Finding the Second Variable by Back-Substitution

Welcome back. In the last lesson, you used substitution to reduce a system to one equation with one variable and solve for that first value. For example, you might have reached $x=4$ or $y=-1$.

This lesson completes the method. You will back-substitute that known value into an original equation, solve for the remaining variable, and write the system’s solution as an ordered pair $(x,y)$. In this module, that is the difference between a partial answer and a complete answer.

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From one known value to a complete solution

Suppose substitution has led you to:

$$x=4$$

This tells you only the $x$-coordinate of the solution. A system needs a value for both variables.

To find $y$, return to one of the original equations. If one equation was already isolated, it is usually the quickest one to use.

For example, consider the system:

$$y=3x-2$$ $$4x+y=28$$

In the previous lesson’s substitution step, you would find:

$$x=4$$

Now back-substitute $x=4$ into the convenient original equation:

$$y=3x-2$$ $$y=3(4)-2$$ $$y=12-2$$ $$y=10$$

Both values are now known:

$$x=4$$ $$y=10$$

So the solution is:

$$(4,10)$$

The order matters. An ordered pair is always written:

$$(x,y)$$

Since $x=4$ and $y=10$, write $(4,10)$, not $(10,4)$.

A useful way to organize your work is to label each value before writing the pair:

Variable	Value
$x$	$4$
$y$	$10$

Then translate the table directly into $(x,y)$.

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The back-substitution move

Back-substitution is simply substitution with a number instead of an algebraic expression.

If you know:

$$x=a$$

and an original equation says:

$$y=mx+b,$$

replace $x$ with $a$ and calculate $y$.

Likewise, if you know:

$$y=b$$

and an equation says:

$$x=py+q,$$

replace $y$ with $b$ and calculate $x$.

The most efficient choice is usually the equation where the unknown variable is already alone. It avoids unnecessary extra equation-solving.

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A full example with a negative value

The substitution diagram below shows the full method. Its lower steps are the focus now: after substitution produces $y=-1$, use that value to determine $x$.

The system is:

$$x-2y=8$$ $$x=5-y$$

After the earlier substitution work, suppose you have found:

$$y=-1$$

Now use the equation that already has $x$ isolated:

$$x=5-y$$

Substitute $-1$ for $y$:

$$x=5-(-1)$$

The parentheses make the negative value clear. Subtracting a negative is the same as adding:

$$x=5+1$$ $$x=6$$

Now identify each coordinate:

$$x=6$$ $$y=-1$$

Therefore, the solution is:

$$(6,-1)$$

Notice an important point: even though $y$ was found first, the ordered pair still begins with $x$. The discovery order does not determine the coordinate order.

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Choosing an equation for back-substitution

You may substitute the known value into either original equation. Both equations should produce the same missing value. But some choices take less work.

Suppose the original system includes:

$$x=2y-2$$ $$x-2y=-2$$

and substitution has already given:

$$y=5$$

Faster choice: use the isolated equation

Because $x$ is alone in the first equation, substitute $y=5$:

$$x=2(5)-2$$ $$x=10-2$$ $$x=8$$

The solution is:

$$(8,5)$$

Valid but longer choice: use the other original equation

You could instead use:

$$x-2y=-2$$

Substitute $y=5$:

$$x-2(5)=-2$$ $$x-10=-2$$

Add $10$ to both sides:

$$x=8$$

This is correct, but it requires an extra step. When choosing where to back-substitute, look first for an equation where the unknown variable is already isolated.

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Write the result in the right form

A complete system solution has three related forms:

$$x=8$$ $$y=5$$ $$(8,5)$$

The equations identify which value belongs to which variable. The ordered pair gives the final answer in the standard coordinate format.

Be especially careful when you find $y$ before $x$. For instance, if your work ends with:

$$y=3$$ $$x=-4,$$

the solution is:

$$(-4,3)$$

not

$$(3,-4).$$

The first position always belongs to $x$; the second always belongs to $y$.

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Common back-substitution mistakes

At this stage, the algebra is usually short, so small mistakes stand out clearly.

Substituting into the wrong kind of equation

Once you find $x=4$, do not substitute it into the one-variable equation that helped you find $x$, such as:

$$7x=28$$

That equation has no $y$, so it cannot help find $y$. Return to one of the two original equations that contains both variables.

Losing a negative sign

If:

$$x=5-y$$

and:

$$y=-1,$$

write:

$$x=5-(-1),$$

not:

$$x=5-1.$$

The value replacing $y$ is negative, so parentheses preserve its sign.

Reversing the ordered pair

If:

$$x=6$$

and:

$$y=-1,$$

then the answer is:

$$(6,-1).$$

The pair $(-1,6)$ represents a different point.

Stopping after one variable

A statement such as

$$y=-1$$

is not yet a complete solution to a two-variable system. Use it to find $x$, then write the pair.

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A dependable finish routine

Whenever you finish the first substitution equation, use this short routine:

1. Record the known variable value clearly, such as $x=4$.
2. Return to an original equation containing both variables.
3. Choose the simplest equation, preferably one with the remaining variable isolated.
4. Replace the known variable with its number, using parentheses for negative values.
5. Solve for the remaining variable.
6. Write the answer as $(x,y)$, regardless of which variable you found first.

This routine keeps the end of a substitution problem organized and makes it much harder to reverse coordinates or lose a negative sign.

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Key takeaways

Back-substitution turns the first variable value into the second one. Once you know one coordinate, substitute that number into an original equation to find the other coordinate.

Remember:

• A system is not completely solved until you know both $x$ and $y$.
• The equation with a variable already isolated is usually the most efficient place to back-substitute.
• Preserve negative values with parentheses.
• Write the final answer in the fixed order $(x,y)$, not in the order you happened to find the values.

Next, you will verify an ordered-pair solution by substituting both values into both original equations.