Linear Inequalities Pattern - Translate

This pattern asks you to translate a real-world scenario into an inequality. You read a word problem about budgets, capacities, goals, or constraints, then pick (or write) the inequality that models it. The core skills are matching quantities to the correct coefficients, choosing the right inequality direction, and handling fixed costs or known values.

 

Two Variables with a Capacity Limit

The simplest version: two types of items contribute to a total that must stay under (or at) a limit. Multiply each item's rate by its count and set up the inequality.

An aquarium hobbyist is stocking a new tank and wants the total length of all fish to be under 80 inches. The hobbyist plans to add neon tetras, which have an average length of 1.5 inches, and guppies, which have an average length of 2 inches. Which inequality represents the possible number of neon tetras, $n$, and guppies, $g$, that can be placed in the tank?

A) $1.5n + 2g > 80$
B) $1.5n + 2g < 80$
C) $2n + 1.5g < 80$
D) $2n + 1.5g > 80$

Total length = $1.5n + 2g$. "Under 80" means $< 80$. Match each fish type to its own length: tetras are 1.5 inches (not 2), guppies are 2 inches (not 1.5). The answer is B. Watch for options that swap the coefficients.

When the problem says "no more than" or "at most," use $\leq$. When it says "at least" or "no less than," use $\geq$.

A small bakery has the capacity to produce no more than 250 baked goods each morning. Which inequality represents this situation, where $m$ is the number of muffins and $c$ is the number of croissants produced?

A) $m + c \geq 250$
B) $m - c \leq 250$
C) $m + c \leq 250$
D) $m - c \geq 250$

Total goods = $m + c$ (add, not subtract). "No more than" means $\leq$. So $m + c \leq 250$. The answer is C. Distractors use subtraction or the wrong direction.

 

Two Variables with a Minimum Goal

Same structure, but the total must meet or exceed a target.

An athlete is creating a meal plan and wants to consume at least 600 calories from a combination of yogurt and granola. Each serving of yogurt contains 150 calories, and each serving of granola contains 120 calories. Which inequality describes the possible servings of yogurt, $y$, and granola, $g$?

A) $150y + 120g \geq 600$
B) $150y + 120g \leq 600$
C) $120y + 150g \geq 600$
D) $120y + 150g \leq 600$

Total calories = $150y + 120g$. "At least 600" means $\geq 600$. The answer is A. Options C and D swap the coefficients — always match each coefficient to the right variable.

 

Fixed Cost Plus Variable Cost

Many problems have a one-time fee plus a per-unit cost, and the total must stay within a budget.

To join a gym, a new member pays a one-time sign-up fee of $50 plus a monthly fee of $15. A person has a budget of at most $200. If $m$ represents the number of months, which inequality describes this?

A) $50 + 15m \leq 200$
B) $15m \leq 200$
C) $15 + 50m \leq 200$
D) $50m \leq 200$

Total cost = fixed fee + variable cost = $50 + 15m$. "At most $200" means $\leq 200$. The answer is A. Option C swaps the fixed and variable amounts. Option B forgets the sign-up fee entirely.

 

One Known Quantity, One Variable

Sometimes one quantity is already a specific number, so you compute its contribution and add it to the variable part.

For a particular meal, a person consumes 30 grams of fat and $g$ grams of protein. Each gram of fat contains 9 calories, and each gram of protein contains 4 calories. The total calories must be at most 800. Which inequality represents this?

A) $4g + 30 \leq 800$
B) $4g + 270 \leq 800$
C) $4g + 270 \geq 800$
D) $4g + 30 \geq 800$

Calories from fat: $30 \times 9 = 270$ (compute the known part). Calories from protein: $4g$. Total: $4g + 270 \leq 800$. The answer is B. Option A uses 30 instead of $30 \times 9 = 270$ — you must multiply the known quantity by its rate.

 

Budget Minus a Fixed Cost

When a fixed expense must be paid first, subtract it from the budget to find what's left for the variable items.

A student committee has a maximum budget of $1,200 for a school dance. The committee must pay a non-refundable fee of $250 to book a DJ. The remaining budget will be spent on snacks ($8 per person) and decorations ($15 per set). Which inequality represents the possible combinations of people, $x$, and decoration sets, $y$?

A) $8x + 15y \leq 950$
B) $8x + 15y \geq 950$
C) $8x + 15y \leq 1200$
D) $8x + 15y \geq 1200$

Remaining after DJ: $1200 - $250 = $950. Variable costs: $8x + 15y$. This must stay within the remaining budget: $8x + 15y \leq 950$. The answer is A. Option C ignores the DJ fee.

 

"X Less Than Y Times Z" Phrasing

Medium questions often test whether you can parse phrases like "at most 10 less than 35 times."

The distance $d$, in miles, that a car can travel on a tank of fuel is at most 10 miles less than 35 times the number of gallons $g$ in the tank. Which inequality represents all possible values of $d$?

A) $d \geq 10 - 35g$
B) $d \geq 35g - 10$
C) $d \leq 35g - 10$
D) $d \leq 10 - 35g$

"35 times the number of gallons" = $35g$. "10 less than" that = $35g - 10$. "At most" means $d \leq 35g - 10$. The answer is C. Read carefully: "10 less than $35g$" is $35g - 10$, not $10 - 35g$.

 

System of Two Inequalities

Some problems give two separate constraints, each producing its own inequality. You need to identify the correct system.

A pet store owner is stocking a display aquarium. The total number of fish cannot exceed 300. The number of saltwater fish will be at most one-fifth the number of freshwater fish. Which system of inequalities represents this, where $f$ is freshwater and $s$ is saltwater?

A) $f + s \geq 300$; $s \geq 5f$
B) $f + s \leq 300$; $s \geq 5f$
C) $f + s \geq 300$; $s \leq \dfrac{1}{5}f$
D) $f + s \leq 300$; $s \leq \dfrac{1}{5}f$

Constraint 1: "Cannot exceed 300" → $f + s \leq 300$. Constraint 2: "At most one-fifth" → $s \leq \dfrac{1}{5}f$. The answer is D. Be careful: "one-fifth of freshwater" is $\dfrac{1}{5}f$, not $5f$. Options A and B flip the fraction.

 

Derived Variable (One Type Costs or Weighs More)

In harder problems, one item's value is defined relative to another, so you need an expression like $(x + 5)$ in the inequality.

A shipping company must ship exactly 80 standard packages and exactly 50 express packages. The total weight must be at least 2,500 kilograms. An express package weighs 5 kilograms more than a standard package. Which inequality represents this, where $x$ is the weight of a standard package?

A) $80x + 50(x + 5) \leq 2500$
B) $80x + 50(x - 5) \geq 2500$
C) $80x + 50(x + 5) \geq 2500$
D) $80x + 50(x - 5) \leq 2500$

Standard: $x$ kg each, so 80 packages weigh $80x$. Express: $(x + 5)$ kg each (5 more, not less), so 50 packages weigh $50(x + 5)$. "At least 2,500" means $\geq$. So $80x + 50(x + 5) \geq 2500$. The answer is C.

 

Compound Inequality with a Variable Bound

The hardest version: the total must fall between two multiples of some base quantity, and you solve for the variable.

A factory's total annual carbon emissions are fixed emissions of $F$ metric tons of CO2, plus variable emissions of 0.02 metric tons per unit produced. The factory aims to keep total emissions at least 1.2 times but no more than 1.5 times its fixed emissions. Which inequality represents all possible values of production output $p$?

A) $\dfrac{0.2}{0.02}F \leq p \leq \dfrac{0.5}{0.02}F$, i.e., $10F \leq p \leq 25F$
B) $10F \leq p \leq 25F$
C) $60F \leq p \leq 75F$
D) $1.2F \leq p \leq 1.5F$

Total emissions = $F + 0.02p$. The goal: $1.2F \leq F + 0.02p \leq 1.5F$.

Left side: $1.2F \leq F + 0.02p \Rightarrow 0.2F \leq 0.02p \Rightarrow p \geq 10F$.

Right side: $F + 0.02p \leq 1.5F \Rightarrow 0.02p \leq 0.5F \Rightarrow p \leq 25F$.

So $10F \leq p \leq 25F$. The answer is B. Option D forgets to isolate $p$ and just states the bounds on the total.

 

Consecutive Integer Problems

Hard questions may describe consecutive integers (or even integers), name one of them with a variable, and ask you to translate a verbal comparison into an inequality.

In a set of four consecutive even integers, the largest integer is represented by $p$. The product of 8 and the smallest integer is at least the value equal to 18 less than the sum of the two middle integers. Which inequality represents this?

A) $8p \geq ((p + 2) + (p + 4)) - 18$
B) $8(p - 6) \leq ((p - 4) + (p - 2)) - 18$
C) $8(p - 6) \geq ((p - 2) + (p - 4)) - 18$
D) $8(p - 6) \geq ((p - 2) + (p - 4)) - 18$

Four consecutive even integers with $p$ as the largest: $p - 6$, $p - 4$, $p - 2$, $p$. (Subtract 2 each step going backward.)

Smallest = $p - 6$. Two middle = $p - 4$ and $p - 2$.

"Product of 8 and the smallest" = $8(p - 6)$.

"18 less than the sum of the two middle" = $(p - 4) + (p - 2) - 18$.

"At least" means $\geq$. So: $8(p - 6) \geq (p - 4) + (p - 2) - 18$.

The answer is D. The key steps: (1) express all integers in terms of $p$ counting backward by 2, and (2) translate "at least" as $\geq$ and "18 less than" as subtracting 18.