Solve {l}{3x_{1}+x_{2}+2x_{3}=8}{2x_{1}+2x_{3}=6}

$$\left. \begin{array}{l}{ 3 x _ { 1 } + x _ { 2 } + 2 x _ { 3 } = 8 }\\{ 2 x _ { 1 } + 2 x _ { 3 } = 6 }\end{array} \right.$$

Solve for x_1, x_2

$$x_{1}=3-x_{3}$$

$$x_{2}=x_{3}-1$$

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.

$$2x_{1}+2x_{3}=6,3x_{1}+x_{2}+2x_{3}=8$$

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.

$$2x_{1}+2x_{3}=6$$

Subtract 2x_{3} from both sides of the equation.

$$2x_{1}=6-2x_{3}$$

Divide both sides by 2.

$$x_{1}=3-x_{3}$$

Substitute 3-x_{3} for x_{1} in the other equation, 3x_{1}+x_{2}+2x_{3}=8.

$$3\left(3-x_{3}\right)+x_{2}+2x_{3}=8$$

Multiply 3 times 3-x_{3}.

$$9-3x_{3}+x_{2}+2x_{3}=8$$

Add 9-3x_{3} to 2x_{3}.

$$x_{2}+9-x_{3}=8$$

Subtract 9-x_{3} from both sides of the equation.

$$x_{2}=x_{3}-1$$

The system is now solved.

$$x_{1}=3-x_{3},x_{2}=x_{3}-1$$