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Linear Systems and Solution Sets · Tutorial 3 of 1000

Translating Verbal Constraints into Equations

Learn to define quantities clearly and translate age, mixture, and cost relationships into equations you can check.

Beginner 9 min read

What You'll Learn

  • Define unknown quantities with units before writing an equation.
  • Translate age differences and changes over time without mixing dates.
  • Represent a mixture balance using the amount of ingredient in each portion.
  • Convert unit prices and item counts into a total-cost equation.
  • Check a translation by substituting values and comparing with the original statement.

From Words to Equations

A verbal constraint describes a relationship among quantities. To translate it, first decide exactly what each unknown represents, including its units and when it is measured. Then turn each relationship in the statement into an equation. The equation should preserve the meaning of the words: it should not add an assumption, omit a quantity, or combine amounts measured in incompatible units.

This tutorial uses age, mixture, and cost statements. In each case, we will write down the relationship before trying to calculate unknown values. A check by substitution is useful: choose or calculate values, put them back into the equation, and then verify that they also satisfy the original wording. When a problem gives several statements, each statement contributes its own relationship; deciding how to solve several relationships together is a separate step.

Translation rule: Define each variable in words, with a unit and a time reference when relevant. Translate each stated relationship into an equation using those definitions. Before using the equation, check that both sides represent the same kind of quantity and use compatible units.

Age Statements: Keep Track of Time

Age statements often involve two different moments: now, and a specified number of years earlier or later. If a person is \(A\) years old now, then in \(t\) years that person will be \(A+t\) years old; \(t\) years ago, the age was \(A-t\). The same time shift must be applied to every person's age when the statement refers to the same future or past moment.

Age-difference theorem: Let \(A\) and \(B\) be two current ages, and let \(d\) be a stated number of years. Saying “the person aged \(A\) is \(d\) years older than the person aged \(B\)” is equivalent to the equation \(A=B+d\).

Proof. By the meaning of “\(d\) years older,” the difference between the two ages is \(d\), so \(A-B=d\). Adding \(B\) to both sides gives \(A=B+d\). Conversely, if \(A=B+d\), subtracting \(B\) from both sides gives \(A-B=d\), which says exactly that the first age exceeds the second by \(d\) years. \(\square\)

Worked Example: Two Ages Now and in Three Years

Lina is four years older than Omar. In three years, their ages will add to thirty. Define \(L\) to be Lina’s age now, in years, and \(O\) to be Omar’s age now, in years. The first sentence gives

$$ L=O+4. $$

For the second sentence, each age increases by three years. Their future ages are \(L+3\) and \(O+3\), so the stated total gives

$$ (L+3)+(O+3)=30. $$

Combining the terms on the left gives \(L+O+6=30\), or \(L+O=24\). To check that the translation matches the words, take \(L=14\) and \(O=10\). The first relationship holds because \(14=10+4\). Three years later the ages are \(17\) and \(13\), and \(17+13=30\). Thus both equations agree with the original statements for these values. This check does not claim that the values were found from both equations; it checks their interpretation.

A time phrase applies to the age at that time, not automatically to the age measured now. For example, “in five years, A will be twice B’s age” translates as \(A+5=2(B+5)\), not \(A+5=2B\). The phrase “twice B’s age” refers to B’s age at the future moment too.

Mixtures: Balance the Amount of Ingredient

A percentage describes a fraction of a whole. If a solution has volume \(V\) and contains a fraction \(c\) of a particular ingredient, the amount of that ingredient is \(cV\). For instance, a solution that is \(15\%\) ingredient has \(c=\frac{15}{100}=\frac{3}{20}\). When portions are combined without losing or creating ingredient, their ingredient amounts add.

Mixture-balance theorem: Suppose two portions have volumes \(V_1,V_2\) and ingredient fractions \(c_1,c_2\). Assume their volumes add and the amount of ingredient is conserved when they are mixed. If the final mixture has fraction \(c\), then its ingredient balance is \(c_1V_1+c_2V_2=c(V_1+V_2)\). This equation is equivalent to saying that the final mixture has fraction \(c\).

Proof. The first portion contains \(c_1V_1\) units of ingredient because its ingredient amount is its fraction times its volume. The second contains \(c_2V_2\) units. Conservation and additivity mean the combined amount is \(c_1V_1+c_2V_2\), while the combined volume is \(V_1+V_2\). By the definition of the final fraction, the combined ingredient amount is also \(c(V_1+V_2)\). Equating these two descriptions gives the displayed equation. In the other direction, if the equation holds and \(V_1+V_2>0\), divide both sides by \(V_1+V_2\). The resulting ratio of ingredient amount to total volume is \(c\), so the final fraction is \(c\). \(\square\)

Worked Example: Mixing Two Solutions

How many litres of a \(35\%\) solution should be added to \(8\) litres of a \(15\%\) solution to obtain a \(25\%\) mixture? Let \(x\) be the volume added, in litres. Use the fractions \(\frac{3}{20}\), \(\frac{7}{20}\), and \(\frac14\) for the three percentages.

The first portion contains \(\frac{3}{20}\cdot 8=\frac{24}{20}=\frac65\) litres of ingredient. The added portion contains \(\frac{7}{20}x\) litres of ingredient. The final volume is \(8+x\) litres, so the target amount of ingredient is \(\frac14(8+x)\) litres. Equating the ingredient amounts gives

$$ \frac65+\frac{7}{20}x=\frac14(8+x). $$

Multiply both sides by \(20\): \(24+7x=5(8+x)=40+5x\). Subtract \(5x\) from both sides to get \(24+2x=40\), then subtract \(24\) to get \(2x=16\), so \(x=8\) litres.

Check: The original portion contributes \(\frac{3}{20}\cdot8=\frac65\) litres of ingredient. The added portion contributes \(\frac{7}{20}\cdot8=\frac{14}{5}\) litres. Their sum is \(\frac65+\frac{14}{5}=\frac{20}{5}=4\) litres. The total volume is \(8+8=16\) litres, and \(\frac14\cdot16=4\) litres is exactly \(25\%\) of that volume.

Cost Statements: Multiply Rate by Count

For a fixed price per item, the cost of a group of identical items is the unit price multiplied by the number of items. The total cost of several kinds of items is the sum of their individual costs. Keep the units visible while translating: dollars per item multiplied by items gives dollars. If the number of items is unknown, use a variable for that count rather than for the total cost unless the total itself is what the statement leaves unknown.

Worked Example: Finding an Unknown Item Count from the Cost

A purchase contains three notebooks costing \(\$\,\frac{12}{5}\) each and some pens costing \(\$\,\frac75\) each. The total cost is \(\$17\). Let \(p\) be the number of pens. The notebooks cost \(3\cdot\frac{12}{5}=\frac{36}{5}\) dollars, and the pens cost \(\frac75p\) dollars. Adding the two costs gives the equation

$$ \frac{36}{5}+\frac75p=17. $$

Multiply by \(5\): \(36+7p=85\). Subtract \(36\) to obtain \(7p=49\), then divide by \(7\) to get \(p=7\) pens.

Check: The notebooks cost \(\frac{36}{5}\) dollars. Seven pens cost \(7\cdot\frac75=\frac{49}{5}\) dollars. The total is \(\frac{36}{5}+\frac{49}{5}=\frac{85}{5}=17\) dollars, as stated. The result is also a whole number, as a count of pens must be.

Sometimes a cost statement gives a fixed fee as well as a price per item. If there is a fee of \(F\) dollars and \(n\) items cost \(r\) dollars each, the total is \(F+rn\), not \(rn\) alone. The equation should reflect every part of the stated charge, and no charge that was not mentioned.

Common Translation Errors

Shifting only one age. A mistaken translation of “in three years their ages total thirty” might be \(L+O+3=30\). That adds three years only once. The correction is \((L+3)+(O+3)=30\), because each of the two people ages by three years. For the checked values \(L=14\) and \(O=10\), the mistaken expression gives \(14+10+3=27\), while the actual future ages sum to \(17+13=30\).

Adding percentages instead of amounts. In the mixture example, a tempting but incorrect equation is \(\frac{3}{20}+\frac{7}{20}x=\frac14\). It treats the initial \(8\)-litre portion as though it were only one litre, and it fails to account for the final volume \(8+x\). The corrected balance is \(\frac{3}{20}\cdot8+\frac{7}{20}x=\frac14(8+x)\): each fraction is multiplied by the volume to which it applies.

Forgetting the unit price. If \(p\) is a count of pens, then \(p\) alone is not their cost in dollars. The cost is \(\frac75p\) dollars when each pen costs \(\frac75\) dollars. A useful check is to ask what the units of every term are: terms being added in a cost equation must all be amounts of money.

A Reliable Translation Routine

Before writing an equation, pause long enough to make each quantity unambiguous. In an age problem, specify whether a variable means a current age or an age at another date. In a mixture problem, identify the volume and ingredient fraction for each portion. In a cost problem, distinguish the number of items from the price of one item and from the total cost.

1
Name the unknown.
Write what it measures, including its unit and, where needed, its time reference.
2
Translate one statement at a time.
Use the defined quantities to express the stated difference, total, amount of ingredient, or cost.
3
Check meaning and units.
Verify that both sides describe the same quantity and that the variables match the words.
4
Substitute to test.
If values are available, put them into the equation and check the original verbal relationship as well.

Translation matters because later calculations depend on the equation representing the situation correctly. Algebra can manipulate an equation without changing its mathematical content, but it cannot repair a relationship that was written incorrectly in the first place. A clear variable definition and a check in the original context make the model easier to trust.

Check Your Understanding

For each question, define the unknowns clearly before writing the requested equation. Check any calculated value by returning to the wording.

  1. Ravi is six years younger than Kim. Write an equation using \(R\) and \(K\) for their current ages.
  2. In four years, two siblings will have a total age of thirty-two. If their current ages are \(a\) and \(b\), write an equation for the statement.
  3. A \(20\%\) solution has volume \(5\) litres. Write an expression for its amount of ingredient, in litres.
  4. Mix \(5\) litres of a \(20\%\) solution with \(x\) litres of a \(40\%\) solution to make a \(30\%\) mixture. Write the ingredient-balance equation.
  5. Each ticket costs \(\$12\), and a group pays \(\$84\) in total. Let \(n\) be the number of tickets and write the cost equation.

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