Question

I need to know if my answer is correct it is the circled one I used systems of equations to solve this one

Answer 1

SOLUTION

STEP1: write out the equations

$\begin{bmatrix}2x-y+3z=1\ldots eq1 \\ 5x+2y-2z=4\ldots eq2 \\ -7x-y-z=-5\ldots eq3\end{bmatrix}$

STEP2: From the first equation, make x the subject of formula

$\begin{gathered} 2x-y+3z=1 \\ 2x=1+y-3z \\ \text{Divide both sides by 2} \\ x=\frac{1+y-3z}{2} \end{gathered}$

STEP3: Substitute the expression for x into equation 2 and 3

for equation 2, we have

$5\mleft\lbrace\frac{1+y-3z}{2}\mright\rbrace+2y-2z=4$

Then for equation 3, we have

$-7\mleft\lbrace\frac{1+y-3z}{2}\mright\rbrace-y-z=-5$

STEP4:Simplify each of the expression above

$\begin{gathered} 5\mleft\lbrace\frac{1+y-3z}{2}\mright\rbrace+2y-2z=4 \\ \frac{5+5y-15z}{2}+2y-2z=4 \\ \text{Multiply the expression above by 2} \\ 5+5y-15z+4y-4z=8 \\ 9y-19z=3\ldots eq4 \end{gathered}$

Similarly for the equation obtain for equation 3, we have

$\begin{gathered} -7\lbrace\frac{1+y-3z}{2}\rbrace-y-z=-5 \\ \frac{-7-7y+21z}{2}-y-z=-5 \\ \text{Multiply through by 2} \\ -7-7y+21z-2y-2z=-10 \\ -9y+19z=-3 \\ \text{Multiply through by -1} \\ 9y-19z=3\ldots\text{eq}5 \end{gathered}$

Step5: Write y interm of z

$\begin{gathered} 9y-19z=3 \\ 9y-19z=3 \\ \text{ since the equation are the same, then} \\ 9y=19z+3 \\ \text{Then} \\ y=\frac{19z+3}{9} \end{gathered}$

Step6: subsitute the expression for y into the equation for x in step 2

$\begin{gathered} x=\frac{1+y-3z}{2} \\ x=\frac{1+\frac{19z+3}{9}-3z}{2} \\ \text{Then} \\ x=\frac{9+19z+3-27z}{18} \\ x=\frac{-8z+12}{18} \\ x=\frac{4(-2z+3)}{18}=\frac{2(-2z+3)}{9} \end{gathered}$

Therefore

$\begin{gathered} y=\frac{19z+3}{9} \\ \text{And } \\ x=\frac{2(-2z+3)}{9} \end{gathered}$