Linear Equation Word Problems

📝 Linear Equation Word Problems

Part 1 of 5 — Translating Words into Algebra

Topics in This Part

🔑 Key Concept: A word problem is just an equation wearing a costume. The whole skill is translation — turning English phrases into algebra one piece at a time. Master that, and every problem becomes a routine equation to solve.

The Translation Dictionary

Most of word-problem difficulty disappears once you memorize how everyday words map to operations.

⚠️ Watch the order with "less than." "$5$ less than a number" is $x - 5$, not $5 - x$. The phrase less than flips the order you read it in. Same with "subtracted from."

Quick examples

Concept Check 🎯

Choosing and Naming the Variable

Every word problem starts the same way: decide what the unknown is and give it a name. Write a short sentence — a let statement — so you never lose track.

🔑 Always write a "Let" statement. Let $x =$ the number of tickets sold. Let $w =$ the width of the rectangle, in feet.

The single most useful trick: when a problem describes two related quantities, name the smaller or simpler one $x$, then write the other in terms of $x$.

Example

"A rope is cut into two pieces. One piece is $3$ feet longer than the other."

• Let $x =$ length of the shorter piece (feet).
• Then the longer piece $= x + 3$.

You now have both quantities described with a single variable — exactly what you need to build one equation.

Name the Second Quantity 🔽

Pick the correct expression for the second quantity, given the variable definition.

Turning a Whole Sentence into an Equation

The word "is" (or equals, gives, results in) is the $=$ sign. Translate the left side, drop the $=$ where "is" appears, then translate the right side.

Example

"When $5$ is added to twice a number, the result is $17$."

$\underbrace{2x + 5}_{\text{twice a number, plus }5} \;=\; \underbrace{17}_{\text{the result}}$

That's a complete linear equation, ready to solve in Part 2.

💡 The 4-step plan you'll use every time:

1. Read and decide what's unknown.
2. Let a variable name it (and name related quantities in terms of it).
3. Translate the sentence into an equation.
4. Solve, then check the answer against the original words.

Build the Equation 🎯

Don't solve yet — just choose the equation that matches the words.

📝 Linear Equation Word Problems

Part 2 of 5 — Number, Consecutive-Integer & Age Problems

🔑 The Idea: These classic problems all use the same engine: name one unknown $x$, write every other quantity in terms of $x$, build one equation, solve, and check. We'll do each type with a full worked example.

Number Problems

Worked Example

"The sum of three times a number and $8$ is $-7$. Find the number."

Let $x =$ the number. Translate: $3x + 8 = -7$. Solve:

$3x + 8 = -7 \;\Rightarrow\; 3x = -15 \;\Rightarrow\; x = -5$

Check: $3(-5) + 8 = -15 + 8 = -7$ ✓

💡 Always finish by plugging the answer back into the original sentence, not just the equation — that catches translation mistakes, the most common kind.

Number Problems 🧮

Set up an equation and solve. Enter the number.

1) Five more than twice a number is $23$. The number $= \,?$ 2) Four less than half a number is $10$. The number $= \,?$ 3) When a number is decreased by $9$, the result is $-2$. The number $= \,?$

Consecutive Integer Problems

Consecutive integers go up by $1$: if the first is $x$, the next are $x+1$, $x+2$, … Consecutive even or odd integers go up by $2$: $x$, $x+2$, $x+4$, … (both even and odd jump by $2$).

Worked Example

"The sum of three consecutive integers is $72$. Find them."

Let $x =$ first integer. Then the three are $x,\; x+1,\; x+2$.

$x + (x+1) + (x+2) = 72$ $3x + 3 = 72 \;\Rightarrow\; 3x = 69 \;\Rightarrow\; x = 23$

So the integers are $23,\, 24,\, 25$. Check: $23+24+25 = 72$ ✓

⚠️ For consecutive even or odd integers, the gap is $2$, not $1$ — even though $x+2$ has the same form for both. The starting value of $x$ is what makes them even or odd.

Set Up Consecutive Integers 🔽

Choose the correct setup for each problem (don't solve).

Solve the Consecutive Integers 🧮

1) The sum of two consecutive integers is $45$. Enter the smaller integer. 2) The sum of three consecutive odd integers is $81$. Enter the middle integer.

Age Problems

Age problems link two people's ages, then often jump forward or backward in time.

Worked Example

"Tara is $3$ times as old as Ben. In $6$ years, the sum of their ages will be $44$. How old is each now?"

Let $x =$ Ben's age now. Then Tara $= 3x$ now. In $6$ years: Ben $= x + 6$, Tara $= 3x + 6$.

$(x + 6) + (3x + 6) = 44 \;\Rightarrow\; 4x + 12 = 44 \;\Rightarrow\; 4x = 32 \;\Rightarrow\; x = 8$

Ben is $8$, Tara is $24$. Check: in $6$ years $14 + 30 = 44$ ✓

⚠️ Add the time change to everyone's age. A common slip is aging one person but not the other.

Age Setup 🎯

📝 Linear Equation Word Problems

Part 3 of 5 — Money, Coins & Percent Problems

🔑 The Idea: When dollars and cents are involved, the key is to track value, not just count. A nickel is worth $5$ cents, so $n$ nickels are worth $5n$ cents. Percent problems lean on a single template: part $=$ percent $\times$ whole.

Coin & Money Problems

The trap: a pile of $10$ coins isn't worth $10$ — it depends on which coins. Separate how many from how much each is worth.

Worked Example

"A jar has $7$ more dimes than nickels. The coins are worth $2.45 in total. How many nickels are there?"

Let $x =$ number of nickels. Then dimes $= x + 7$. Value in cents ($2.45 $= 245$ cents):

$\underbrace{5x}_{\text{nickels}} + \underbrace{10(x + 7)}_{\text{dimes}} = 245$ $5x + 10x + 70 = 245 \;\Rightarrow\; 15x = 175 \;\Rightarrow\; x = \tfrac{175}{15}$

Hmm — that isn't a whole number, so let's recheck with cleaner numbers below. The method is what matters: (count) × (value each) summed = total value.

⚠️ Keep all money in the same unit. Mixing dollars and cents is the #1 coin-problem error. Convert $2.45 to $245$ cents (or write everything in dollars: $0.05x + 0.10(x+7) = 2.45$).

A Clean Worked Example

"A jar has $5$ more dimes than nickels and is worth $3.50. How many nickels?"

Let $x =$ nickels, so dimes $= x + 5$. In cents ($3.50 $= 350$):

$5x + 10(x + 5) = 350$ $5x + 10x + 50 = 350 \;\Rightarrow\; 15x = 300 \;\Rightarrow\; x = 20$

So there are $20$ nickels and $25$ dimes. Check: $20(5) + 25(10) = 100 + 250 = 350$ cents, i.e. $3.50 ✓

Coin Drill 🧮

Work in cents to stay clean.

1) $x$ nickels and twice as many dimes are worth $125$ cents. Find $x$ (number of nickels). 2) A box has $x$ quarters and $x$ dimes (equal numbers), worth $175$ cents total. Find $x$.

Percent Problems

Every basic percent problem fits one template:

$\text{part} = \text{percent (as a decimal)} \times \text{whole}$

Convert the percent to a decimal first: $25\% = 0.25$, $7\% = 0.07$, $150\% = 1.5$.

Three flavors

💡 Tip: "of" almost always means multiply, and "is" means equals — so "$30\%$ of $80$" becomes $0.30 \times 80$.

Percent Concept Check 🎯

Percent Drill 🧮

Use part $=$ percent $\times$ whole. Enter numbers only (no $\%$ sign or $).

1) $12$ is $30\%$ of what number? 2) A $45 shirt has $20\%$ added in tax. What is the total cost, in dollars?

📝 Linear Equation Word Problems

Part 4 of 5 — Distance, Rate & Time + Geometry & Perimeter

🔑 The Idea: Motion problems all run on one formula, $d = rt$. Geometry word problems run on a perimeter or area formula. In both, you draw or tabulate the quantities, write each in terms of $x$, then build one equation.

Distance = Rate × Time

$d = r \cdot t$

Build a small table — one row per moving object — and fill in two of the three columns; the third is the product (or the unknown).

Worked Example

"Two cars leave the same point going opposite directions. One drives $55$ mph, the other $45$ mph. After how many hours are they $300$ miles apart?"

Let $t =$ time in hours (same for both).

Going opposite ways, the distances add to $300$:

$55t + 45t = 300 \;\Rightarrow\; 100t = 300 \;\Rightarrow\; t = 3 \text{ hours}$

Check: in $3$ h, car A goes $165$ mi, car B goes $135$ mi; $165 + 135 = 300$ ✓

Motion Setup 🎯

Motion Drill 🧮

1) A cyclist rides at $12$ mph. How many hours to travel $54$ miles? (Decimal ok.) 2) Two buses leave a station in opposite directions at $40$ mph and $50$ mph. After how many hours are they $270$ miles apart?

Geometry & Perimeter Problems

Pull in the right formula, then express each dimension in terms of one variable.

Worked Example

"A rectangle's length is $4$ cm more than its width, and its perimeter is $36$ cm. Find the dimensions."

Let $w =$ width. Then length $\ell = w + 4$.

$2(w + 4) + 2w = 36 \;\Rightarrow\; 2w + 8 + 2w = 36 \;\Rightarrow\; 4w = 28 \;\Rightarrow\; w = 7$

Width $= 7$ cm, length $= 11$ cm. Check: $2(11) + 2(7) = 22 + 14 = 36$ ✓

Geometry Setup 🔽

Geometry Drill 🧮

1) A rectangle's length is $3$ ft more than its width, and the perimeter is $26$ ft. Find the width, in feet. 2) A triangle's angles are $x$, $x$, and $x + 30$ degrees. Find $x$.

📝 Linear Equation Word Problems

Part 5 of 5 — Mixed Mastery & Exit Quiz

You can now (1) translate words into algebra, (2) handle number, consecutive-integer, and age problems, (3) work money and percent problems, and (4) set up distance and geometry problems. Time to mix them all together.

The Universal Game Plan

⚠️ Two recurring traps:

• "Less than" / "subtracted from" flip the order ($5$ less than $x$ is $x - 5$).
• Answer what's asked. If you let $x$ be the width but the problem wants the length, don't stop at $x$ — convert.

💡 Keep units consistent (all cents, or all dollars; all hours, etc.) and the algebra stays clean.

Spot the Setup 🔽

Match each problem to its correct equation (don't solve).

Mixed Practice 🧮

Solve each fully and enter the requested value.

1) The sum of three consecutive integers is $90$. Enter the largest. 2) A coat costs $120 after a $20\%$ discount was taken off the original price. What was the original price, in dollars? 3) A rectangle's length is $2$ cm more than $3$ times its width. Its perimeter is $44$ cm. Find the width, in cm.

Mixed Concept Check 🎯

Exit Quiz ✅

Answer all three to finish the lesson.