Solving One-Variable Linear Equations

Good to see you again. Last time, you distinguished an absolute change from a percentage change and learned why a movement from $62\%$ to $65\%$ is a change of 3 percentage points. This lesson turns to the algebraic operation that often sits behind quantitative questions: finding an unknown value from a relationship.

You will solve linear equations in one unknown by preserving equality at every step, using inverse operations, and checking the result by substitution. These skills will soon support formula rearrangement, market relationships, and later regression work. Plan on about 30 minutes.

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An equation is a balanced claim

An equation says that two expressions have the same value. Solving it means finding the value of the unknown that makes the claim true.

For example, in

$$3x+2=17$$

the question is: what number can replace $x$ so that the left side equals 17?

The central idea is that an equality behaves like a balanced scale. If you change one side, you must make the same change to the other side. Otherwise, the balance is destroyed.

Mathematically, if

$$A=B$$

then each of the following produces an equivalent equation:

$$A+c=B+c$$ $$A-c=B-c$$ $$cA=cB$$ $$\frac{A}{c}=\frac{B}{c}\text{for }c\ne 0$$

“Equivalent” means the new equation has exactly the same solution as the original one. The purpose of these operations is to isolate the variable: leave it alone on one side of the equals sign.

A linear equation in one unknown can typically be written in the form

$$ax+b=c$$

where $a\ne 0$. “Linear” means that the variable is only to the first power: expressions such as $x^2$, $\sqrt{x}$, and $1/x$ are not part of a linear equation.

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Undo operations in reverse order

To isolate a variable, undo whatever has been done to it. The operation that undoes another is its inverse operation:

Operation on the variable	Inverse operation
Add a number	Subtract that number
Subtract a number	Add that number
Multiply by a nonzero number	Divide by that number
Divide by a nonzero number	Multiply by that number

Consider a one-step equation:

$$x+9=15$$

The variable has had 9 added to it. Subtract 9 from both sides:

$$\begin{array}{rl}x+9& =15\\ x+9-9& =15-9\\ x& =6\end{array}$$

The final line, $x=6$, is the solution. It says the unknown number is 6.

Now consider a coefficient:

$$-4y=28$$

Here, $y$ is multiplied by $-4$. Divide both sides by $-4$:

$$\begin{array}{rl}-4y& =28\\ \frac{-4y}{-4}& =\frac{28}{-4}\\ y& =-7\end{array}$$

Keep the sign attached to the coefficient. Dividing a positive number by a negative number produces a negative result.

Two-step equations

Many practical equations have both a coefficient and a constant. Suppose an unknown quantity $q$ satisfies

$$0.8q+6=30$$

First undo the addition of 6:

$$\begin{array}{rl}0.8q+6& =30\\ 0.8q& =24\end{array}$$

Then undo multiplication by $0.8$, either by dividing by $0.8$ or multiplying by its reciprocal:

$$\begin{array}{rl}q& =\frac{24}{0.8}\\ q& =30\end{array}$$

The order matters. The expression $0.8q+6$ means “multiply $q$ by $0.8$, then add 6.” To reverse it, first remove 6, then remove the factor $0.8$.

Fractions work the same way. For example:

$$\begin{array}{rl}\frac{3}{4}x-5& =10\\ \frac{3}{4}x& =15\\ x& =15\cdot \frac{4}{3}\\ x& =20\end{array}$$

Multiplying by $\frac{4}{3}$ works because it is the reciprocal of $\frac{3}{4}$:

$$\frac{3}{4}\cdot \frac{4}{3}=1$$

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A reliable method for longer equations

For multi-step equations, do not rely on the informal rule that terms “move across the equals sign and change sign.” That shortcut can be useful once understood, but its real meaning is always an operation performed on both sides. Writing the operation makes errors easier to catch.

Use this sequence:

1. Simplify each side. Distribute through parentheses and combine like terms.
2. Put variable terms on one side. Add or subtract the same variable term from both sides.
3. Put constants on the other side. Add or subtract the same constant from both sides.
4. Make the coefficient 1. Divide by the remaining coefficient.
5. Check by substitution in the original equation.

Suppose a market relationship is represented by

$$4+2(3q-5)=5q+13$$

The unknown $q$ might represent an output level. Begin by simplifying the left side. The factor 2 must multiply every term in the parentheses:

$$\begin{array}{rl}4+2(3q-5)& =5q+13\\ 4+6q-10& =5q+13\\ 6q-6& =5q+13\end{array}$$

Now remove $5q$ from both sides:

$$\begin{array}{rl}6q-6-5q& =5q+13-5q\\ q-6& =13\end{array}$$

Finally, add 6 to both sides:

$$\begin{array}{rl}q-6+6& =13+6\\ q& =19\end{array}$$

So the solution is

$$q=19$$

The check uses the original equation, not a later simplified version:

$$4+2(3(19)-5)=5(19)+13$$ $$4+2(52)=108$$ $$95+13=108$$

Both sides equal 108, so $q=19$ is verified.

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When an equation does not have one answer

Most equations in this lesson lead to one numerical solution. But after simplifying, some equations reveal a different situation.

Consider:

$$3(x+2)=3x+6$$

Distribute:

$$3x+6=3x+6$$

Subtracting $3x$ from both sides leaves

$$6=6$$

This statement is always true. Therefore, every real number is a solution. The equation is an identity: both sides were simply two forms of the same expression.

Now compare:

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

Subtract $2x$ from both sides:

$$4=-1$$

This is never true. Therefore, the equation has no solution.

These outcomes are useful warnings. If the variable disappears, do not try to divide by zero or force a numerical answer. Instead, inspect the remaining statement:

What remains after simplifying	Meaning
A true statement, such as $6=6$	Infinitely many solutions
A false statement, such as $4=-1$	No solution
A statement such as $5x=20$	One solution remains to be found

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Checking and diagnosing errors

Substitution is the most dependable check. If you claim that $x=-5$ solves

$$-3x+5=20$$

substitute carefully, using parentheses around a negative value:

$$-3(-5)+5=15+5=20$$

The two sides agree, so the solution is correct.

Several common errors are worth actively watching for:

• Changing only one side. Equality is preserved only when the same operation is applied to both sides.
• Dropping a negative sign. Treat $-x$ as $(-1)x$, especially when dividing.
• Incomplete distribution. In $2(a-3)$, the factor 2 multiplies both $a$ and $-3$.
• Combining unlike terms. You may combine $6x-5x=x$, but not $6x-5$.
• Checking a transformed equation only. Checking the original equation confirms that no mistake entered during the work.

A clear layout is not cosmetic. Writing one equality per line and recording the operation prevents most algebra errors before they become difficult to trace.

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Key takeaways

A solution to an equation is a value that makes the equation true. Solving a linear equation means producing equivalent equations until the variable is isolated.

The governing principle is simple:

$$\text{Whatever you do to one side, do to the other side.}$$

For routine equations, simplify first, gather variable terms, gather constants, divide by the final coefficient, and check by substitution. A variable that disappears can indicate either infinitely many solutions or no solution, depending on whether the remaining statement is true or false.

Next, you will use this same balance logic more flexibly: rearranging formulas to isolate a specified variable.