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MK · lesson 5 of 14 · 39 min

Solving Linear Equations

Solving for an unknown is a core skill on Mathematics Knowledge, and this lesson treats it as a reliable routine rather than guesswork. It starts by separating expressions from equations and picturing equality as a balance, then builds through inverse operations, two-step equations, negative coefficients, distribution, combining like terms and variables on both sides.

Video lesson: Linear Equations

Video transcript

0:00 Welcome! So, you're prepping for the ASVAB and maybe feeling a little uneasy about the math section? Well, you are definitely not alone. I'm so glad you're here for this explainer, because today we are going to completely demystify linear equations. The mathematics knowledge subtest relies heavily on high school algebra principles, specifically moving things around and isolating variables. And here's the kicker. No calculators allowed. Zero. But hey, don't worry, conquering these equations is literally just about following a steady logical step-by-step process with nothing but your pencil and scratch paper. Once you start seeing the patterns, it honestly becomes really satisfying. Here's our clear roadmap for today. We'll start with equation basics and rules, then move on to handling fractions and decimals. After that, word problems and formulas, followed by some step-by-step worked examples. Finally, we'll wrap up with costly mistakes to avoid, and a quick check yourself final practice. It's a logical, totally manageable path. Alright, let's kick things off with section 1, equation basics and rules. Before we start solving anything, we really need to understand exactly what we're looking at.

1:01 Notice the clear distinction here between an expression and an equation. An expression, like 4x minus 9, is basically just a mathematical phrase. It doesn't have an equal sign. You can simplify it, sure, but you can't actually solve it. An equation, on the other hand, makes a bold claim of balance. It says 4x minus 9 is exactly equal to 11. That equal sign? It's everything. It tells us we have a real claim that we can actively solve by finding the specific value of x that makes the whole statement true. Think of an equation like a perfectly balanced scale. Whatever you do to one side, you have got to do to the other to keep that balance. We solve these using inverse operations, which is really just a fancy way of saying we undo the math in reverse order. So take a look at step 1. 4x minus 5 equals 23. To undo that subtraction, we just add 5 to both sides. That brings us nicely to step 2. 4x equals 28. Now to undo the multiplication of 4 times x, we divide both sides by 4.

2:03 And that gives us our final answer right there in step 3. x equals 7. Simple, perfectly balanced, and totally step by step. Handling fractions and decimals. Now, this is where things can look a little intimidating, but seriously, don't let it psych you out. Those scary looking fractions and decimals can be wiped out right at the start. This right here illustrates one of my absolute favorite shortcuts in algebra. When you see multiple fractions, like in step 1 where we have x over 2 plus x over 3 equals 10, all you need to do is find the least common denominator. For 2 and 3, well, that's 6. Now in step 2, multiply every single term on both sides by that LCD of 6. And like magic, in step 3, the fractions are completely gone. We're just left with 3x plus 2x equals 60. Combine your like terms to get 5x equals 60, divide by 5, and boom, in step 4, x equals 12. No calculator, no messy fraction addition, just a clean whole number equation.

3:04 And guess what? The exact same trick works for decimals. You just multiply every term by a power of 10. Word problems and formulas. We're shifting our focus to translating plain English into math equations, which is an absolutely vital skill for the ASVAB. The crucial point here is just learning the vocabulary of algebra. Let's break down this phrase. 5 less than 3 times a number is 22. First, we assign a letter. Let's just use n to represent our unknown number. So 3 times a number becomes 3n. The word is translates directly to an equal sign. Now here is the tricky part. 5 less than actually means we subtract 5 at the very end, kind of working in reverse order. So step 1 gives us our perfectly translated equation. 3n minus 5 equals 22. From there, we just balance the scale. Add 5 to get 3n equals 27 in step 2, and then divide by 3 to get n equals 9 in step 3. Easy, right? All right, let's roll up our sleeves for section 4.

4:04 Step by step worked examples. We're going to meticulously execute seven specific examples exactly as they appear on the test to really build up your confidence. Let's get into it. Here's worked example 1. Step 1. Negative 2x plus 9 equals 1. First up, we subtract 9 from both sides, which brings us to step 2. Negative 2x equals negative 8. Now whenever the coefficient is negative, you have to divide by that exact negative number. Sign included. So we divide both sides by negative 2. Remember, a negative divided by a negative is a positive. So step 3 gives us x equals 4. Let's check out worked example 2, which brings parentheses into the mix. Step 1. 6 times the quantity x minus 2 plus 4 equals 22. Our very first move is to distribute that six to everything inside those parentheses. That gives us step 2. 6x minus 12 plus 4 equals 22. Now before you start doing anything across the equals sign, clean up your own house first.

5:06 Combine your like terms on the left. Negative 12 plus 4 is negative 8, leading to step 3. 6x minus 8 equals 22. Add 8 to both sides to get 6x equals 30, divide by 6, and there's step 4. x equals 5. Worked example 3 reinforces that beautiful fraction clearing rule we talked about earlier. Step 1. x over 4 minus 2 equals x over 6. The lowest common denominator for 4 and 6 is 12. So step 2, we multiply every single term by 12. And please remember, even that constant 2 gets multiplied. This gives us step 3. 3x minus 24 equals 2x. Subtract 2x from both sides, and you get x minus 24 equals 0. Add 24, and you're done. Step 4. x equals 24. Now for worked example 4, where we're dealing with a fractional coefficient. Step 1. 2 fifths x plus 3 equals 11. First things first, keep the balance by subtracting 3 from both sides. That leaves us with step 2.

6:10 2 fifths x equals 8. Here is a super fast shortcut. To get rid of a fraction that's multiplied by a variable, just multiply both sides by its reciprocal. Basically flip the fraction upside down. So in step 3, we multiply both sides by 5 over 2. The fraction on the left turns into a 1, and on the right, 8 times 5 halves equals 20. Step 4. x equals 20. Worked example 5 highlights two very special but super important edge cases. Sometimes when you're solving, the variable terms just cancel out completely on both sides. If you're left with a totally false statement, like 2 equals 5, that is a contradiction. It simply means there is no solution. But if you're left with a universally true statement, like 4x minus 12 equals 4x minus 12, well, that's called an identity. It means there are infinitely many solutions. Literally any real number you plug in for x is going to work. Let's look at worked example 6, which is all about rearranging a literal formula. This looks tricky because it's almost entirely letters, but the secret is to just treat those other letters exactly like numbers.

7:12 Step 1 shows the formula for the area of a triangle, a equals 1 half b times h, and we want to solve for h. First, let's clear out that fraction by multiplying both sides by 2, giving us step 2. 2a equals bh. Now, since b is multiplying h, we undo that by dividing both sides by b. That cleanly leaves us with step 3, h equals 2a over b. Finally, worked example 7 ties a little geometry back into our algebra. Imagine an isosceles triangle with 2 equal sides of length x, and a third side that's 6 units longer. The perimeter, which is just the total distance around the outside, is 39. So step 1 is building that equation. x plus x plus the quantity x plus 6 equals 39. Combine those like terms on the left side for step 2. 3x plus 6 equals 39. Subtract 6 to get step 3, 3x equals 33, divide by 3, and step 4 reveals that x equals 11. Okay, section 5, costly mistakes, and review. You've seen the right way to do things, but let's lean in a bit here because we need to look closely at the specific traps that constantly drain points from test takers on the ASVAB.

8:20 Here is perhaps the single most common error of all. When distributing a number across parentheses, like 3 times the quantity x plus 4, a lot of people rush and only multiply the first term. They write down 3x plus 4. That is definitely wrong. You have to distribute that 3 to every single thing inside the parentheses. The right way is 3 times x plus 3 times 4, which gives you 3x plus 12. Don't leave part of your product behind. Think of this as your mental diagnostic checklist to avoid losing easy points. First, dropping negative signs when dividing. Always, always keep the sign attached to the coefficient. Second, applying operations to only one side. Remember our balance scale? What you do to one side, you absolutely must do to the other. And finally, missing the substitution check. Honestly, this one is mandatory. When you get an answer, take a second to plug it back into the original equation. If both sides equal the same number, you know with 100% certainty you got it right. It takes seconds and it is entirely foolproof.

9:23 We have reached the final section. Section 6, check yourself final practice. I want to invite you to test out this new calculator-free confidence with two actual practice questions, straight from our source material. So grab that pencil and scratch paper. Here's question 1. What is the solution of negative 4x minus 3 equals 21? Is it A, negative 4.5, B, negative 6, C, 6, or D, 4.5? I'll give you a second to pause the explainer if you want to work the problem out on your paper right now. All right, let's check it. The correct answer is B, negative 6. Let's walk through it together. Step 1 is our equation, negative 4x minus 3 equals 21. We undo the subtraction by adding 3 to both sides, which hands us step 2, negative 4x equals 24. Finally, we divide both sides by negative 4. Remember that rule we just talked about? Keep the negative sign. 24 divided by negative 4 gives us step 3, x equals negative 6.

10:25 Outstanding. Now for the final question, a classic word translation challenge. 7 more than twice a number is 31. What is the number? Is it A, 8.5, B, 19, C, 12, or D, 24? Take one last quick pause to translate this English phrase into an equation and solve it. And the correct answer is C, 12. Let's build the equation in step 1. Twice a number is 2n, and 7 more than means we add 7 at the end. Is means equals. So we get 2n plus 7 equals 31. Subtract 7 from both sides for step 2, getting 2n equals 24. Divide by 2, and step 3 gives us n equals 12. Just look at what you've accomplished here today. Step by step, maintaining balance, and clearing out obstacles like fractions and negative signs. Think about how empowering it is to walk into a testing room knowing you can completely deconstruct and beat a math test with nothing more than pure logic, careful steps, and a simple pencil. You've got the tools now. So how many more points are you going to rack up on the ASVAB using these exact methods?

11:32 Keep practicing, and thanks for joining me.

Visual lesson

Lesson slides

15 slides
Linear Equations – slide 1 of 15

Lesson Goal and Exam Relevance

In this lesson you will solve one-variable linear equations and check every answer. You will handle signed and negative coefficients, parentheses, fractions, decimals and variables on both sides, and you will recognize equations with no solution or with infinitely many solutions. You will also turn word problems and formulas into equations and solve them.

Equation solving is one of the most heavily tested Mathematics Knowledge skills. The questions are designed to be done without a calculator, so a clean, reliable step-by-step method is what earns points.

Expressions Versus Equations

An expression is a mathematical phrase such as . It has no equals sign, so it can be simplified or evaluated but not solved. An equation such as

claims that two expressions have the same value. Solving means finding the value of the variable that makes that claim true. Here x = 5 works, because

A linear equation has the variable only to the first power, never squared, and usually has exactly one solution.

Equality as a Balance

Think of an equation as a balanced scale: the left side and the right side weigh the same. You can add, subtract, multiply or divide by the same amount on both sides, and the scale stays balanced. The one exception is that you may not multiply or divide both sides by zero. For example, if

subtracting 6 from both sides leaves

Whatever you do to one side, you must do to the other, or the new equation will no longer have the same solution.

Balance scale for x + 6 = 15: an x block and 6 unit weights on the left, 15 on the right, with 6 weights outlined and tagged −6 on each pan; after subtracting 6 from both sides, x balances 9 units, so x = 9.

Inverse Operations and Two-Step Equations

Undo operations in reverse order using their opposites: subtraction undoes addition, and division undoes multiplication. For

first add 5 to get

then divide by 4 to get

Building 4x − 5 from x by multiplying by 4 then subtracting 5; a connector shows 4x − 5 = 23, and undoing in reverse order, 23 plus 5 is 28 and 28 divided by 4 gives x = 7.

Negative Coefficients and Signed Arithmetic

When the coefficient is negative, divide by that negative number, sign included. For

subtract 4 to get

Divide both sides by −3:

Check:

Distribution Before Collecting Terms

If the variable is inside parentheses, distribute first. For

write

so
and

Combine Like Terms

Before isolating the variable, combine like terms on each side separately. For example:

The x terms give 5x and the constants give 4, so the equation becomes

Subtract 4 to get

then divide by 5:

Combining first keeps the number of steps small and reduces sign errors.

Variables on Both Sides

When x appears on both sides, move all variable terms to one side and all constants to the other. For

subtract 3x from both sides to get

Add 4 to get

so

Moving the smaller variable term avoids a negative coefficient.

Equations With Parentheses on Both Sides

Distribute on each side, then gather terms. For

Subtract 2x and add 6:

Check: both sides equal 36.

Fractional Coefficients

When the variable has a fractional coefficient, multiply both sides by the reciprocal of that fraction. For example:

Multiply both sides by :

The reciprocal turns the coefficient into 1 in a single step.

Clearing Several Denominators

With more than one fraction, multiply every term on both sides by the least common denominator. For

the LCD is 6. Multiplying each term gives

so
and

Fractions on Both Sides

When a single fraction equals another, you can cross-multiply. For

write

then
so

Keep binomial numerators in parentheses.

Decimal Coefficients

Decimals can be cleared by multiplying every term by a power of ten. For

multiply by 10 to get

then
and

Use 100 if any term has two decimal places.

No Solution and Infinitely Many Solutions

Sometimes the variable terms cancel completely. If what remains is false, such as

reducing to

the equation is a contradiction and has no solution. If what remains is always true, such as

reducing to

the equation is an identity and every real number is a solution. In both cases, the two sides have the same x coefficient once simplified.

Equivalent Equations and Reversibility

Each legal step, adding, subtracting, multiplying or dividing both sides by the same nonzero number, produces an equivalent equation: one with exactly the same solution set. Because each step can be undone, you can travel from the original equation to the answer and back without gaining or losing solutions. That is why the final line, such as

tells you the solution of the first line. Multiplying by zero is not reversible, which is why it is not allowed: it turns every equation into

Translate Words Into a Linear Equation

Assign a letter to the unknown, then translate phrases piece by piece: is means equals, of often means multiply, more than means add, and less than means subtract in reverse order. For example, five less than three times a number is 22 becomes

then
and
The phrase "five less than three times a number is 22" mapped piece by piece to 3n − 5 = 22, with "five less than" becoming − 5 placed after 3n; n = 9.

Consecutive Integer Relationships

Consecutive integers differ by 1, so call them n, n + 1 and n + 2. Consecutive even or odd integers differ by 2: n, n + 2, n + 4. If three consecutive integers add to 72, then

so n = 23 and the integers are 23, 24 and 25.

Number line from 22 to 26 with 23, 24 and 25 marked as n, n + 1 and n + 2, each one step of +1 apart; 23 + 24 + 25 = 72.

Perimeter as a One-Variable Equation

Perimeter is the total distance around a figure. For a rectangle,

Write both dimensions in terms of one variable. If the length is 3 more than the width and the perimeter is 54, then

so

That gives w = 12 and a length of 15.

Rectangle drawn to scale with width w = 12 and length w + 3 = 15; perimeter 15 + 12 + 15 + 12 = 54 from 2(w + 3) + 2w = 54.

Rearranging a Literal Linear Formula

A literal equation has several letters. To solve for one of them, treat the other letters like numbers and use the same inverse operations. From

divide both sides by r to get

From

subtract 2l and divide by 2 to get

Solutions With Fractions or Decimals

Not every answer is a whole number. If

then

which equals 1.5. If

then:

This is about 2.67. Leave answers as reduced fractions unless the choices are decimals, and convert carefully when matching an answer choice.

Choose an Efficient Solving Route

There is often a shortcut. If one side is a multiple of the coefficient outside the parentheses, divide first instead of distributing:

Dividing both sides by 5 gives

so

with no distribution needed. When fractions appear, clear them at the start. When x is on both sides, move the smaller x term so the coefficient stays positive. On multiple-choice items, you can also plug answer choices into the equation, starting with a middle value.

Substitution as a Mandatory Check

Always check by substituting your answer into the original equation, not a later step, because an error in an early step will carry through. If both sides give the same number, the answer is right. For example, if you solve

and get x = 5, compute

It matches. A check takes seconds and catches most sign and arithmetic slips.

Common Error Diagnostic

If your check fails, look for these causes in order. Did you distribute to every term inside the parentheses? Did you flip every sign after a minus in front of parentheses? Did you multiply every term, including constants, when clearing denominators? Did you keep the sign when dividing by a negative coefficient? Did you apply the operation to both sides? Fix the first failure you find, then check again.

Worked Example 1 — Two-Step Signed Equation

Problem: Solve

Step 1: Get the x-term alone. Undo the + 9 by subtracting 9 from both sides.

Step 2: Undo the multiplication by −2. Divide both sides by −2. A negative divided by a negative is positive.

Answer: x = 4.

Check: Put 4 back into the original equation.

It matches the right side, so x = 4 is correct.

Worked Example 2 — Distribute Then Solve

Problem: Solve

Step 1: Distribute the 6. Multiply it by each term inside the parentheses:

6 × x = 6x
and
6 × 2 = 12

Step 2: Combine the plain numbers on the left side:

−12 + 4 = −8

Step 3: Undo the subtraction by adding 8 to both sides, so 6x is alone.

Step 4: Undo the multiplication by dividing both sides by 6.

Answer:

Check: Put 5 back in. First

5 − 2 = 3
then
6 × 3 = 18
and
18 + 4 = 22

It matches.

Worked Example 3 — Clear Denominators

Problem: Solve

Step 1: Get rid of the fractions. The smallest number that both 4 and 6 divide into is 12, so multiply every term on both sides by 12.

Step 2: Simplify each term:

12 ÷ 4 = 3

so the first term is 3x;

12 × 2 = 24
12 ÷ 6 = 2

so the right side is 2x.

Step 3: Move the x terms to one side by subtracting 2x from both sides.

Step 4: Add 24 to both sides to get x alone.

Answer:

Check: Left side:

24 ÷ 4 = 6
and
6 − 2 = 4

Right side:

24 ÷ 6 = 4

Both sides equal 4.

Worked Example 4 — Fractional Coefficients

Problem: Solve

Step 1: Undo the + 3 first. Subtract 3 from both sides:

11 − 3 = 8

Step 2: To undo multiplying by , multiply both sides by its flip (reciprocal), . On the left,

which leaves just x.

Step 3: Multiply:

8 × 5 = 40
then
40 ÷ 2 = 20

Answer:

Check: Two-fifths of 20 means

20 ÷ 5 = 4
then
4 × 2 = 8

Add 3:

8 + 3 = 11

It matches.

Worked Example 5 — Contradiction Versus Identity

Problem: Decide how many solutions each equation has: (a)

and (b)

Step 1: Distribute in (a):

2 × 3x = 6x
and
2 × 1 = 2

Step 2: Subtract 6x from both sides. The x terms disappear.

This statement is false, and no value of x can fix it. So (a) has no solution (a contradiction).

Step 3: Distribute in (b):

3 × x = 3x
and
3 × 4 = 12

Then combine

3x + x = 4x

Step 4: Both sides are exactly the same, so the equation is true for every x (an identity).

Answer: (a) has no solution; (b) has infinitely many solutions.

Check: Try x = 0 in (b):

3(−4) + 0 = −12
and
4(0) − 12 = −12

Equal. In (a), x = 0 gives 2 on the left and 5 on the right. Not equal.

Worked Example 6 — Formula Rearrangement

Problem: Solve

for h. Then find h when A = 30 and b = 12.

Triangle drawn to scale with base b = 12, a dashed altitude marked h = ? meeting the base at a right angle, and area A = 30.

Step 1: Clear the fraction by multiplying both sides by 2. On the right, 2 × = 1.

Step 2: h is being multiplied by b, so divide both sides by b to get h alone.

Step 3: Substitute A = 30 and b = 12. First

2 × 30 = 60

Step 4: Divide:

60 ÷ 12 = 5

Answer:

and h = 5 when A = 30 and b = 12.

Check: Put the numbers into the original formula: one half of 12 × 5 is one half of 60, which is 30. That equals A.

Worked Example 7 — Geometry Application

Problem: An isosceles triangle has two equal sides of length x, and the third side is 6 longer than x. The perimeter is 39. Find the sides.

Isosceles triangle with the two equal sides marked x (tick marks), the third side x + 6, and perimeter P = 39.

Step 1: The perimeter is the sum of all three sides. Write that as an equation.

Step 2: Combine like terms:

x + x + x = 3x

Step 3: Subtract 6 from both sides:

39 − 6 = 33

Step 4: Divide both sides by 3:

33 ÷ 3 = 11

Step 5: Find the third side:

x + 6 = 11 + 6 = 17

Answer: The sides are 11, 11 and 17.

Check:

11 + 11 = 22
and
22 + 17 = 39

That is the given perimeter, and 17 is indeed 6 more than 11.

Isosceles triangle drawn to scale with two equal sides x = 11 and third side x + 6 = 17; perimeter 39.

Mistakes That Cost MK Points

Applying an operation to only one side breaks the balance; always do the same thing to both sides. Distributing to the first term only, as in writing as , loses part of the product. After a minus sign in front of parentheses, every sign inside must flip, not just the first. When clearing fractions, multiply every term, including whole-number constants. Dividing by a negative coefficient but dropping its sign gives the opposite of the correct answer. Stopping at

or

without interpreting it misses the no-solution or identity case. Finally, choosing an answer without substituting it back skips the easiest error check you have.

Quick Review: Linear Equations

A linear equation sets two expressions equal and can be solved for its variable. Keep the equation balanced by doing the same operation to both sides. Simplify each side first: distribute, flip signs after a leading minus, and combine like terms. Collect variable terms on one side and constants on the other, then divide by the coefficient, sign included. Clear fractions by multiplying every term by the least common denominator, and clear decimals by multiplying by a power of ten. If the variable cancels and a false statement remains, there is no solution; if a true statement remains, every number is a solution. Translate word problems carefully, rearrange formulas with the same inverse operations, and always check your answer in the original equation.

Quick review

What is the balance rule for solving an equation?
Do the same thing to both sides: add, subtract, multiply or divide both sides by the same number. Never multiply both sides by zero (that wipes out the solution, leaving 0 = 0), and never divide by zero, which is undefined.
In what order do you solve a two-step equation such as 5x + 3 = 38?
Undo the operations in reverse order using inverse operations: subtract 3 first to get 5x = 35, then divide by 5 to get x = 7. Check: 5(7) + 3 = 38.
Solving −4x = 28: what is the common mistake?
Dropping the negative sign. Divide both sides by −4, sign included: x = −7. Writing x = 7 gives the opposite of the correct answer. Check: −4(−7) = 28.
How do you solve with x on both sides, e.g. 9x − 2 = 4x + 18?
Subtract the smaller x term from both sides to keep a positive coefficient: subtract 4x to get 5x − 2 = 18, add 2 to get 5x = 20, so x = 4.
How do you clear fractions or decimals from an equation?
Multiply every term on both sides by the LCD (or a power of ten for decimals), constants included. For , multiply by 12: 4x + 3x = 84, so x = 12.
What does it mean if the variable cancels out completely?
A false leftover such as 5 = 9 means the equation is a contradiction with no solution. A true leftover such as 3 = 3 means it is an identity: every real number is a solution.
Translate: four less than twice a number is 18.
'Is' means equals and 'less than' subtracts in reverse order: 2n − 4 = 18. Then 2n = 22 and n = 11.
How do you solve the formula P = 2l + 2w for l?
Treat the other letters as numbers and use inverse operations: subtract 2w, then divide by 2, giving .

What next?

← Previous: Algebraic Expressions

About this lesson

Fractions and decimals get extended attention: clearing several denominators, fractions on both sides and decimal coefficients. You will also learn to spot equations with no solution or infinitely many, translate words into equations, handle consecutive integers and perimeter problems, and rearrange a literal formula. Substitution is presented as a mandatory final check, and a common-error diagnostic helps you catch mistakes before you commit.

Seven worked examples show each technique in a realistic setting, and a section on mistakes that cost MK points closes the lesson. Since MK feeds the AFQT, this is high-value study time. Expect about 39 minutes, then measure your progress with the lesson test.

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