Solve the system 3x + 2y + 5z = 7, x + 2y + 5z = 4, -x + 3y + z = 2, 3x + 4y + 5z = 32 by the substitution method.

Step-by-step solution

1. 1

   Notice the system is overdetermined. There are four equations but only three unknowns. A system like this usually has no solution: the fourth equation is one constraint more than the unknowns can absorb, so part of the work is deciding whether the equations are consistent at all.

2. 2

   Solve equation (2) for x. From $x+2y+5z=4$ we get $x=4-2y-5z$. Equation (2) is the natural choice because $x$ has coefficient $1$ there, so no fractions appear.

3. 3

   Substitute into the other three equations. In (1): $3(4-2y-5z)+2y+5z=7$ collapses to $4y+10z=5$, call it (A). In (3): $-(4-2y-5z)+3y+z=2$ gives $5y+6z=6$, call it (B). In (4): $3(4-2y-5z)+4y+5z=32$ gives $2y+10z=-20$, call it (C).

4. 4

   Solve the pair (A) and (C). Subtracting (C) from (A) eliminates $z$: $(4y+10z)-(2y+10z)=5-(-20)$, so $2y=25$ and $y=\frac{25}{2}$. Then (C) gives $25+10z=-20$, so $z=-\frac92$.

5. 5

   Test those values in equation (B). $5\left(\frac{25}{2}\right)+6\left(-\frac92\right)=\frac{125}{2}-27=\frac{71}{2}=35.5$, but (B) demands $6$. The three reduced equations cannot hold at once.

6. 6

   Conclude and confirm by rank. Because a consequence of the system contradicts another equation, the system is inconsistent. The coefficient matrix has rank $3$ while the augmented matrix has rank $4$, which is exactly the Rouche-Capelli criterion for no solution. Concretely, $(1)-(2)$ forces $x=\frac32$ and $(4)-(1)$ forces $y=\frac{25}{2}$, then (2) forces $z=-\frac92$; those values give $-\frac32+\frac{75}{2}-\frac92=\frac{63}{2}=31.5$ in equation (3), not $2$.