Solve {l}{100=15x_{1}-5x_{2}}{80=20x_{4}+5x_{1}}

$$\left. \begin{array} { l } { 100 = 15 x _ { 1 } - 5 x _ { 2 } } \\ { 80 = 20 x _ { 4 } + 5 x _ { 1 } } \end{array} \right.$$

Solve for x_1, x_2

$$x_{1}=16-4x_{4}$$

$$x_{2}=28-12x_{4}$$

Steps Using Substitution

Consider the second equation. Swap sides so that all variable terms are on the left hand side.

$$20x_{4}+5x_{1}=80$$

Subtract 20x_{4} from both sides.

$$5x_{1}=80-20x_{4}$$

Consider the first equation. Swap sides so that all variable terms are on the left hand side.

$$15x_{1}-5x_{2}=100$$

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.

$$5x_{1}=80-20x_{4},15x_{1}-5x_{2}=100$$

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.

$$5x_{1}=80-20x_{4}$$

Divide both sides by 5.

$$x_{1}=16-4x_{4}$$

Substitute 16-4x_{4} for x_{1} in the other equation, 15x_{1}-5x_{2}=100.

$$15\left(16-4x_{4}\right)-5x_{2}=100$$

Multiply 15 times 16-4x_{4}.

$$240-60x_{4}-5x_{2}=100$$

Subtract 240-60x_{4} from both sides of the equation.

$$-5x_{2}=60x_{4}-140$$

Divide both sides by -5.

$$x_{2}=28-12x_{4}$$

The system is now solved.

$$x_{1}=16-4x_{4},x_{2}=28-12x_{4}$$