 Reggie

2021-02-09

Solve the following system of equations. (Write your answers as a comma-separated list. If there are infinitely many solutions, write a parametric solution using t and or s. If there is no solution, write NONE.)
${x}_{1}+2{x}_{2}+6{x}_{3}=6$
${x}_{1}+{x}_{2}+3{x}_{3}=3$
$\left({x}_{1},{x}_{2},{x}_{3}\right)=$? Clara Reese

Step 1
Given to solve the system of equations.
The system of equations can be solved using elimination.
To eliminate ${x}_{­3}$, the second equation is multiplied by 2 and subtracted from the first equation.

${x}_{1}+2{x}_{2}+6{x}_{3}=6$

${x}_{1}+{x}_{2}+3{x}_{3}=3$

$\left({x}_{1}+2{x}_{2}+6{x}_{3}\right)-2\left({x}_{1}+{x}_{2}+3{x}_{3}\right)=6-2\left(3\right)$

${x}_{1}+2{x}_{2}+6{x}_{3}-2{x}_{1}-2{x}_{2}-6{x}_{3}=6-6-{x}_{1}=0$

${x}_{1}=0$

Step 2

Plugging the value of ${x}_{1}$ in the first and second equations:

It is seen that the equations after plugging the value of ${x}_{1}$ are same. Hence, there are infinitely many solutions that satisfy the equation

${x}_{2}+3{x}_{3}=3$

Hence, let ${x}_{3}=t.$

So the value of ${x}_{2}$ is given by:

Hence, the solution to the system of equations is given by

$\left({x}_{1},{x}_{2},{x}_{3}\right)=\left(0,3–3t,t\right)$

$\left(0\right)+2{x}_{2}+6{x}_{3}=6⇒2{x}_{2}+6{x}_{3}=6⇒{x}_{2}+3{x}_{3}=3$

$\left(0\right)+{x}_{2}+3{x}_{3}=3⇒{x}_{2}+3{x}_{3}=3$

${x}_{2}+3{x}_{3}=3$

${x}_{2}+3\left(t\right)=3$

${x}_{2}+3t=3$

${x}_{2}=3-3t$

Step 3

Result:

$\left({x}_{1},{x}_{2},{x}_{3}\right)=\left(0,3–3t,t\right)$ Jeffrey Jordon