Type the correct answer in each box. If necessary, use / for the fraction bar. Find the solution to this system of equations. Ity = 1 21-y+z = 1 1+2y+z IL y = = Reset Next

Type the correct answer in each box If necessary use for the fraction bar Find the solution to this system of equations Ity 1 21yz 1 12yz IL y Reset Next class=

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ANSWER

• x = 1/3

,

• y = 2/3

,

• z = 1

EXPLANATION

There are many methods to solve a linear system of equations, but in this case we have to use the substitution method -which consists in clearing one variable as a function of the other/s and replace in another equation. For a system of three variables such as this one, we have to do this twice:

1° clear x from the first equation:

[tex]\begin{gathered} x+y=1 \\ x=1-y \end{gathered}[/tex]

Replace x by this expression in the second equation:

[tex]\begin{gathered} 2x-y+z=1 \\ 2(1-y)-y+z=1 \end{gathered}[/tex]

Note that now we have two variables, y and z. Before the next step we have to rewrite the equation above so that we only have one y:

[tex]\begin{gathered} 2\cdot1-2y-y+z=1 \\ 2-3y+z=1 \end{gathered}[/tex]

2° clear y from the equation above:

[tex]\begin{gathered} -3y+z=1-2 \\ -3y=-1-z \\ y=\frac{-(1+z)}{3} \\ y=\frac{1+z}{3} \end{gathered}[/tex]

And replace y by this expression in the last equation. Note that the third equation also contains x, so we have to replace first x as a function of y like in the first step:

[tex]\begin{gathered} x+2y+z=\frac{8}{3} \\ (1-y)+2y+z=\frac{8}{3} \end{gathered}[/tex]

Rewrite it so we only se one y:

[tex]\begin{gathered} 1-y+2y+z=\frac{8}{3} \\ 1+y+z=\frac{8}{3} \end{gathered}[/tex]

And now we replace y by the expression we found in the second step:

[tex]1+\frac{1+z}{3}+z=\frac{8}{3}[/tex]

So now we have one equation with one variable. Let's solve for z:

[tex]\begin{gathered} 1+\frac{1}{3}+\frac{z}{3}+z=\frac{8}{3} \\ \frac{4}{3}+\frac{4z}{3}=\frac{8}{3} \\ \frac{4z}{3}=\frac{8}{3}-\frac{4}{3} \\ \frac{4z}{3}=\frac{4}{3} \\ z=1 \end{gathered}[/tex]

We have that z = 1.

The next steps are to back replace and find the other variables. Remember that in the second step we had y as a function of z:

[tex]y=\frac{1+z}{3}[/tex]

Replace z = 1 and solve:

[tex]y=\frac{1+1}{3}=\frac{2}{3}[/tex]

y = 2/3

And finally, replace y = 2/3 in the expression of the first step, where we found x as a function of y:

[tex]x=1-y=1-\frac{2}{3}=\frac{1}{3}[/tex]

and we got x = 1/3

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