Solve {l}{0.0001x_{1}+0.5u_{2}=0.5}{0.4x_{1}-0.3x_{2}=0.1}

$$\left. \begin{array} { l } { 0.0001 x _ { 1 } + 0.5 u _ { 2 } = 0.5 } \\ { 0.4 x _ { 1 } - 0.3 x _ { 2 } = 0.1 } \end{array} \right.$$

Solve for x_1, x_2

$$x_{1}=5000-5000u_{2}$$

$$x_{2}=\frac{19999-20000u_{2}}{3}$$

Steps Using Substitution

To solve a pair of equations using substitution, first solve one of the equations for one of the variables. Then substitute the result for that variable in the other equation.

$$0.0001x_{1}+\frac{u_{2}}{2}=0.5,0.4x_{1}-0.3x_{2}=0.1$$

Pick one of the two equations which is more simple to solve for x_{1} by isolating x_{1} on the left hand side of the equal sign.

$$0.0001x_{1}+\frac{u_{2}}{2}=0.5$$

Subtract \frac{u_{2}}{2} from both sides of the equation.

$$0.0001x_{1}=\frac{1-u_{2}}{2}$$

Multiply both sides by 10000.

$$x_{1}=5000-5000u_{2}$$

Substitute 5000-5000u_{2} for x_{1} in the other equation, 0.4x_{1}-0.3x_{2}=0.1.

$$0.4\left(5000-5000u_{2}\right)-0.3x_{2}=0.1$$

Multiply 0.4 times 5000-5000u_{2}.

$$2000-2000u_{2}-0.3x_{2}=0.1$$

Subtract 2000-2000u_{2} from both sides of the equation.

$$-0.3x_{2}=2000u_{2}-1999.9$$

Divide both sides of the equation by -0.3, which is the same as multiplying both sides by the reciprocal of the fraction.

$$x_{2}=\frac{19999-20000u_{2}}{3}$$

The system is now solved.

$$x_{1}=5000-5000u_{2},x_{2}=\frac{19999-20000u_{2}}{3}$$