Problem 13 - College Algebra Essentials

Question

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

Step-by-Step Solution

Answer

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

Step 1 - 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)

Step 2 - 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 \(x = -4\)

Step 3 - 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 \(z=14\).