Problem 2 - College Algebra Essentials

Question

Determine if the given ordered triple is a solution of the system. $$\begin{aligned} &(5,-3,-2)\\\ &\left\\{\begin{array}{rr} x+y+z= & 0 \\ x+2 y-3 z= & 5 \\ 3 x+4 y+2 z= & -1 \end{array}\right. \end{aligned}$$

Step-by-Step Solution

Answer

No, the ordered triple (5,-3,-2) is not a solution of the system.

Step 1: Substitute into the first equation

Substitute \(x = 5\), \(y = -3\), and \(z = -2\) into the first equation \(x + y + z = 0\). The result is \(5 - 3 - 2\), which equals \(0\). Therefore, the triple satisfies the first equation.

Step 2: Substitute into the second equation

Substitute the values into the second equation \(x + 2y - 3z = 5\). This gives \(5 + 2(-3) - 3(-2) = 7\), which does not equal to \(5\). Therefore, the triple does not satisfy the second equation.

Step 3: Conclusion

Even though we could substitute into the third equation, we don't need to. A solution to a system of equations must satisfy all the equations in the system simultaneously. Because the triple does not satisfy the second equation, it cannot be a solution to the system regardless of whether it satisfies the third equation. Hence, the ordered triple (5,-3,-2) is not a solution of the system.