Respuesta :
SOLUTION
The system of equation given are
[tex]\begin{bmatrix}5x-3y+4z=5\ldots\text{equation}1 \\ -4x+2y-3z=-5\ldots\text{equation}2 \\ -x+5y+7z=20\ldots\text{equation}3\end{bmatrix}[/tex]To solve the equation by elimination method, we need to eliminate each variable one after the other.
To eliminate x,
Step1: Multiply equation3 by 5 and add equation 1 and 3
[tex]\begin{gathered} 5\times eq3=5(-x+5y+7z=20)\rightarrow-5x+25y+35z=100 \\ \text{Then Eq1+Eq3 will be } \\ +5x-3y+4z=5 \\ -5x+25y+35z=100 \end{gathered}[/tex]By addtion, x will be eliminated. we have
[tex]22y+39z=105\ldots Eq4[/tex]Step2: Multipy equation 3 by 4 and subtract from equation2 and 3
[tex]\begin{gathered} 4\times eq3=4(-x+5y+7z=20)\rightarrow-4x+20y+28y=80 \\ \text{Then Eq2 -Eq3} \\ -4x+2y-3z=-5 \\ -4x+20y+28y=80 \end{gathered}[/tex]After the subtraction of equation2-equation3, we have
[tex]-18y-31z=-85\ldots Eq5[/tex]Step3: Solve equation 5 and 4 to obtain the value of x and y
[tex]\begin{gathered} -18y-31z=-85\ldots eq5 \\ 22y+39z=105\ldots eq4 \end{gathered}[/tex]Then
[tex]\begin{gathered} \text{Multiply equation 5 by -1, we have } \\ 18y+31z=85\ldots eq5 \\ 22y+39z=105\ldots eq4 \end{gathered}[/tex]Eliminate y in equation 4 and 5, we have
[tex]\begin{gathered} 22\times eq5=22(18y+31z=85)\rightarrow396y+682z=1870 \\ 18\times eq4=18(22y+39z=105)\rightarrow396y+702z=1890 \\ \text{Then} \\ 396y+682z=1870 \\ 396y+702z=1890 \end{gathered}[/tex]Subtract the last two equation obtain in the last lone we have,
y is eliminated
[tex]\begin{gathered} -20z=-20 \\ \text{divide both sides by -20} \\ -\frac{20z}{-20}=-\frac{20}{-20} \\ \text{Then} \\ z=1 \end{gathered}[/tex]
Hence, Z=1
We now eliminate z, by using the coefficient of z, we have
[tex]\begin{gathered} 22y+39z=105\ldots eq4 \\ 18y+31z=85\ldots eq5 \\ \text{Multiply eq4 by31 and eq5 by 39, we have } \\ 31(22y+39z=105)\rightarrow682y+1209z=3255 \\ 39(18y+31z=85)\rightarrow702y+1209z=3315 \end{gathered}[/tex]Hence
[tex]\begin{gathered} 682y+1209z=3255 \\ 702y+1209z=3315 \\ \text{subtract the equation above, we have } \\ -20y=-60 \\ \text{divide both sides by -20, we have } \\ -\frac{20y}{-20}=-\frac{60}{-20} \\ y=3 \end{gathered}[/tex]Hence, y=3
Put the value of y and z into any of the equation to find x, we have
[tex]\begin{gathered} \text{ using equation 1, we have } \\ 5x-3y+4z=5 \\ z=1,y=3 \\ 5x-3(3)+4(1)=5 \\ 5x-9+4=5 \\ 5x-5=5 \end{gathered}[/tex]Add 5 to both sides we obtain
[tex]\begin{gathered} 5x-5+5=5+5 \\ 5x=10 \\ \text{divide both sides by 5, we have } \\ \frac{5x}{5}=\frac{10}{5} \\ \text{Then} \\ x=2 \end{gathered}[/tex]Hence, x=2
Therefore
Answer is x=2,y=3, z=1
