Solving Linear Systems by Substitution

Hello. In the previous lesson, you practiced solving a one-variable equation by undoing operations in reverse order. That skill is exactly what substitution needs: after we replace one variable with an equivalent expression, a system becomes a familiar one-variable equation.

In this lesson, you will learn the central substitution move: replace an isolated variable in one equation with its equivalent expression in the other equation, then solve for the one variable left. We will stop once that first variable value is found; the next lesson will use it to find the second value and write the complete ordered-pair solution.

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Why substitution is valid

A system of linear equations asks for values of $x$ and $y$ that make both equations true at the same time.

Suppose one equation says

$$y=2x$$

This is not merely a hint about $y$. It is a rule: wherever $y$ appears, you may replace it with $2x$, because $y$ and $2x$ have the same value for any solution of the system.

For example, consider:

$$y=2x$$ $$x+y=24$$

The second equation contains $y$. Since the first equation tells us that $y=2x$, replace $y$ in the second equation with $2x$:

$$x+2x=24$$

Now the equation has only one variable:

$$3x=24$$ $$x=8$$

At this point, substitution has done its main job: it turned two equations with two variables into one equation with one variable.

The key idea is:

Use the equation with an isolated variable as a replacement rule in the other equation.

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Watch the substitution happen

The video’s example begins with:

$$2y=x+7$$ $$x=y-4$$

Because the second equation has already isolated $x$, substitute $y-4$ for every $x$ in the first equation:

$$2y=y-4+7$$

Simplify the constants:

$$2y=y+3$$

Subtract $y$ from both sides:

$$y=3$$

Notice what did not happen: we did not replace both variables, and we did not substitute into the equation that already said $x=y-4$. We used that equation as the replacement rule and inserted it into the other equation.

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The reliable substitution routine

When one equation already has a variable isolated, such as

$$x=\text{an expression involving }y$$

or

$$y=\text{an expression involving }x,$$

follow this routine:

1. Identify the isolated variable and its expression.
   For example, from $y=3x-2$, the replacement is $3x-2$ for $y$.

2. Choose the other equation.
   This is the equation where you will make the replacement.

3. Replace every occurrence of the isolated variable.
   If $y=3x-2$, write $3x-2$ wherever the other equation has $y$.

4. Use parentheses around an expression with more than one term.
   This is especially important if the expression is multiplied or subtracted.

5. Simplify and solve the resulting one-variable equation using the inverse-operation method from the previous lesson.

Here is a compact example:

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

Replace $y$ with $3x$ in the second equation:

$$4x+3x=28$$

Combine like terms:

$$7x=28$$

Divide both sides by $7$:

$$x=4$$

For this lesson, $x=4$ is the result we were seeking. In the next lesson, you will substitute $x=4$ back into an original equation to determine $y$.

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Parentheses protect the whole replacement

Parentheses are not decoration. They show that the entire expression replaces one variable.

Consider this system:

$$x=4+y$$ $$2x-y=7$$

The replacement expression for $x$ is $4+y$. Substitute it into the second equation:

$$2(4+y)-y=7$$

The parentheses matter because the $2$ multiplies all of $4+y$. Now simplify:

$$8+2y-y=7$$ $$8+y=7$$

Subtract $8$ from both sides:

$$y=-1$$

A common incorrect substitution would be:

$$2(4)+y-y=7$$

That changes $2(4+y)$ into $8+y$, but it should be $8+2y$. The coefficient $2$ must multiply both terms inside the parentheses.

Use parentheses automatically whenever the replacement has two or more terms:

$$x=-3y+5$$ $$4x+y=19$$

becomes

$$4(-3y+5)+y=19$$

Even if you could sometimes omit the parentheses without changing the result, keeping them makes your work safer and easier to read.

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A worked substitution example

The substitution diagram below shows the full method. For now, focus on the middle stage: replacing $x$ with the expression $-10+3y$ and solving the resulting equation for $y$.

Suppose the isolated equation has already been provided:

$$x=-10+3y$$

and the other equation is

$$3x+2y=14$$

Because $x=-10+3y$, every $x$ in the second equation can be replaced by $-10+3y$:

$$3(-10+3y)+2y=14$$

Now distribute the $3$:

$$-30+9y+2y=14$$

Combine like terms:

$$-30+11y=14$$

Add $30$ to both sides:

$$11y=44$$

Divide both sides by $11$:

$$y=4$$

The important transition is this one:

$$3x+2y=14$$ $$3(-10+3y)+2y=14$$

Only $x$ was replaced, because only $x$ had an isolated expression. The $2y$ stayed exactly as it was.

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Three errors to catch before they spread

1. Replacing the wrong variable

If you know

$$y=5x-1,$$

then replace $y$ with $5x-1$, not $x$.

For example,

$$2x+y=11$$

becomes

$$2x+(5x-1)=11.$$

The $x$ terms remain because the replacement rule was about $y$, not $x$.

2. Forgetting parentheses

If

$$y=x-1$$

and the other equation contains $-3y$, write:

$$-3(x-1),$$

not

$$-3x-1.$$

The correct distribution is:

$$-3(x-1)=-3x+3.$$

The negative coefficient multiplies both terms of the replacement expression.

3. Solving before simplifying

After substitution, first simplify enough to create a clear one-variable equation. For instance:

$$5x+2(x-3)=17$$

should become

$$5x+2x-6=17$$

then

$$7x-6=17.$$

Only then use inverse operations:

$$7x=23$$ $$x=\frac{23}{7}.$$

Fractions are valid solutions. Do not change the method just because the answer is not a whole number.

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A short self-check while you work

Before you begin solving, look at your substituted equation and ask yourself these three things:

• Does it contain only one variable letter?
• Did I replace the correct variable every time it appeared?
• If the replacement had multiple terms, did I place it in parentheses?

If the answer to all three is yes, you have set up substitution correctly. The rest is the one-variable equation solving you already know.

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Key takeaways

Substitution relies on equality: if

$$x=4+y,$$

then $4+y$ can replace $x$ anywhere in the other equation. That replacement eliminates one variable, leaving a one-variable equation to simplify and solve.

The essential habits are:

• use the already isolated equation as a replacement rule;
• substitute into the other equation;
• replace the correct variable everywhere it occurs;
• protect multi-term replacements with parentheses;
• solve the resulting equation using inverse operations.

Next, you will take the variable value found through substitution, plug it back into an original equation, find the second variable, and express the system’s solution as an ordered pair.