\pi \right)^{-1}-13\left(3+\pi \right)^{-1}\pi$$

Nalutas na ang system.

$$x=16\left(\pi +3\right)^{-1} y=-16\left(3+\pi \right)^{-1}+16\left(3+\pi \right)^{-1}\pi z=9\left(3+\pi \right)^{-1}-13\left(3+\pi \right)^{-1}\pi$$