Solve {l}{4u_{1}-4u_{2}=-1/100}{-4u_{1}+10u_{2}-6u_{3}=0}{18u_{3}-6u_{2}=1/200}

$$\left\{ \begin{array} { l } { 4 u _ { 1 } - 4 u _ { 2 } = - \frac { 1 } { 100 } } \\ { - 4 u _ { 1 } + 10 u _ { 2 } - 6 u _ { 3 } = 0 } \\ { 18 u _ { 3 } - 6 u _ { 2 } = \frac { 1 } { 200 } } \end{array} \right.$$

Solve for u_1, u_2, u_3

$$u_{1}=-\frac{11}{2400}\approx -0.004583333$$

$$u_{2}=-\frac{1}{480}\approx -0.002083333$$

$$u_{3}=-\frac{1}{2400}\approx -0.000416667$$

Short Steps Using Substitution

Solve 4u_{1}-4u_{2}=-\frac{1}{100} for u_{1}.

$$u_{1}=u_{2}-\frac{1}{400}$$

Substitute u_{2}-\frac{1}{400} for u_{1} in the equation -4u_{1}+10u_{2}-6u_{3}=0.

$$-4\left(u_{2}-\frac{1}{400}\right)+10u_{2}-6u_{3}=0$$

Solve the second equation for u_{2} and the third equation for u_{3}.

$$u_{2}=-\frac{1}{600}+u_{3} u_{3}=\frac{1}{3}u_{2}+\frac{1}{3600}$$

Substitute -\frac{1}{600}+u_{3} for u_{2} in the equation u_{3}=\frac{1}{3}u_{2}+\frac{1}{3600}.

$$u_{3}=\frac{1}{3}\left(-\frac{1}{600}+u_{3}\right)+\frac{1}{3600}$$

Solve u_{3}=\frac{1}{3}\left(-\frac{1}{600}+u_{3}\right)+\frac{1}{3600} for u_{3}.

$$u_{3}=-\frac{1}{2400}$$

Substitute -\frac{1}{2400} for u_{3} in the equation u_{2}=-\frac{1}{600}+u_{3}.

$$u_{2}=-\frac{1}{600}-\frac{1}{2400}$$

Calculate u_{2} from u_{2}=-\frac{1}{600}-\frac{1}{2400}.

$$u_{2}=-\frac{1}{480}$$

Substitute -\frac{1}{480} for u_{2} in the equation u_{1}=u_{2}-\frac{1}{400}.

$$u_{1}=-\frac{1}{480}-\frac{1}{400}$$

Calculate u_{1} from u_{1}=-\frac{1}{480}-\frac{1}{400}.

$$u_{1}=-\frac{11}{2400}$$

The system is now solved.

$$u_{1}=-\frac{11}{2400} u_{2}=-\frac{1}{480} u_{3}=-\frac{1}{2400}$$