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Chapter 5: Problem 13

Solve each system. $$\left\\{\begin{aligned} 2 x+y &=2 \\ x+y-z &=4 \\ 3 x+2 y+z &=0 \end{aligned}\right.$$

Short Answer

Expert verified
The solution to the system of equations is \(x = -4\), \(y = 10\), and \(z = 14\).

Step by step solution

01

- Elimination of variable

Eliminate variable \(z\) from equation (2) and equation (3). To do this, add equation (2) and equation (3). The result is \(4x + 3y = 4\). This will serve as equation (4)
02

- Further elimination of variable

Now, to bring the system down to two variables, eliminate variable \(y\) from equation (1) and equation (4). To do this, multiplying equation (1) by 3 and subtract the result from equation (4) will yield a result of \(-2x=8\). Then, solve this for \(x\) to get \(x = -4\)
03

- Back substitution

Substitute \(x\) back into equations (1) and (3) to find the remaining values. Substituting \(x = -4\) into equation (1) gives \(y = 10\). Substituting \(x=-4\) and \(y=10\) into equation (3) gives \(z=14\).

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Most popular questions from this chapter

Find the partial fraction decomposition for \(\frac{2}{x(x+2)}\) and use the result to find the following sum: $$ \frac{2}{1 \cdot 3}+\frac{2}{3 \cdot 5}+\frac{2}{5 \cdot 7}+\dots+\frac{2}{99 \cdot 101} $$

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