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

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=(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(1+y-3z)/(2)\mright\rbrace+2y-2z=4`

Then for equation 3, we have

`-7\mleft\lbrace(1+y-3z)/(2)\mright\rbrace-y-z=-5`

STEP4:Simplify each of the expression above

`\begin{gathered} 5\mleft\lbrace(1+y-3z)/(2)\mright\rbrace+2y-2z=4 \\ (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(1+y-3z)/(2)\rbrace-y-z=-5 \\ (-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=(19z+3)/(9) \end{gathered}`

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

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

Therefore

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