Linear equations in one variable

Learning objectives

What you should be able to do

Solve and create linear equations; use fractions, decimals, and parentheses reliably; rearrange formulas; interpret a solution in context; solve for a requested expression; and distinguish one solution, no solution, and infinitely many solutions, including equations with unknown constants.

Equality, structure, and valid operations

An expression, such as $3x + 8,$ names a quantity. An equation, such as $3x + 8 = 20,$ asserts that two quantities are equal. A solution is a value that makes the original equation true. Solving means finding all such values, not merely producing a number after some arithmetic.

A linear equation in x can be simplified to $ax + b = cx + d,$ where a, b, c, d are constants. A constant may be an unfamiliar letter; what matters is that it does not vary with x. Terms such as $x 2$ and $1/x$ are not linear in x.

Valid move	Why it preserves the solutions
Add or subtract the same quantity on both sides	Equal quantities change by equal amounts.
Multiply both sides by a nonzero constant	The operation is reversible by division.
Divide both sides by a nonzero constant	The operation is reversible by multiplication.
Distribute and combine like terms	These rewrite a quantity without changing its value.
Swap the two sides	Equality is symmetric: $A = B$ means $B = A.$

The condition matters. Multiplying an equation by zero destroys information. Dividing by zero is undefined. When the proposed divisor contains an unknown parameter, check whether it could equal zero before using it.

A dependable solving routine

Simplify: distribute carefully and combine like terms on each side.

Collect: put variable terms on one side and constants on the other.

Isolate: divide by the variable’s coefficient, provided it is nonzero.

Finish: answer the quantity requested, then substitute into the original relationship.

Signs live with terms. In $8 - 3(x - 2),$ the factor distributed through the parentheses is −3, so the result is $8 - 3x + 6,$ not $8 - 3x - 6.$ When subtracting a whole expression, parentheses preserve that meaning: $P - (Q + R) = P - Q - R.$

Combining like terms. $4x + 3x = 7x,$ but $4x + 3$ cannot be combined into $7x.$ A coefficient tells how many copies of a variable occur; a constant is a different kind of term.

Interpret the last line. If the variable disappears, do not invent a value for it. A true statement such as $11 = 11$ means every allowed x works. A false statement such as $-7 = 1$ means none works. Section 1.4 develops this fully.

Fractions, decimals, and the actual target

Clear numeric denominators by multiplying every term. For denominators 3 and 4, a convenient multiplier is 12. If an equation has three terms, all three must be multiplied, including a constant standing alone. Clearing denominators is an operation on the entire equation, not permission to cancel pieces of a sum.

For example, the two expressions $\frac{x+2}{3}$ and $\frac{x}{3}$ are different: only the first divides the entire numerator by 3. Parentheses are especially important when entering fractions in a calculator.

Decimals are exact when the given decimals are exact. Multiplying a money equation by 100 can replace dollars with cents. Keep enough precision to preserve the given relationship; do not round a coefficient before solving. A result such as $65/7$ is often better kept as a fraction than converted to a rounded decimal.

Solve for the target, not automatically for the variable

Before doing algebra, underline the requested quantity. If the equation repeatedly contains $2x - 1$ and the question asks for $2x - 1,$ treat that expression as one unit. If the target is twice that expression, find the unit first and multiply by 2. Unnecessary expansion can make a short problem long.

Rearranging a formula. Choose the letter to isolate and treat every other letter as a constant for that calculation. Undo additions or subtractions around the target before undoing multiplication. State any nonzero condition introduced by division. A symbolic expression is not a numeric answer until values are supplied.

Translate a context before calculating

A useful model has a defined variable, consistent units, and an equation expressing the actual relationship. Letting h mean hours is more informative than writing “let x be the answer.” It prevents a rate in dollars per hour from being multiplied by minutes.

Language or structure	Mathematical translation
Fixed fee plus a per-unit charge	$total = fixed fee + rate \times number of units$
A quantity decreased by r%	$new quantity = (1 - r/100) times original quantity$
A quantity increased by r%	$new quantity = (1 + r/100) times original quantity$
“a less than b”	$b - a,$ not $a - b$
“a is k times b”	$a = kb;$ define what each letter counts

Check the context as well as the equation. An algebraic solution may be negative, fractional, or outside a model’s time interval. Such a value can be mathematically correct but not usable in the situation. A number of people must be a nonnegative integer; a measured length may be a positive fraction. Do not assume an integer unless the problem justifies it.

Solution counts and unknown constants

For any equation that simplifies to $ax + b = cx + d,$ collecting terms gives

$(a - c)x = d - b.$

This one line explains all three outcomes.

Condition	Result after collecting	Solution set
$a \neq c$	$nonzero coefficient \timesx = d - b$	Exactly one solution, $x = (d - b)/(a - c).$
$a = c$ and $b = d$	$0x = 0$	Every real number; infinitely many solutions.
$a = c$ and $b \neq d$	$0x = a nonzero constant$	No solution.

Why coefficient comparison works. For the two sides to agree for every x, their rates of change must match and their constant terms must match. Matching only the coefficients of x is not enough: $5x + 1 = 5x + 9$ is never true. Matching only the constants is not enough either: $2x + 7 = 3x + 7$ has the single solution $x = 0.$

Parameters are fixed but unknown. In $k(2x - 3) = 8x + 5,$ each chosen value of k defines a different equation in x. A question asking for no solution asks you to choose the equation whose variable terms cancel while its constants disagree. A question supplying a particular solution instead lets you substitute that value and solve for k.

An efficient decision tree for parameter questions

Given a solution? Substitute it into the original equation.

Asked for no or infinitely many solutions? Expand and compare the total coefficients and constants.

Asked for a general formula? Isolate the target, state the nonzero denominator condition, and consider the excluded case separately when needed.

Graph interpretation. View the two sides as $y = ax + b$ and $y = cx + d.$ One shared point means one solution; distinct parallel lines mean none; the same line means every x. This is the same classification you will later use for systems. A graph’s finite window is not enough to prove that two lines never meet; compare slopes or eliminate algebraically.

Final-target discipline. Write the answer with its meaning: “6 hours,” “original price $60,” or “ $2x - 1 = 7.$ ” A correct value of x earns no benefit when a different quantity was requested. For numeric response entry, submit only the required number, using the directions in the review section.

15 worked examples

1.01. Undo operations without losing a negative sign

What value of x satisfies 7 − 3x = 22? A) −5 B) 5 C) −29/3 D) 29/3

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Recognize the structure. Subtract 7 before dividing by the coefficient −3.

Work it through. Subtracting 7 from both sides gives $-3x = 15.$ Divide both sides by −3:

$\frac{15}{−3}$

Answer: A, −5.

Check. $7 - 3(-5) = 7 + 15 = 22,$ so the original equation is true.

Avoid the trap. The coefficient of x is −3, not 3. A positive right side does not imply a positive solution.

1.02. Distribute before collecting variable terms

If 4(2x − 3) = 3x + 18, what is x?

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Recognize the structure. Both terms inside the parentheses must be multiplied by 4.

Work it through. Expand and collect:

$8x - 12 = 3x + 18 5x = 30 x = 6.$

The subtraction of $3x$ and addition of 12 can be done in either order, as long as each operation is applied to both sides.

Answer: 6.

Check. The left side is $4(12 - 3) = 36;$ the right side is $18 + 18 = 36.$

Avoid the trap. $4(2x - 3)$ is $8x - 12,$ not $8x - 3.$ Distribution changes every term inside the parentheses.

1.03. Clear every denominator at once

What value of x satisfies $\frac{x − 2}{3}$ + $\frac{x + 1}{4}$ = 5?

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Recognize the structure. The least common multiple of 3 and 4 is 12. Multiply the entire equation by 12.

Work it through. The equation becomes

$4(x - 2) + 3(x + 1) = 60.$

Expand to get $4x - 8 + 3x + 3 = 60,$ or $7x - 5 = 60.$ Thus $7x = 65$ and $x = 65/7.$

Answer: $\frac{65}{7}$

Check. The two fractions become $17/7$ and $18/7,$ whose sum is $35/7 = 5.$

Avoid the trap. The right side must also be multiplied by 12. Keeping it equal to 5 would create a different equation.

1.04. The expression is already the unknown you need

If 5(2x − 1) − 7 = 28, what is the value of 2x − 1? A) 4 B) 7 C) 8 D) 35

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Recognize the structure. Treat $2x - 1$ as a single quantity instead of expanding it.

Work it through. Add 7 to both sides: $5(2x - 1) = 35.$

Dividing by 5 gives $2x - 1 = 7,$ which is exactly the requested quantity. There is no need to continue to $x = 4.$

Answer: B, 7.

Another route. Solving for x and substituting back also works, but adds an unnecessary step.

Avoid the trap. Choice A is the value of x, not the value asked for. Choice D stops one operation too early.

1.05. Make a decimal equation exact and manageable

If 0.35x + 1.8 = 0.20x + 5.4, what is x?

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Recognize the structure. The decimals are exact. Multiplying by 100 removes every decimal point.

Work it through. Rewrite the equation as $35x + 180 = 20x + 540.$ Subtract $20x$ and then 180:

$15x = 360 x = 24.$

Answer: 24.

Check. $0.35(24) + 1.8 = 10.2$ and $0.20(24) + 5.4 = 10.2.$

Another route. Subtract directly to get $0.15x = 3.6,$ then divide. Choose the route that is less likely to produce an arithmetic error for you.

Avoid the trap. Multiplying only the decimal coefficients and not the constants fails to preserve equality.

1.06. Recognize an identity

How many real solutions does 3(2x + 5) − 4 = 6x + 11 have?
A) Zero B) One C) Two D) Infinitely many

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Recognize the structure. Simplify both sides before trying to isolate x.

Work it through. The left side is $6x + 15 - 4 = 6x + 11.$ Thus the equation is

$6x + 11 = 6x + 11.$

Subtracting $6x + 11$ leaves $0 = 0,$ true for every real value of x. There are no denominator or context restrictions to remove any values.

Answer: D, infinitely many real solutions.

Avoid the trap. $0 = 0$ does not mean $x = 0.$ It means the equation places no restriction on x.

Key takeaway. If two simplified expressions are identical, they agree for every allowed input.

1.07. A disappearing variable can reveal a contradiction

How many real solutions does 5(x − 2) + 3 = 5x + 1 have?

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Recognize the structure. Compare the constants after the variable terms cancel.

Work it through. Expanding gives $5x - 10 + 3 = 5x + 1,$ so $5x - 7 = 5x + 1.$ Subtract $5x:$

$-7 = 1.$

This is false regardless of x. Therefore no real number makes the original equation true.

Answer: No real solutions.

Check. The left side is always 8 less than the right side. Changing x moves both sides by the same amount, so it cannot close that gap.

Avoid the trap. The disappearance of x does not automatically mean infinitely many solutions. The remaining statement must be true.

1.08. Translate a fixed fee and an hourly rate

A bicycle rental costs a fixed fee of $18 plus $7.50 per hour. A customer’s total charge is $63, with no other charges. How many hours was the bicycle rented?

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Recognize the structure. Total cost equals the fixed charge plus the hourly charge.

Work it through. Let h be the number of hours. Then

$18 + 7.50h = 63 7.50h = 45 h = 6.$

The units agree: dollars per hour multiplied by hours gives dollars.

Answer: 6 hours.

Check. Six hours costs $45 in hourly charges; adding $18 gives $63.

Avoid the trap. Dividing 63 by 7.50 would incorrectly treat the fixed fee as an hourly charge. Remove the fixed part first.

1.09. Reverse a discount without discounting the wrong quantity

A jacket is sold at a 20% discount from its original price. A $6 shipping charge is then added. The total is $54. What was the original price of the jacket, in dollars?

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Recognize the structure. The discount applies to the original jacket price, not to the shipping charge.

Work it through. Let p be the original price. The discounted jacket costs $0.80p,$ so

$0.80p + 6 = 54 0.80p = 48 p = 60.$

Answer: 60.

Check. Twenty percent of 60 is 12. The jacket costs $60 - 12 = 48$ after the discount, and $48 + 6 = 54.$

Avoid the trap. Adding 20% to 48 does not reverse a 20% discount. The two percentages would use different starting amounts. Divide by the remaining fraction, 0.80.

1.10. Rearrange a formula for a different variable

The variables F and C satisfy F = $\frac{9}{5}$ C + 32. Which expression gives C in terms of F?
A) $\frac{5F − 32}{9}$ B) $\frac{9(F − 32)}{5}$ C) $\frac{5(F − 32)}{9}$ D) $\frac{5(F + 32)}{9}$

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Recognize the structure. Undo the outer addition before undoing multiplication by $9/5.$

Work it through. Subtract 32, giving $F - 32 = (9/5)C.$ Multiply both sides by $5/9:$

$\frac{5}{9}$

Answer: C, $\frac{5(F − 32)}{9}$

Check. At $F = 32,$ the original equation requires $C = 0,$ which the answer produces.

Avoid the trap. The factor $5/9$ multiplies the entire quantity $F - 32.$ It does not multiply F alone.

1.11. Use a given solution to determine a coefficient

The equation a(3x − 2) = 4x + 10 has the solution x = 2. What is the value of a?

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Recognize the structure. A known solution must make the original equation true. Substitute before expanding.

Work it through. With $x = 2,$ the equation becomes

$a(6 - 2) = 8 + 10 4a = 18.$

Therefore $a = 18/4 = 9/2.$

Answer: $\frac{9}{2}$ or 4.5.

Check. For $a = 9/2$ and $x = 2,$ both sides equal 18. The coefficient of x after collecting is $3a - 4 = 19/2 \neq 0,$ so the equation indeed has a unique solution.

Avoid the trap. The unknown being requested is a, not x. The information $x = 2$ is something to use, not something to solve again.

1.12. Force cancellation, then verify the contradiction

For what value of k does k(2x − 3) = 8x + 5 have no solution?

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Recognize the structure. No solution requires equal coefficients of x but unequal constant terms.

Work it through. Expand to get $2kx - 3k = 8x + 5.$ The coefficients match when

$2k = 8 k = 4.$

For this value, the equation is $8x - 12 = 8x + 5,$ which reduces to $-12 = 5,$ a contradiction. If $k \neq 4,$ the coefficient $2k - 8$ is nonzero and there is one solution instead.

Answer: 4.

Avoid the trap. Setting the coefficients equal is only the first test. Always inspect the constants to distinguish no solution from infinitely many solutions.

1.13. Match two constants in an identity

The equation a(x − 4) + b = 7x + 9 is true for every real number x. What is a + b?

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Recognize the structure. An identity must have matching variable coefficients and matching constants.

Work it through. The left side expands to $ax - 4a + b.$ Matching coefficients gives $a = 7.$ Matching constants then gives $-4(7) + b = 9 b = 37.$

Thus $a + b = 7 + 37 = 44.$

Answer: 44.

Another route. Use two convenient inputs. At $x = 4,$ $b = 37.$ At $x = 0,$ $-4a + b = 9,$ so $a = 7.$ Two distinct inputs determine the two coefficients of these linear expressions.

Avoid the trap. Do not match b directly with 9: the left side’s full constant term is $-4a + b.$

1.14. A repeated expression reveals a shorter route

If 4(3x − 2) + 5 = 2(3x − 2) + 23, what is the value of 6x − 4?

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Recognize the structure. The target is twice the repeated expression: $6x - 4 = 2(3x - 2).$

Work it through. Subtract $2(3x - 2)$ from both sides and subtract 5:

$2(3x - 2) = 18.$

The left side is exactly $6x - 4,$ so the requested value is already isolated.

Answer: 18.

Another route. Let $u = 3x - 2.$ Then $4u + 5 = 2u + 23,$ so $u = 9$ and the target $2u = 18.$

Avoid the trap. The value 9 is only the repeated unit; the question asks for twice that unit. Expanding is valid, but it hides the useful structure.

1.15. A symbolic denominator carries a condition

The constants p and q satisfy p ≠ q. If p(x − 4) = q(x + 1), which expression equals x?
A) $\frac{4p − q}{p − q}$ B) $\frac{4p + q}{p − q}$ C) $\frac{4p + q}{p + q}$ D) $\frac{p − q}{4p + q}$

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Recognize the structure. Collect the x terms, factor out x, and use the stated nonzero condition.

Work it through. Expanding and rearranging gives

$px - 4p = qx + q (p - q)x = 4p + q.$

Since $p - q \neq 0,$ division is valid: $x = (4p + q)/(p - q).$

Answer: B, $\frac{4p + q}{p − q}$

Key takeaway. The condition is not decorative. If $p = q,$ the equation instead becomes $0 = 5p:$ it has no solution when $p = q \neq 0,$ and every real x works when $p = q = 0.$

Mastery check and error prevention

Try these without notes. Attempt the exit check below, then reveal each answer and explanation.

Topic practice

Before leaving this section

Check that you can distribute a negative sign, multiply every term when clearing denominators, distinguish a true identity from a contradiction, and explain why a division by an unknown constant is allowed. When an answer is wrong, locate the first invalid line rather than restarting blindly.