Problem 14 - College Algebra Essentials

Question

Solve each system. $$\left\\{\begin{aligned} x+3 y+5 z &=20 \\ y-4 z &=-16 \\ 3 x-2 y+9 z &=36 \end{aligned}\right.$$

Step-by-Step Solution

Answer

The solution to the system of equations is \(x = 4\), \(y = 0\), and \(z = 4\).

Step 1: Simplify the Second Equation

Isolate \(y\) in the second equation: \(y = 4z -16\)

Step 2: Substitute \(y\) into the First and Third Equations

Substitute \(y = 4z-16\) into the first equation to get a new equation: \(x + 3(4z-16) + 5z = 20\) or \(x+17z-48 = 20\), which simplifies to \(x = -17z+68\). Do the same for the third equation to get: \(3x -2(4z-16) + 9z = 36\) or \(3x + z + 32 = 36\), which simplifies to \(3x = -z +4\)

Step 3: Substitute \(x\) into the Equation Found in Step 2

Substitution \(x = -17z+68\) into the equation \(3x = -z + 4\) gives: \(3(-17z + 68)=-z + 4\), which simplifies to \(51z - 204 = -z + 4\), leading to \(52z = 208\), so \(z = 4\)

Step 4: Solve for \(x\) and \(y\)

With \(z = 4\), we can find \(x\) by substituting \(z = 4\) into \(x = -17z + 68\), obtaining \(x = 4\). For \(y\), substitute \(z = 4\) into \(y = 4z -16\) to get \(y = 0\).