Solving Equations with Two Unknowns

This is a guide on solving equations with two unknowns.

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When faced with two linear equations involving the same two variables, the goal is to determine the values of those variables that satisfy both equations simultaneously. Such a pair of equations is called a system of simultaneous linear equations. Consider the following system as an example:

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

The task is to find values for $x$ and $y$ that make both equations true at the same time. One of the most straightforward and reliable techniques for accomplishing this is known as the elimination method. The central idea of elimination is to combine the two equations in such a way that one of the variables is removed, leaving a single equation in one variable that can be solved directly. Once that variable is known, it can be substituted back into either of the original equations to find the other variable.

Applying the Elimination Method

In the system above, the variable $x$ appears with a coefficient of 2 in the first equation and a coefficient of 3 in the second equation. To eliminate $x$, the coefficients of $x$ in the two equations must be made equal in magnitude. This is accomplished by multiplying each equation by a suitable number. Specifically, the first equation can be multiplied by 3, and the second equation can be multiplied by 2. This yields:

$$6x+3y=3$$$$6x-4y=8$$

Now that the coefficients of $x$ are identical in both equations, subtracting the second equation from the first will cause the $x$-terms to cancel. It is important to subtract each side of the second equation from the corresponding side of the first equation. Carrying out this subtraction gives:

$$3y-(-4y)=3-8$$

Simplifying the left-hand side:

$$3y+4y=7y$$

And simplifying the right-hand side:

$$3-8=-5$$

Thus:

$$7y=-5$$

Dividing both sides by 7 produces:

y=−57y=−75​

With the value of $y$ now known, the next step is to determine $x$. This can be done by substituting y=−57y=−75​ into either of the original equations. Using the first equation, $2x+y=1$, the substitution gives:

2x−57=12x−75​=1

Adding 5775​ to both sides:

2x=1+57=77+57=1272x=1+75​=77​+75​=712​

Dividing both sides by 2:

x=127÷2=1214x=712​÷2=1412​

Thus the solution is:

x=1214,y=−57x=1412​,y=−75​

The fraction 12141412​ can be reduced to lowest terms by dividing the numerator and denominator by their greatest common divisor, which is 2. This gives x=67x=76​. Both 12141412​ and 6776​ are correct representations of the same value; reducing to lowest terms is simply a matter of convention and convenience.

An Alternative Approach: Eliminating $y$ First

The order in which the variables are eliminated is a matter of choice. As a variation, one could eliminate $y$ first instead of $x$. To do this, multiply the first equation by 2 and leave the second equation unchanged:

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

Adding the two equations together causes the $y$-terms to cancel:

$$4x+3x=6$$

This simplifies to:

$$7x=6$$

Dividing both sides by 7 gives:

x=67x=76​

This is the same value for $x$ that was found previously. To find $y$, substitute x=67x=76​ into the first original equation, $2x+y=1$:

y=1−2x=1−127=77−127=−57y=1−2x=1−712​=77​−712​=−75​

Once again, the solution is x=67x=76​ and y=−57y=−75​, confirming that either elimination strategy leads to the same result.

Systems with No Solution

Not every system of two linear equations has a solution. There are cases in which no pair of values for $x$ and $y$ can satisfy both equations simultaneously. For example, consider the system:

$$2x-y=6$$$$2x-y=7$$

The left-hand sides of these two equations are identical, yet the right-hand sides differ. It is impossible for the same expression $2x-y$ to equal both 6 and 7 at the same time. Therefore, this system has no solution.

A similar situation arises with the system:

$$2x-y=5$$$$6x-3y=7$$

Observe that the second equation can be divided by 3 to obtain:

2x−y=732x−y=37​

This means that any solution of $6x-3y=7$ must also satisfy 2x−y=732x−y=37​. However, the first equation requires $2x-y=5$. Since 7337​ and 5 are not equal, no pair of values can satisfy both equations at once. Consequently, this system also has no solution.

These examples illustrate the two typical ways in which a system of two linear equations can fail to have a solution: either the equations contradict each other directly, or they represent parallel relationships that never intersect. The purpose here is not to develop a comprehensive theory of when solutions exist or do not exist, but rather to become comfortable with the basic techniques for solving systems that do have solutions. In practice, most simple systems encountered will have a unique solution unless they are constructed in a way that is essentially equivalent to one of the contradictory examples above.

Verifying the Solution

When a solution is found by the elimination method, it is always possible to verify that the values obtained actually satisfy both original equations. This verification can be performed explicitly each time. Substituting x=67x=76​ and y=−57y=−75​ into the first equation:

2(67)+(−57)=127−57=77=12(76​)+(−75​)=712​−75​=77​=1

And into the second equation:

3(67)−2(−57)=187+107=287=43(76​)−2(−75​)=718​+710​=728​=4

Both equations are satisfied, confirming that the solution is correct. While it is possible to prove in general that the elimination method always produces a valid solution when one exists, such a proof requires setting up convenient notation and using general letters for the coefficients. That level of abstraction is not necessary here. The primary objective is to develop a simple, efficient, and reliable approach to solving systems of two equations in two unknowns.

Summary of Key Steps

• Multiply one or both equations by suitable constants so that the coefficients of one variable become equal in magnitude.

• Add or subtract the equations to eliminate that variable, leaving a single equation in one unknown.

• Solve the resulting equation for the remaining variable.

• Substitute the value found back into either original equation to determine the other variable.

• Verify the solution by substituting both values into both original equations.

Conclusion

Simultaneous linear equations in two variables arise naturally in a wide variety of problems, including those that might otherwise be posed in terms of a single variable. The elimination method provides a systematic and efficient way to find the unique solution when it exists. By mastering this technique, one gains a practical tool for solving problems involving two unknown quantities. The method is straightforward, requires only basic arithmetic, and can be applied consistently to any system of two linear equations that has a solution.