Question

4
type the correct answer in each box. if necessary, use / for the fraction bar.
find the solution to this system of equations.
$x+y=1$
$2x - y + z = 1$
$x + 2y + z = \frac{8}{3}$
$x = \square$
$y = \square$
$z = \square$

Explanation:

Step1: Isolate $x$ from first equation

$x = 1 - y$

Step2: Subtract second from third equation

$(x + 2y + z) - (2x - y + z) = \frac{8}{3} - 1$
$x + 2y + z - 2x + y - z = \frac{5}{3}$
$-x + 3y = \frac{5}{3}$

Step3: Substitute $x=1-y$ into new equation

$-(1 - y) + 3y = \frac{5}{3}$
$-1 + y + 3y = \frac{5}{3}$
$4y = \frac{5}{3} + 1 = \frac{8}{3}$
$y = \frac{8}{3} \times \frac{1}{4} = \frac{2}{3}$

Step4: Solve for $x$ using $y$

$x = 1 - \frac{2}{3} = \frac{1}{3}$

Step5: Solve for $z$ using $x,y$

Substitute $x=\frac{1}{3}, y=\frac{2}{3}$ into $2x - y + z = 1$:
$2\times\frac{1}{3} - \frac{2}{3} + z = 1$
$\frac{2}{3} - \frac{2}{3} + z = 1$
$z = 1$

Answer:

$x = \frac{1}{3}$
$y = \frac{2}{3}$
$z = 1$