Respuesta :
Answer:
The solution to the simultaneous equation is;
[tex]\begin{gathered} x=-3 \\ y=4 \\ z=2 \end{gathered}[/tex]Explanation:
We want to solve the simultaneous equation below using elimination method;
[tex]\begin{gathered} 3x+4y+2z=11\ldots\ldots\ldots.1 \\ 2x+3y-z=4\ldots\ldots\ldots\ldots\text{.}.2 \\ 5x+5y-3z=-1\ldots\ldots\ldots\ldots\text{.}.3 \end{gathered}[/tex]Firstly, let us multiply equation 2 by 2 and add to equation 1 to eliminate z;
[tex]\begin{gathered} 2x+3y-z=4\ldots\ldots\ldots\ldots\text{.}.2 \\ \text{multiply by 2, we have;} \\ 4x+6y-2z=8\ldots\ldots\ldots\ldots\text{.}.4 \\ \text{adding to equation 1, we have;} \\ 4x+6y-2z=8 \\ +\left(3x+4y+2z\right?=11 \\ 7x+10y=19\ldots\ldots\ldots5 \end{gathered}[/tex]Secondly, let us multiply equation 2 by 3 and subtract equation 3 from it to eliminate z;
[tex]\begin{gathered} 2x+3y-z=4\ldots\ldots\ldots\ldots\text{.}.2 \\ \text{Multiply by 3, we have;} \\ 6x+9y-3z=12\ldots\ldots\ldots6 \\ \text{subtracting equation 3 from it;} \\ 6x+9y-3z=12 \\ -(5x+5y-3z=-1) \\ x+4y=13\ldots\ldots\ldots7 \end{gathered}[/tex]Now, we need to eliminate either x or y from equation 5 and 7;
multiply equation 7 by 7 and subtract equation 5 from it;
[tex]\begin{gathered} x+4y=13\ldots\ldots\ldots7 \\ \times7 \\ 7x+28y=91\ldots\ldots\ldots8 \\ -(7x+10y=19) \\ 18y=72 \\ y=\frac{72}{18} \\ y=4 \end{gathered}[/tex]let now substitute y=4 into equation 7;
[tex]\begin{gathered} x+4y=13\ldots\ldots\ldots7 \\ x+4(4)=13 \\ x+16=13 \\ x=13-16 \\ x=-3 \end{gathered}[/tex]Then substituting x and y into equation 2 to get z;
[tex]\begin{gathered} 2x+3y-z=4\ldots\ldots\ldots\ldots\text{.}.2 \\ 2(-3)+3(4)-z=4 \\ -6+12-z=4 \\ 6-z=4 \\ z=6-4 \\ z=2 \end{gathered}[/tex]The solution to the simultaneous equation is;
[tex]\begin{gathered} x=-3 \\ y=4 \\ z=2 \end{gathered}[/tex]