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Solve by elimination: 14. Solve the system 3x + 4y + 2z = 11 2x + 3y - 2=4 5x + 5y - 3z = -1

User Joey Harwood
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1 Answer

23 votes
23 votes

Answer:

The solution to the simultaneous equation is;


\begin{gathered} x=-3 \\ y=4 \\ z=2 \end{gathered}

Step-by-step explanation:

We want to solve the simultaneous equation below using elimination method;


\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}

Firstly, let us multiply equation 2 by 2 and add to equation 1 to eliminate z;


\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}

Secondly, let us multiply equation 2 by 3 and subtract equation 3 from it to eliminate z;


\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}

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;


\begin{gathered} x+4y=13\ldots\ldots\ldots7 \\ *7 \\ 7x+28y=91\ldots\ldots\ldots8 \\ -(7x+10y=19) \\ 18y=72 \\ y=(72)/(18) \\ y=4 \end{gathered}

let now substitute y=4 into equation 7;


\begin{gathered} x+4y=13\ldots\ldots\ldots7 \\ x+4(4)=13 \\ x+16=13 \\ x=13-16 \\ x=-3 \end{gathered}

Then substituting x and y into equation 2 to get z;


\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}

The solution to the simultaneous equation is;


\begin{gathered} x=-3 \\ y=4 \\ z=2 \end{gathered}

User Oscar Albert
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