Q.Solve the system of equations $\dfrac{2}{x} + \dfrac{3}{y} + \dfrac{10}{z} = 4$, $\dfrac{4}{x} - \dfrac{6}{y} + \dfrac{5}{z} = 1$, $\dfrac{6}{x} + \dfrac{9}{y} - \dfrac{20}{z} = 2$.

The unknowns sit in the denominators, so substitute $a=\dfrac1x,\ b=\dfrac1y,\ c=\dfrac1z$ to linearize:

Write this as $AX=B$ with $A=\begin{pmatrix}2 & 3 & 10\\ 4 & -6 & 5\\ 6 & 9 & -20\end{pmatrix}$, $B=\begin{pmatrix}4\\ 1\\ 2\end{pmatrix}$.

Determinant: $\det(A)=2(75)-3(-110)+10(72)=150+330+720=1200$.

Adjoint and inverse:

$\operatorname{adj}(A)=\begin{pmatrix}75 & 150 & 75\\ 110 & -100 & 30\\ 72 & 0 & -24\end{pmatrix},\qquad A^{-1}=\frac{1}{1200}\operatorname{adj}(A).$ …

Why substitute

The unknowns $x,y,z$ only ever appear as $\dfrac1x,\dfrac1y,\dfrac1z$, so the system is really linear in disguise. Let $a=\dfrac1x,\ b=\dfrac1y,\ c=\dfrac1z$; the system becomes

Now this is exactly the kind of system the matrix method is built for.

Step 1: Write as $AX=B$.

Step 2: Find $\det(A)$ by expanding along the first row.

Since $\det(A)=1200\neq0$, $A$ is invertible.

Step 3: Find the cofactors of $A$.

So the cofactor matrix is $\begin{pmatrix}75 & 110 & 72\\ 150 & -100 & 0\\ 75 & 30 & -24\end{pmatrix}$.

Step 4: Find $\operatorname{adj}(A)$ (transpose of the cofactor matrix) and $A^{-1}$. …

Method: Substitution to Linearize, then Solve the System

This method handles a system where the unknowns appear only as reciprocals ($1/x$, $1/y$, $1/z$) — recognize the disguised linear system and solve it with standard elimination or the matrix method.

Steps

Step 1: Spot the reciprocal pattern

If every equation contains the unknowns only as $\tfrac1x,\tfrac1y,\tfrac1z$ (never $x,y,z$ themselves), the system is linear in the reciprocals, not in $x,y,z$. Rename $a=\tfrac1x,\ b=\tfrac1y,\ c=\tfrac1z$.

Step 2: Rewrite as an ordinary linear system in $a,b,c$

Step 3: Solve for $a,b,c$

Use elimination (multiply equations to cancel one variable at a time) or write the system as $AX=B$ and solve $X=A^{-1}B$ via the adjoint method — whichever is faster for the given coefficients. Either route is valid; elimination is usually quicker when the coefficients are small integers with an obvious combination that cancels a variable. …

Common Mistakes

Mistake 1: Forgetting to invert back to $x,y,z$ after solving for $a,b,c$

Why it's wrong: the substitution $a=1/x$ etc. is only a tool to make the system linear — leaving the final answer as $a=\tfrac12,b=\tfrac13,c=\tfrac15$ doesn't answer the question, which asks for $x,y,z$. Correct approach: always the last step is $x=1/a,\ y=1/b,\ z=1/c$ — never submit the reciprocal values as the final answer.

Mistake 2: Sign errors while eliminating a variable between three equations …