If $3x - 4y + z = 7$; $2x - z + 3y = 19$; $x + 2y + 2z = 24$, then what is the value of $z$?

### Concept & Formula Solving a 3-variable system of linear equations requires systematic elimination. The strategy is to eliminate one variable completely using two pairs of equations, reducing the problem to a standard 2-variable system. ### Step-by-Step Solution **Given:** 1. $3x - 4y + z = 7$ 2. $2x + 3y - z = 19$ (Note: terms rearranged for alignment) 3. $x + 2y + 2z = 24$ **Calculation:** * **Step 1: Eliminate $z$ from Eqs (1) and (2)** Simply add Equation 1 and Equation 2 because they have $+z$ and $-z$: $(3x + 2x) + (-4y + 3y) + (z - z) = 7 + 19$ $5x - y = 26 \implies y = 5x - 26$ --- (Eq A) * **Step 2: Eliminate $z$ from Eqs (2) and (3)** Multiply Equation 2 by 2 so its $z$ term matches Equation 3: $2(2x + 3y - z = 19) \implies 4x + 6y - 2z = 38$ Add this to Equation 3 ($x + 2y + 2z = 24$): $(4x + x) + (6y + 2y) + (-2z + 2z) = 38 + 24$ $5x + 8y = 62$ --- (Eq B) * **Step 3: Solve for $x$ and $y$** Substitute Eq A ($y = 5x - 26$) into Eq B: $5x + 8(5x - 26) = 62$ $5x + 40x - 208 = 62$ $45x = 270 \implies x = 6$ Find $y$: $y = 5(6) - 26 = 30 - 26 = 4$ * **Step 4: Solve for $z$** Substitute $x=6, y=4$ into Equation 1: $3(6) - 4(4) + z = 7$ $18 - 16 + z = 7 \implies 2 + z = 7 \implies z = 5$ ### Exam Strategy & Shortcut If you are only asked for $z$, and the algebra looks dense, check if you can cleverly multiply the equations to cancel $x$ and $y$ simultaneously. If no obvious multiplier exists, execute systematic elimination rapidly. Never mix up the pairs of equations you are eliminating from. ### Common Pitfall Sign errors are the #1 reason for missed marks here. When multiplying an entire equation (like Eq 2 by 2), students often forget to multiply the constant on the right side of the equals sign (e.g., writing 19 instead of 38). ### Final Answer **Therefore, the correct answer is 5.**