Solving Equations with Three Unknowns

This is a guide on solving equations with three unknowns.

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When working with a system of three linear equations involving three variables, the objective is to determine the values of those variables that satisfy all equations simultaneously. The method presented here follows a systematic process of successive elimination, in which one variable is removed at a time until a single equation in one unknown remains. Once that value is found, it can be substituted back into the earlier equations to determine the remaining unknowns. This approach is both reliable and efficient, provided that the order of elimination is chosen thoughtfully to simplify the arithmetic.

Consider the following system of equations:

$$3x+2y+4z=1$$$$-x+y+2z=2$$$$x-3y+z=-1$$

The goal is to find the values of $x$, $y$, and $z$ that satisfy all three equations at once. To begin, it is helpful to examine the equations and decide which variable can be eliminated most conveniently. In this case, adding the second and third equations eliminates $x$ immediately, because the coefficients of $x$ in those two equations are opposites of one another. Carrying out this addition yields:

$$y-3y+3z=2-1$$

which simplifies to:

$$-2y+3z=1$$

This newly obtained equation involves only $y$ and $z$, reducing the complexity of the problem.

Next, the goal is to eliminate $x$ from another pair of equations so that a second equation in only $y$ and $z$ can be formed. To accomplish this, the second equation is multiplied by 3 and then added to the first equation. This operation produces:

$$2y+3y+4z+6z=1+6$$

which simplifies to:

$$5y+10z=7$$

At this stage, two equations involving only $y$ and $z$ have been obtained. These two equations form a smaller system that can be solved using the same elimination strategy applied previously. The two equations are:

$$-2y+3z=1$$$$5y+10z=7$$

To eliminate $y$, the first of these two equations is multiplied by 5 and the second is multiplied by 2, and then the results are added together. This gives:

$$15z+20z=5+14$$

which simplifies to:

$$35z=19$$

Solving for $z$ produces:

z=1935z=3519​

With the value of $z$ now known, it becomes possible to determine $y$ by substituting this value back into one of the equations involving only $y$ and $z$. Using the equation $-2y+3z=1$, the following steps are carried out:

2y=3z−1=3⋅1935−12y=3z−1=3⋅3519​−1

This becomes:

57−3535=22353557−35​=3522​

Dividing by 2 then yields the value of $y$:

y=1135y=3511​

Finally, with both $y$ and $z$ determined, the value of $x$ can be found by substituting these values into any one of the original three equations. Using the third original equation, $x-3y+z=-1$, the following calculation is performed:

$$x=-1+3y-z$$

Substituting the known values gives:

x=−3535+3335−1935x=−3535​+3533​−3519​

which simplifies to:

x=−9135x=−3591​

However, this fraction can be reduced by dividing both the numerator and denominator by the common factor of 7, resulting in:

x=−135x=−513​

It is worth noting that the final solution may be expressed in equivalent fractional forms depending on the stage at which simplification is applied. In the present case, the solution to the system is:

• x=−2135x=−3521​, which reduces to −35−53​

• y=1135y=3511​

• z=1935z=3519​

To summarize, the process of solving a system of three linear equations in three unknowns involves selecting an efficient order of elimination, systematically removing one variable at a time, solving the resulting single-variable equation, and then back-substituting to find the remaining variables. This method is general and can be applied to a wide variety of linear systems, though the arithmetic may vary depending on the coefficients involved. With practice, the selection of the most convenient elimination order becomes increasingly intuitive, allowing for both accuracy and efficiency in obtaining the final solution.