Applications of Linear Equations

(a) $600   (b) $3$ years

Both parts use the simple-interest formula

where $P$ is the principal, $r$ is the annual rate written as a decimal, and $t$ is the time in years.

(a) Here $P=2500$, $t=4$, and the rate $6\mathrm{\%}$ must first be converted to a decimal:

Substitute:

Work from left to right. First $2500\times 0.06=150$, which is the interest for one year. Then

The interest is $600.

Forgetting to turn $6\mathrm{\%}$ into $0.06$ and multiplying by $6$ instead is the classic error; it would give an answer a hundred times too large.

(b) Now the unknown is $t$. Here $I=450$, $P=3000$, and $r=5\mathrm{\%}=0.05$. Substitute into $I=Prt$:

Simplify the numbers that are known: $3000\times 0.05=150$, so

Divide both sides by $150$:

The money must be invested for $3$ years.

Check. $3000\times 0.05\times 3=150\times 3=450$. Correct.

(a) ${59}^{\circ }$F   (b) $-{15}^{\circ }$C

(a) Here $C$ is known and $F$ is wanted, so substitute $C=15$ directly:

Compute the product first, following the order of operations:

Then add:

So ${15}^{\circ }$C is ${59}^{\circ }$F.

(b) Now $F$ is known and $C$ is wanted, so substituting turns the formula into an equation for $C$:

Subtract $32$ from both sides:

To undo multiplication by ${\displaystyle \frac{9}{5}}$, multiply both sides by ${\displaystyle \frac{5}{9}}$:

So $5^{\circ }$F is $-{15}^{\circ }$C.

Check. Put $C=-15$ back into the formula: ${\displaystyle \frac{9}{5}}(-15)+32=-27+32=5$. Correct.

Notice that the two parts are not symmetrical: in (a) you evaluate a formula, in (b) you solve an equation. Substituting the known value first, then solving, works in both directions.

(a) $3000 per year   (b) $V=28000-3000t$   (c) $13{,}000   (d) $6$ years

(a) The truck loses value from $28{,}000 down to $4000, a total loss of

In a linear model this loss is spread evenly over the $8$ years, so each year the truck loses

(b) After $t$ years the truck has lost $3000t$ dollars from its original price, so

(c) Substitute $t=5$:

The truck is worth $13{,}000 after five years.

(d) Now the value is known and the time is wanted, so set $V=10000$:

Subtract $28000$ from both sides:

Divide both sides by $-3000$. Dividing a negative by a negative gives a positive:

The truck is worth $10{,}000 after $6$ years.

Check. $28000-3000(6)=28000-18000=10000$. Correct. The answer is also reasonable: $6$ years lies between the $5$ years of part (c), where the truck was worth more, and the $8$ years at which it is worth only $4000.

(a) $30$ meals   (b) $50$ meals

Write the cost and revenue for $x$ meals.

Cost: the fixed $135 plus $2.50 for each meal, so

Revenue: $7.00 for each meal, so

(a) Break even means $C=R$:

Subtract $2.50x$ from both sides to gather the unknown on the right:

Divide both sides by $4.50$:

Thirty meals must be sold.

Check. Cost of $30$ meals: $2.50(30)+135=75+135=210$ dollars. Revenue: $7(30)=210$ dollars. They match.

(b) Profit is revenue minus cost, so a profit of $90 means

that is,

The minus sign applies to the whole bracket, so both terms inside it change sign:

Combine the $x$ terms, since $7-2.50=4.50$:

Add $135$ to both sides:

Divide both sides by $4.50$:

Fifty meals must be sold.

Check. Revenue $7(50)=350$; cost $2.50(50)+135=125+135=260$; profit $350-260=90$. Correct.

The number $4.50$ appearing in both parts is the profit made on one extra meal: $7.00 comes in and $2.50 goes straight back out.

At $14:00$ (2 p.m.), $300$ km from the depot.

Choose the unknown. Let $t$ be the number of hours the van has been travelling when it catches up. Because the lorry started two hours earlier, the lorry has then been travelling $t+2$ hours.

Write each distance using distance $=$ speed $\times$ time.

Lorry: $50(t+2)$ kilometres.

Van: $75t$ kilometres.

Write the equation. Catching up means the two vehicles are at the same place, so they have covered the same distance from the depot:

Solve. Expand the right side:

Subtract $50t$ from both sides:

Divide both sides by $25$:

So the van catches up $4$ hours after it left. It left at $10:00$, so the time is $14:00$.

The distance. Use the van's expression:

Check with the lorry. The lorry has been travelling $4+2=6$ hours, covering $50\times 6=300$ km. The two distances agree, so the answer is right.

The trap here is to give both vehicles the same travel time. They meet at the same place and at the same clock time, but the lorry has been on the road two hours longer.

$24$ kilometres.

Choose the unknown. Let $d$ be the one-way distance in kilometres. The same $d$ is driven in each direction.

Write each time. Rearranging distance $=$ speed $\times$ time gives

Outward trip: ${\displaystyle \frac{d}{40}}$ hours. Return trip: ${\displaystyle \frac{d}{60}}$ hours.

Write the equation. The two times add to $1$ hour:

Clear the fractions. The denominators are $40$ and $60$; the smallest number both divide into is $120$. Multiply every term by $120$:

Since $120÷40=3$ and $120÷60=2$, this becomes

Combine like terms:

Divide both sides by $5$:

The distance is $24$ kilometres.

Check. Outward: ${\displaystyle \frac{24}{40}}=0.6$ hours. Return: ${\displaystyle \frac{24}{60}}=0.4$ hours. Total $0.6+0.4=1$ hour. Correct.

Note that it would be wrong to average the speeds to $50$ km/h and conclude the distance is $25$ km each way. More time is spent at the slower speed, so the average speed is below $50$; in fact it is ${\displaystyle \frac{48\text{km}}{1\text{h}}}=48$ km/h.

$10$ kg of the $12 beans and $10$ kg of the $18 beans.

Choose the unknown. Let $x$ be the number of kilograms of the $12 beans. Since the blend must weigh $20$ kg altogether, the amount of the $18 beans is $20-x$ kilograms. Using $20-x$ instead of a second letter keeps this to one equation.

Count the money. The value of a quantity of beans is (price per kg) $\times$ (kilograms).

Cheap beans: $12x$ dollars.

Expensive beans: $18(20-x)$ dollars.

Finished blend: $15\times 20=300$ dollars.

Write the equation. The value of the two parts equals the value of the blend:

Solve. Expand the bracket, watching the sign on the second term:

so

Combine the $x$ terms: $12x-18x=-6x$. Hence

Subtract $360$ from both sides:

Divide both sides by $-6$:

So $10$ kg of the $12 beans, and $20-10=10$ kg of the $18 beans.

Check. Value: $12(10)+18(10)=120+180=300$ dollars for $20$ kg, which is ${\displaystyle \frac{300}{20}}=15$ dollars per kilogram. Correct.

The answer is an equal split because $15 sits exactly halfway between $12 and $18. That is a useful way to see in advance that the answer is reasonable.

$4.5$ litres.

Identify what is conserved. Only water leaves the tank; the salt stays behind. So the amount of salt is the same before and after.

Salt at the start.

Let $x$ be the number of litres that evaporate. The liquid remaining measures $12-x$ litres, and it is to be $8\mathrm{\%}$ salt, so the salt in it measures $0.08(12-x)$ litres.

Write the equation.

Solve. Divide both sides by $0.08$ to undo the multiplication:

Compute the left side. Multiplying numerator and denominator by $100$ turns it into a whole-number division:

So

Add $x$ to both sides and subtract $7.5$ from both sides:

So $4.5$ litres of water must evaporate.

Check. What remains is $12-4.5=7.5$ litres, still containing $0.6$ litres of salt. Its concentration is

that is $8\mathrm{\%}$. Correct.

Note the direction of the change: removing water makes the solution stronger, so the remaining volume must be smaller than $12$ litres — and $7.5$ is. If your answer made the volume grow, you have set the problem up backwards.

$4000 at $4\mathrm{\%}$ and $4000 at $7\mathrm{\%}$.

Choose the unknown. Let $x$ be the number of dollars placed in the 4% account. The rest of the 8000 dollars goes into the 7% account, so that account receives $8000-x$ dollars.

Write the interest from each account. With $t=1$ year the formula $I=Prt$ becomes simply $I=Pr$.

From the $4\mathrm{\%}$ account: $0.04x$ dollars.

From the $7\mathrm{\%}$ account: $0.07(8000-x)$ dollars.

Write the equation. The two amounts of interest total $440:

Solve. Expand the bracket:

so

Combine the $x$ terms: $0.04x-0.07x=-0.03x$. Hence

Subtract $560$ from both sides:

Divide both sides by $-0.03$:

So 4000 dollars went into the 4% account, and $8000-4000=4000$ dollars into the 7% account.

Check. Interest earned: $0.04(4000)=160$ dollars and $0.07(4000)=280$ dollars. Together $160+280=440$ dollars. Correct.

$6\mathrm{\%}$

Start from the simple-interest formula and substitute everything that is known:

Multiply the two known numbers on the right. Since multiplication may be done in any order,

so the equation becomes

Divide both sides by $3600$:

Simplify the fraction. Both numbers are divisible by $36$: $216÷36=6$ and $3600÷36=100$, so

Convert to a percentage. A decimal is turned into a percentage by multiplying by $100$:

so the rate is $6\mathrm{\%}$.

Check. $1200\times 0.06\times 3=72\times 3=216$ dollars, which is the interest given. Correct.

An answer such as "$0.06\mathrm{\%}$" would be a hundred times too small; the decimal $0.06$ and the percentage $6\mathrm{\%}$ are the same number written two ways.

Concentrations do not add. The mixture is $20\mathrm{\%}$ acid.

What is wrong. A percentage is not an amount; it is a ratio of acid to total liquid. Adding two ratios does not produce the ratio of the combined mixture, because the total volume grows as well. What can legitimately be added is the acid itself, measured in litres.

A quick sanity check shows the student's answer must be wrong: mixing two solutions, one $25\mathrm{\%}$ and one $15\mathrm{\%}$, cannot possibly give something stronger than the stronger of the two. The result has to lie somewhere between $15\mathrm{\%}$ and $25\mathrm{\%}$, and $40\mathrm{\%}$ is outside that range.

The correct computation.

Acid in the first solution:

Acid in the second solution:

Total acid in the mixture:

Total volume of the mixture:

Concentration of the mixture:

which is $20\mathrm{\%}$.

Why $20\mathrm{\%}$ makes sense. The two solutions were mixed in equal amounts, so the result sits exactly halfway between $15\mathrm{\%}$ and $25\mathrm{\%}$ — and $20$ is halfway between $15$ and $25$. Percentages may be averaged like this only when the volumes being mixed are equal; the reliable method in every case is to count the acid in litres.

Yes: $-40$ degrees on both scales.

Set up the condition. We want the two readings to be the same number, so we want $F=C$. In the conversion formula

replace $F$ by $C$, since we are asking for the case where they are equal:

This is now an ordinary equation in the single unknown $C$.

Solve it. Subtract ${\displaystyle \frac{9}{5}}C$ from both sides:

Write $C$ as ${\displaystyle \frac{5}{5}}C$ so that the two terms have the same denominator:

So the equation is

To undo multiplication by $-{\displaystyle \frac{4}{5}}$, multiply both sides by $-{\displaystyle \frac{5}{4}}$:

Interpret. There is exactly one such temperature, namely $-40$.

Check. Put $C=-40$ into the formula:

The Fahrenheit reading is also $-40$, so the two thermometers agree there.

The equation had only one solution, so $-40$ is the only temperature at which the two scales give the same number. At every other temperature the readings differ.

Deriving the break-even number. Break-even means cost equals revenue:

Subtract $cx$ from both sides to gather the $x$ terms together:

Both terms on the right contain $x$, so pull it out by the distributive property:

Provided $p-c$ is not zero, divide both sides by it:

Why the denominator is $p-c$ and not $p$. Selling one item brings in $p$ dollars, but making that item first costs $c$ dollars. Only the difference $p-c$ is left over to pay off the fixed cost $F$. Dividing by $p$ alone would pretend that the whole selling price is available to cover the fixed cost, which ignores the money spent on materials, and would give a break-even number that is too small.

When does a break-even point exist? The division is legal only when $p-c\ne 0$, and the answer is only meaningful when the number of items is positive. Since $F$ is a positive fixed cost, the fraction ${\displaystyle \frac{F}{p-c}}$ is positive exactly when

So each item must sell for more than it costs to make.

The two failing cases are worth naming. If $p=c$, the equation reduces to $F=0$, which is false for a genuine fixed cost, so the business never breaks even no matter how much it sells. If , every item sold loses money, and selling more only makes the loss larger.

A numerical illustration. With $F=600$, $c=4$ and $p=10$, the correct answer is ${\displaystyle \frac{600}{10-4}}=100$ items: cost $4(100)+600=1000$ and revenue $10(100)=1000$, which match. The wrong formula ${\displaystyle \frac{600}{10}}=60$ items gives cost $4(60)+600=840$ against revenue $10(60)=600$ — a loss of $240, not a break-even.

After $7.5$ years, when each is worth $6000.

Write a value expression for each machine. Let $t$ be the number of years from today.

Machine A starts at $15{,}000 and loses $1200 per year, so after $t$ years its value is

Machine B starts at $9000 and loses $400 per year, so

Write the equation. The values are equal when

Solve. Add $1200t$ to both sides, so that the $t$ terms end up on the side where the coefficient is larger and stays positive:

Combine the $t$ terms: $-400t+1200t=800t$. So

Subtract $9000$ from both sides:

Divide both sides by $800$:

The values are equal after $7.5$ years.

Find the common value. Use machine B's expression, since the numbers are smaller:

Check with machine A.

Both give $6000, so the answer is correct.

Is the answer reasonable? Machine A starts $6000 ahead but loses $800 more each year than machine B does. To give away a $6000 lead at $800 per year takes $6000÷800=7.5$ years, which agrees with the algebra. A fractional answer is perfectly acceptable here: $7.5$ years simply means half way through the eighth year.