Using linear systems to model real situations, particularly mixture problems with one counting equation and one value equation. Includes building the verbal model and labels, choosing the efficient solution method from the coefficients, clearing decimals, and checking the answer against the situation rather than only the algebra.
Subject: Algebra 1 · 65 slides · symbolic lesson
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Title
Algebra 1 · Chapter 7 — Systems of Linear Equations and Inequalities
Linear Systems and Problem Solving
Objectives
Five outcomes, each one you can test yourself on with a pencil and no answer key.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 7 Systems of Linear Equations and Inequalities — Lesson 7.4 Linear Systems and Problem Solving §7.4, pp. 409-414 — the lesson these objectives are drawn from
Warm-up
You now have three ways to solve a system. This lesson moves the difficulty to setting one up.
Discussion prompt
A store sold 7 violins for 1600 dollars. One kind costs 200 dollars and the other 300. What two facts does that give you, and what does each become?
Hint: Count the objects, then count the money.
Answer:
\[ x + y = 7 \qquad 200x + 300y = 1600 \]
The first sentence counts violins and the second counts dollars, so each gives one equation. Two unknowns need two facts, and the situation has supplied exactly two.
Concept
A situation with two unknown quantities and two independent facts about them can be modelled as a linear system. Once the system is written, any of the three methods will solve it.
mixture problem — A problem combining two kinds of item, usually giving one equation counting the items and another counting their total value or content.
The Study Tip notes that mixture problems typically have one equation of the form x plus y equals an amount.
Figure (svg): A situation supplying two facts, each becoming one equation
McDougal Littell Algebra 1: Concepts and Skills, Ch. 7 Systems of Linear Equations and Inequalities — Lesson 7.4 Linear Systems and Problem Solving §7.4, pp. 409-409
Section
Section 1
Concept
Write the two relationships as sentences first, label every quantity with a symbol and a unit, and only then write the algebraic model. This is the procedure from Lessons 1.6 and 5.5 applied to two equations.
Each fact in the situation produces exactly one equation.
Figure (svg): The verbal model, labels and algebraic model of a mixture problem
McDougal Littell Algebra 1: Concepts and Skills, Ch. 7 Systems of Linear Equations and Inequalities — Lesson 7.4 Linear Systems and Problem Solving §7.4, pp. 409-409 — Example 1, with its verbal model, labels and algebraic model
Picture it
Words, labels with units, then algebra.
Figure (svg): The verbal model, labels and algebraic model of a mixture problem
The middle stage attaches the units, and the units are what make the second equation checkable — dollars per violin times violins gives dollars, matching the total.
Worked example
This is Example 1 from the textbook.
\[ \text{Seven violins sold for } 1600 \text{ dollars. One kind costs } 200 \text{ and the other } 300. \text{ How many of each?} \]
Write the counting equation
Why: The two numbers add to seven.
\[ x + y = 7 \]
Write the value equation
Why: Each price times its count, added.
\[ 200 x + 300 y = 1600 \]
Choose a method
Why: The first equation has coefficients of one, so substitute.
\[ x = 7 - y \]
Solve
Why: 1400 plus 100y is 1600, so y is two and x is five.
\[ (5, 2) \]
Figure (svg): A situation supplying two facts, each becoming one equation
\[ x = 5, \; y = 2 \]
Verify: check both facts in the situation
Why: Five plus two is seven violins, and a thousand plus six hundred is sixteen hundred dollars. Both sentences of the problem are satisfied, which is a stronger check than substituting into the equations alone — it tests the model as well as the arithmetic.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 7 Systems of Linear Equations and Inequalities — Lesson 7.4 Linear Systems and Problem Solving §7.4, pp. 409-409
Matching
Each fact becomes one equation.
Match the pairs
Why: The counting sentences give equations with coefficients of one, and the value sentences give equations whose coefficients are prices or concentrations. Every mixture problem in the lesson has exactly this pair of shapes.
Worked example
Guided Practice 1. The same structure with different numbers.
\[ \text{Thirty-two pairs of jeans sold for } 1050 \text{ dollars, at } 30 \text{ and } 35 \text{ dollars a pair.} \]
Write the counting equation
Why: The two counts add to thirty-two.
\[ x + y = 32 \]
Write the value equation
Why: Thirty x plus thirty-five y is 1050.
\[ 30 x + 35 y = 1050 \]
Substitute
Why: Thirty times thirty-two minus y, plus thirty-five y.
\[ 960 + 5 y = 1050 \]
Solve
Why: Five y is ninety, so y is eighteen and x is fourteen.
\[ (14, 18) \]
Figure (svg): The solution to Worked example the jeans problem shown as a ladder of expressions, one row per algebraic move
\[ x = 14, \; y = 18 \]
Verify: check both facts
Why: Fourteen plus eighteen is thirty-two pairs, and four hundred and twenty plus six hundred and thirty is one thousand and fifty dollars. Both match, and the counts are whole positive numbers, which they must be for pairs of jeans.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 7 Systems of Linear Equations and Inequalities — Lesson 7.4 Linear Systems and Problem Solving §7.4, pp. 410-410
Trap
Seven violins sold for sixteen hundred dollars, at two hundred and three hundred dollars each.
Write 200x + 300y = 1600 and try to solve it
Why: The money equation contains all the interesting numbers, so it looks like the problem.
One equation with two unknowns has infinitely many solutions — a whole line of them. The count of seven is the second fact, and without it the answer is not pinned down.
\[ x + y = 7 \quad \text{and} \quad 200x + 300y = 1600 \]
Look for a second fact, and count the unknowns against the facts
Why: Two unknowns need two independent facts, and a situation with two unknowns will have supplied two.
If only one fact can be found, the problem as stated does not have a unique answer, which is itself worth saying.
Faded example
Price times count, twice, added.
Fill in the blanks
\text200 200 \text300 300, \text___ 1600: \quad ___x + ___y = 1600
Why: Each coefficient is a price in dollars per violin and each variable a count in violins, so every term is in dollars and can be compared with the total. A units check confirms the equation before any solving.
Elimination
The situation has two unknown counts.
Eliminate the wrong options
How many independent facts must the problem supply?
Survives elimination: A
Why: Two unknowns need two independent facts, matching the two blanks a system has. This is the same count as in Lesson 5.1, where a line needed two pieces of information — and it is why a situation offering only one fact does not have a unique answer.
Socratic
The equations are short and the sentences are long.
Discussion prompt
Explain what the verbal model and the labels protect against when a problem has two equations rather than one. Then say which stage catches a wrongly paired price and count.
Hint: Think about which errors are invisible in the finished equations.
Answer:
With two equations there are two structures to get right, and it is easy to write a plausible pair that describes a different situation — swapping which price belongs to which count, or adding counts where values should be added. The verbal model fixes both structures before any symbols appear, so the equations are transcriptions rather than inventions.
The labels stage catches a mispaired price and count, because each label carries a unit and a description: price of type A is two hundred dollars per violin, number of type A is x violins. Writing them next to each other makes the pairing explicit, and the units then confirm that their product is in dollars.
Section
Section 2
Concept
Mixture problems have a recognisable shape: one equation adds the two quantities and has coefficients of one, and the other multiplies each by a price, rate or percentage.
The Study Tip in the textbook names this pattern explicitly.
Figure (svg): Two columns describing the two equations of a mixture problem
McDougal Littell Algebra 1: Concepts and Skills, Ch. 7 Systems of Linear Equations and Inequalities — Lesson 7.4 Linear Systems and Problem Solving §7.4, pp. 409-409 — the Study Tip describing the shape of mixture problems
Picture it
Counting against valuing.
Figure (svg): Two columns describing the two equations of a mixture problem
Recognising the shape means you know what to look for in the problem: one sentence about how much altogether, and one about how much it is worth or contains.
Worked example
Reading a problem for its two facts.
\[ \text{A } 10 \text{-pound mix of peanuts and cashews sells for } 5.32 \text{ a pound; peanuts are } 3.60 \text{ and cashews } 7.90. \]
Find the counting fact
Why: The two weights add to ten pounds.
\[ p + c = 10 \]
Find the value fact
Why: Ten pounds at 5.32 is 53.20 in total.
\[ \text{total } 53.20 \]
Write the value equation
Why: Each price times its weight.
\[ 3.60 p + 7.90 c = 53.20 \]
Note the shape
Why: Coefficients of one in the first, prices in the second.
Figure (svg): Two columns describing the two equations of a mixture problem
\[ p + c = 10, \quad 3.60p + 7.90c = 53.20 \]
Verify: check the total price was computed correctly
Why: The problem gives a price per pound and a weight, so the total is their product: ten times 5.32 is 53.20 dollars. Writing 5.32 on the right instead would compare dollars per pound with dollars, which the units immediately reject.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 7 Systems of Linear Equations and Inequalities — Lesson 7.4 Linear Systems and Problem Solving §7.4, pp. 410-410
Sorting
Counting equations have coefficients of one.
Sort into buckets
Sort each equation by its role in a mixture problem.
Every problem in the lesson has exactly one of each. Spotting which sentence gives which is most of the modelling work.
Worked example
Guided Practice 2, solved with the counting equation.
\[ \text{Solve } \; p + c = 10 \; \text{ and } \; 3.60p + 7.90c = 53.20. \]
Solve the counting equation
Why: p equals ten minus c.
\[ p = 10 - c \]
Substitute
Why: 3.60 times the bracket, plus 7.90c.
\[ 36 + 4.30 c = 53.20 \]
Solve for c
Why: 4.30c is 17.20, so c is four.
\[ c = 4 \]
Back-substitute
Why: p is ten minus four.
\[ p = 6 \]
Figure (svg): The solution to Worked example solve the peanut problem shown as a ladder of expressions, one row per algebraic move
\[ p = 6, \; c = 4 \]
Verify: check both facts against the situation
Why: Six plus four is ten pounds, and 21.60 plus 31.60 is 53.20 dollars. Both hold, and the mixture's price of 5.32 sits between the two ingredient prices — which it must, since a mixture cannot cost more than its dearest ingredient or less than its cheapest.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 7 Systems of Linear Equations and Inequalities — Lesson 7.4 Linear Systems and Problem Solving §7.4, pp. 410-410
Error analysis
The student modelled the peanut and cashew mixture.
Annotate
On: \( \begin{aligned} p + c &= 10 \\ 3.60p + 7.90c &= 5.32 \end{aligned} \)
Whenever a problem gives a price per unit and an amount, the total is their product. Checking that both sides of an equation carry the same units catches this before any solving.
Faded example
Price per unit times the number of units.
Fill in the blanks
10 \text53.20 5.32 \text___: \quad \text___ = 10 \times 5.32 = ___
Why: A price per pound has to be multiplied by the number of pounds to give a total in dollars. Putting 5.32 on the right of the value equation would compare dollars with dollars per pound, which the units reject.
Prediction
Peanuts cost 3.60 and cashews 7.90 a pound.
Predict first
What can you say about the price per pound of any mixture of the two?
Correct: It lies between 3.60 and 7.90.
\[ 3.60 < 5.32 < 7.90 \quad \text{and } 5.32 \text{ is nearer } 3.60 \]
Why: A mixture cannot cost less per pound than its cheapest ingredient nor more than its dearest, so its price is somewhere between them. It equals the plain average only when the two amounts are equal, and here 5.32 sits nearer the peanut price, which correctly suggests more peanuts than cashews. That is a free sanity check on the answer before any algebra.
Socratic
Its position between the two prices carries information.
Discussion prompt
Explain why a mixture price nearer one ingredient's price means more of that ingredient. Then say what the price would be for a fifty-fifty mixture of peanuts and cashews.
Hint: Think of the mixture price as a weighted average.
Answer:
The mixture's price per pound is the total cost divided by the total weight, which is a weighted average of the two prices with the weights being the amounts used. A weighted average sits nearer the value carrying the larger weight, so a price close to 3.60 means the cheap ingredient dominates.
A fifty-fifty mixture would cost the plain average, five and three quarters a pound. Since 5.32 is below that, the mixture must be more than half peanuts — and indeed it is six pounds to four. That estimate takes seconds and would have caught an answer with the two amounts swapped.
Section
Section 3
Concept
Once the system is written, choose the method from the coefficients. A counting equation has coefficients of one, so solving it for one variable and substituting costs nothing.
Mixture problems nearly always contain a counting equation, so substitution is usually efficient.
Figure (svg): The coefficients of a system pointing to a solution method
McDougal Littell Algebra 1: Concepts and Skills, Ch. 7 Systems of Linear Equations and Inequalities — Lesson 7.4 Linear Systems and Problem Solving §7.4, pp. 409-409 — Example 1's remark that the coefficients are 1, so substitution is used
Picture it
Coefficients of one, so isolate there.
Figure (svg): The coefficients of a system pointing to a solution method
The value equation is where the awkward numbers live, so it is the one to substitute into rather than the one to rearrange.
Worked example
The textbook chooses substitution for Example 1 and says why.
\[ \text{Solve } \; x + y = 7 \; \text{ and } \; 200x + 300y = 1600. \]
Scan the coefficients
Why: The first equation's are both one.
Solve it for x
Why: x equals seven minus y.
\[ x = 7 - y \]
Substitute into the value equation
Why: Two hundred times the bracket, plus 300y.
\[ 1400 + 100 y = 1600 \]
Solve and back-substitute
Why: y is two, so x is five.
\[ (5, 2) \]
Figure (svg): The coefficients of a system pointing to a solution method
\[ x = 5, \; y = 2 \]
Verify: try linear combinations instead
Why: Multiplying the counting equation by negative two hundred gives negative 200x minus 200y equals negative 1400, and adding gives 100y equals 200 — the same answer in the same number of lines. Both methods are comfortable here, which is common when a counting equation is present.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 7 Systems of Linear Equations and Inequalities — Lesson 7.4 Linear Systems and Problem Solving §7.4, pp. 409-409
Elimination
The system is x + y = 90 and 0.2x + 0.5y = 36.
Eliminate the wrong options
Which equation do you solve for a variable?
Survives elimination: A
Why: The counting equation has coefficients of one, so x equals ninety minus y in a single subtraction. Rearranging the concentration equation instead would introduce decimals into every subsequent line for no benefit.
Worked example
Both coefficients are one, so either choice is free.
\[ \text{In } x + y = 7, \text{ should you solve for } x \text{ or } y? \]
Note both coefficients are one
Why: Neither requires division.
Look at the value equation
Why: Its coefficients are two hundred and three hundred.
Choose to keep the smaller multiplication
Why: Isolating x means multiplying the bracket by two hundred.
Note the difference is small
Why: Either route works comfortably.
Figure (svg): The solution to Worked example which variable to isolate shown as a ladder of expressions, one row per algebraic move
\[ x = 7 - y \quad \text{or} \quad y = 7 - x \]
Verify: work it the other way
Why: Isolating y gives 200x plus 300 times seven minus x, which is 2100 minus 100x equals 1600, so x is five. The same answer with slightly larger intermediate numbers, which is the entire difference between the two routes.
Trap
\[ 200x + 300y = 1600 \]
Solve this equation for x, since it is the one with the real numbers in it
Why: The value equation looks like the substantial one, so it gets the attention.
\[ x = \tfrac{1600 - 300y}{200} = 8 - 1.5y \]
It works and it introduces a decimal coefficient for no reason. The counting equation would have given x equals seven minus y in one subtraction.
\[ x + y = 7 \;\Longrightarrow\; x = 7 - y \]
Isolate in the equation with coefficients of one
Why: That is what makes substitution cheap, and a counting equation always provides it.
The value equation is where the substitution goes, not where the rearranging happens.
Faded example
One subtraction, no division.
Fill in the blanks
x + y = 90 \;\Longrightarrow\; x = 90 - y
Why: Subtracting y from both sides isolates x with no division at all. That is why a counting equation makes substitution the efficient method for almost every mixture problem.
Sorting
Read the coefficients.
Sort into buckets
Sort each system by the method that is less work.
Every mixture problem in this lesson lands in the first column, which is why the textbook uses substitution throughout. Systems arriving without a counting equation are usually not mixture problems at all.
Socratic
The pattern is too consistent to be an accident.
Discussion prompt
Explain why a mixture problem's first equation almost always has coefficients of one. Then say what a problem would look like if it did not.
Hint: Ask what the first sentence of such a problem says.
Answer:
The counting equation says that the two amounts add up to a known total — seven violins, ten pounds, ninety millilitres. Adding amounts means coefficients of one, because each unit of each ingredient contributes exactly one unit to the total. The coefficient is one because nothing is being scaled.
A problem without it would have to give two value facts instead — say the total cost and the total weight of the same mixture, with both prices and both densities known. That happens, and then both equations have awkward coefficients and linear combinations becomes the natural method. Recognising which shape you have is the point of the scan.
Section
Section 4
Concept
Percentages cannot be added directly. Each has to be converted into an actual quantity — twenty per cent of x millilitres is 0.2x millilitres of acid — and those quantities do add.
Forty per cent of ninety millilitres is thirty-six millilitres of acid.
Figure (svg): Percentages converted into amounts of acid
McDougal Littell Algebra 1: Concepts and Skills, Ch. 7 Systems of Linear Equations and Inequalities — Lesson 7.4 Linear Systems and Problem Solving §7.4, pp. 410-410 — Example 2, Solve a Mixture Problem
Picture it
Three concentrations, three amounts.
Figure (svg): Percentages converted into amounts of acid
The mixture's forty per cent had to be turned into thirty-six millilitres before it could sit on the right of the equation, which is the step that makes the units match.
Worked example
This is Example 2 from the textbook.
\[ \text{Mix a } 20\% \text{ acid and a } 50\% \text{ acid to get } 90 \text{ mL of } 40\% \text{ acid. How much of each?} \]
Write the counting equation
Why: The two volumes add to ninety.
\[ x + y = 90 \]
Convert the percentages
Why: 0.2x, 0.5y, and forty per cent of ninety.
\[ 0.2 x + 0.5 y = 36 \]
Clear the decimals
Why: Multiply the second equation by ten.
\[ 2 x + 5 y = 360 \]
Substitute and solve
Why: Two times ninety minus y, plus 5y, gives 3y equals 180.
\[ (30, 60) \]
Figure (svg): Percentages converted into amounts of acid
\[ x = 30, \; y = 60 \]
Verify: check the acid content
Why: Six millilitres of acid from the first and thirty from the second gives thirty-six, which is forty per cent of ninety. And thirty plus sixty is ninety millilitres in total. Both facts of the situation hold.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 7 Systems of Linear Equations and Inequalities — Lesson 7.4 Linear Systems and Problem Solving §7.4, pp. 410-410
Faded example
Each percentage acts on its own amount.
Fill in the blanks
20\% \text0.2 x: 36x \qquad 50\% \text___ y: 0.5y \qquad 40\% \text___ 90: ___
Why: Each percentage is multiplied by the amount it applies to, giving three quantities of acid in millilitres. Only then can they be compared and added, which is what makes the equation's two sides carry the same units.
Worked example
An optional step that removes a class of errors.
\[ \text{Rewrite } \; 0.2x + 0.5y = 36 \; \text{ without decimals.} \]
Find the multiplier
Why: One decimal place, so multiply by ten.
\[ x 10 \]
Multiply every term
Why: Including the right-hand side.
\[ 2 x + 5 y = 360 \]
Check the equivalence
Why: The pair (30, 60) satisfies both versions.
Note the benefit
Why: Every later line is in whole numbers.
Figure (svg): An equation with decimal coefficients multiplied to clear them
\[ 2x + 5y = 360 \]
Verify: test one point in both versions
Why: At (30, 60) the original gives six plus thirty, which is thirty-six, and the cleared version gives sixty plus three hundred, which is three hundred and sixty. Both hold, so the two equations describe the same line — which is what multiplying through preserves.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 7 Systems of Linear Equations and Inequalities — Lesson 7.4 Linear Systems and Problem Solving §7.4, pp. 410-410
Trap
A 20% solution and a 50% solution are mixed to give 40%.
Write 20 + 50 = 40 or 0.2x + 0.5y = 0.4
Why: The percentages are the numbers given, so they look like the quantities to combine.
Percentages of different amounts cannot be added. What adds is the actual acid: 0.2x millilitres and 0.5y millilitres give the mixture's thirty-six millilitres.
\[ 0.2x + 0.5y = 0.4(90) = 36 \]
Convert every percentage into an amount before adding
Why: A percentage is a rate and needs an amount to act on.
The units settle it: millilitres of acid on both sides, rather than a per cent compared with millilitres.
Elimination
Ninety millilitres of a 40% mixture.
Eliminate the wrong options
Which is the acid content of the mixture?
Survives elimination: A
Why: Forty per cent of ninety millilitres is thirty-six millilitres of acid. A percentage is a rate and produces a quantity only after being multiplied by the amount it acts on, which is the step this whole section turns on.
Prediction
Mixing 20% and 50% acid to get 40%.
Predict first
Before solving, which solution would you expect more of?
Correct: The 50%, since 40 is nearer 50 than 20.
\[ x = 30, \; y = 60 \quad \text{twice as much of the } 50\% \]
Why: The mixture's concentration is a weighted average of the two, so it sits nearer the concentration with the larger share. Forty is two thirds of the way from twenty to fifty, which correctly predicts twice as much of the stronger solution — sixty millilitres against thirty. That estimate takes seconds and would catch an answer with the two amounts swapped.
Socratic
The equation is correct with them.
Discussion prompt
Say what is gained by multiplying a decimal equation through by a power of ten, and what would be lost by getting the multiplier wrong. Then say what the same step was called in Lesson 5.4.
Hint: Think about the arithmetic in every later line.
Answer:
Every subsequent line works in whole numbers rather than decimals, which removes a whole class of place-value slips. Nothing is lost, because multiplying an equation through by a non-zero number leaves its solutions unchanged — the two versions describe the same line.
Getting the multiplier wrong, or applying it to only some terms, produces a different line and a wrong answer. In Lesson 5.4 the same step was called clearing fractions, done to put a standard-form equation into integer coefficients — the operation is identical and only the reason for wanting it differs.
Section
Section 5
Concept
Check the answer against both facts of the original situation, and then ask whether the values are possible — counts must be whole and positive, volumes non-negative.
The algebra does not know that violins come in whole numbers.
Figure (svg): An answer checked against both facts of the original situation
McDougal Littell Algebra 1: Concepts and Skills, Ch. 7 Systems of Linear Equations and Inequalities — Lesson 7.4 Linear Systems and Problem Solving §7.4, pp. 409-414 — the checks accompanying Examples 1 and 2
Picture it
Counting and valuing.
Figure (svg): An answer checked against both facts of the original situation
Reading each check as a sentence catches a model set up wrongly, which a bare substitution into your own equations would confirm rather than expose.
Worked example
The check for Example 1, read as sentences.
\[ \text{Check that } 5 \text{ and } 2 \text{ violins fit the situation.} \]
Check the count
Why: Five plus two is seven violins.
Check the value
Why: Five at two hundred and two at three hundred.
\[ 1000 + 600 \]
Add
Why: Sixteen hundred dollars.
Check plausibility
Why: Both counts are whole and positive.
Figure (svg): An answer checked against both facts of the original situation
\[ 5 + 2 = 7, \quad 1000 + 600 = 1600 \]
Verify: say what the plausibility check adds
Why: A pair could satisfy both equations and still be impossible — two and a half violins, or a negative count. The equations cannot detect that, so the last check is about the situation rather than the algebra, and it is the one the model needs a human for.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 7 Systems of Linear Equations and Inequalities — Lesson 7.4 Linear Systems and Problem Solving §7.4, pp. 409-409
Sorting
The variables count violins sold.
Sort into buckets
Sort each candidate answer by whether it could describe a real sale.
The pair with a zero in it is possible: selling none of one kind is a real outcome, and zero is a whole number. Zero and negative are different cases, which is worth keeping distinct.
Worked example
A well-posed problem can have an impossible solution.
\[ \text{A shop sells } 5 \text{ items for } 100 \text{ dollars, at } 30 \text{ and } 10 \text{ dollars each. How many of each?} \]
Write the system
Why: The counts add to five; the values add to a hundred.
\[ x + y = 5, 30 x + 10 y = 100 \]
Solve
Why: Substituting gives 20x plus fifty equals a hundred.
\[ x = 2.5 \]
Back-substitute
Why: y is also two and a half.
\[ (2.5, 2.5) \]
Interpret
Why: Half an item cannot be sold, so no such sale is possible.
Figure (svg): An answer rejected because it is impossible in the situation
\[ x = y = 2.5: \text{ impossible} \]
Verify: confirm the algebra is not at fault
Why: Substituting two and a half into both equations gives five items and a hundred dollars, so the pair genuinely solves the system. The impossibility is a fact about the situation rather than an error, and reporting it is the correct answer to the question as asked.
Trap
\[ x = 2.5 \text{ violins} \]
Report two and a half violins, since the algebra says so
Why: The equations were solved correctly and the pair checks out.
Violins come in whole numbers, so the model has produced an answer the situation cannot realise. Saying so is the right response rather than rounding.
No whole number of each kind gives those totals, so the situation as described cannot occur.
Read every answer back into the situation before reporting it
Why: The algebra does not know what the variables count.
Rounding would produce a pair failing one of the two facts, which is worse than reporting the impossibility.
Faded example
Counting and valuing, as sentences.
Fill in the blanks
5 + 2 = 7 \text1600 \qquad 5(200) + 2(300) = ___ \text___
Why: Both facts of the situation are recovered, so the model and the arithmetic are both confirmed. Reading each line as a sentence about violins and dollars is what makes it a check on the model rather than only on the algebra.
Elimination
The algebra gives 2.5 items of each kind.
Eliminate the wrong options
What is the right response?
Survives elimination: A
Why: The system has a solution and the situation does not, which is a real and reportable finding. Rounding breaks one of the given facts, and the honest answer is that the described sale cannot have happened as stated.
Socratic
Both substitutions passed.
Discussion prompt
Say what a substitution check confirms and what it cannot, in a modelling problem. Then say which errors only a reading of the situation will catch.
Hint: Ask what the equations know about violins.
Answer:
A substitution confirms that the pair satisfies the equations you wrote. It cannot confirm that those were the right equations, and it knows nothing about what the variables count — so a fractional or negative answer passes it without complaint.
Only reading the situation catches a mispaired price and count, a total computed as a rate rather than an amount, or an answer that is arithmetically fine and physically impossible. Those are model errors rather than algebra errors, which is why the last step of every modelling problem is a sentence rather than a substitution.
Comparison
Fill the blanks from memory before you scroll back.
Comparison matrix
| Counting equation | Value equation | |
|---|---|---|
| Coefficients | both 1 | prices, rates or concentrations |
| What it says | the amounts add to a total | the values or contents add to a total |
| Its role in solving | isolate a variable here | substitute into here |
The two equations have different shapes and different jobs. Recognising which is which decides both how to model the problem and how to solve it.
Pattern
Whether the problem counts objects, weights or volumes, the same five moves cover it.
Step five has two halves. The substitutions test the arithmetic, and the plausibility question tests the model, which is the part the algebra cannot do for you.
OpenStax Elementary Algebra 2e, §5.4 Solve Applications with Systems of Equations §5.4
Check
Price per unit times the number of units.
Check your understanding
Twelve items sell for 5 dollars each and eight for 9 dollars each. What is the total?
Answer: A
Why: Sixty dollars from the first group and seventy-two from the second gives one hundred and thirty-two dollars. Each price is multiplied by its own count before the two are added.
Check
A percentage acts on an amount.
Check your understanding
How much acid is in 90 mL of a 40% solution?
Answer: A
Why: Four tenths of ninety millilitres is thirty-six millilitres. A percentage produces a quantity only after being multiplied by the amount it applies to.
Check
Scan the coefficients.
Check your understanding
For x + y = 40 and 12x + 20y = 640, which method is efficient?
Answer: A
Why: The counting equation has coefficients of one, so x equals forty minus y in one subtraction and the substitution introduces no fractions. Solving the value equation for a variable would mean dividing by twelve or twenty.
Real world
A coffee shop blends a 12 dollar a kilogram bean with an 18 dollar a kilogram bean to make 50 kilograms of a blend selling at 15 dollars 60 a kilogram.
Discussion prompt
Model the blend as a system, solve it, and check both facts. Then say what the blend price tells you about the answer before you solve.
Hint: One equation weighs and the other values.
Answer:
\[ x + y = 50 \qquad 12x + 18y = 780 \]
\[ x = 50 - y \;\Longrightarrow\; 600 + 6y = 780 \;\Longrightarrow\; y = 30, \; x = 20 \]
So twenty kilograms of the cheaper bean and thirty of the dearer. Checking: the weights add to fifty kilograms, and two hundred and forty plus five hundred and forty is seven hundred and eighty dollars, which is fifty times fifteen sixty.
The blend price of 15.60 sits above the midpoint of fifteen, so more of the dearer bean was predictable before any algebra — and thirty against twenty is indeed a three-to-two split in its favour. That estimate is worth making first, because it catches an answer with the two amounts swapped.
Commit first
Answer, then rate your confidence honestly. Confident and wrong is the combination worth finding.
Predict first
Mixing a 20% and a 50% acid solution, what equation describes the acid content?
Correct: 0.2x + 0.5y = 0.4(90).
\[ 0.2x + 0.5y = 36 \;\xrightarrow{\times 10}\; 2x + 5y = 360 \]
\[ x = 30, \; y = 60 \]
Why: Each percentage has to be multiplied by the volume it acts on before anything is added, giving 0.2x and 0.5y millilitres of acid against the mixture's thirty-six. The third option adds the percentages directly, which is the standard error: percentages of different amounts are not comparable quantities. The fourth counts total volume rather than acid, which is the other equation of the system.
Explain it
They can solve a system and freeze when the problem is a paragraph.
Discussion prompt
In no more than four sentences, explain how to turn a mixture problem into a system. Then tell them the check that catches a wrong model rather than wrong arithmetic.
Hint: Two unknowns, two facts.
Answer:
A usable answer: name the two unknown amounts, then look for two facts in the paragraph. Almost always one sentence says how much altogether, which gives you x plus y equals the total, and another says how much it costs or contains, which gives you each price or percentage times its amount. Those two sentences are your two equations.
The check that matters is reading your answer back as sentences about the situation: did the counts add to the total, and did the money add to the total? If a percentage was added instead of an amount, or a price paired with the wrong count, those sentences will not come out right even though your algebra was fine.
Exit ticket
Name the weakest spot before you close the deck. That is the one worth ten minutes tonight.
Predict first
Which of these would you least want to be handed cold on a quiz tomorrow?
Correct: Whichever you picked is the right answer — and each one has a specific fix.
Why: Finding the facts is fixed by counting the unknowns and then looking for that many sentences. Percentages are fixed by multiplying each one by the amount it acts on before anything is added. The choice of equation is fixed by isolating in the one with coefficients of one. Plausibility is fixed by asking what the variables count and whether the answer could occur. Pick yours and do five of that kind tonight rather than twenty mixed ones.
Connect it up
Do this on paper. It is worth more than rereading the slides.
Draw it
At the top of a page write a mixture problem of your own in words, involving two items at different prices and a known total of each kind of thing. Underneath, write the verbal model as two sentences, then the labels with a symbol and a unit for every quantity, then the two equations, boxing the counting equation and circling the value equation. Solve the system by whichever method the coefficients suggest, writing beside your first line why you chose it. Check the answer underneath by writing both facts back out as sentences about the situation, not as substitutions. In the lower half, write a percentage mixture problem, convert all three percentages into amounts, write the system, clear the decimals, and solve. Finally, in the margin, write down one answer that would be algebraically valid and impossible in your situation, and say why.
Both of your checks should read as true sentences about the objects, not just as equalities. If a check is true as arithmetic but describes something that could not happen, the model rather than the algebra is what needs another look.
Recap
Five things, and the last is the one the algebra cannot do for you.
| If the question says | Your first move is |
|---|---|
| Two kinds of item and a total count | Write x + y = the total |
| Prices per item and a total cost | Write price times count for each |
| A percentage of a mixture | Multiply it by the amount it acts on |
| One equation has coefficients of 1 | Isolate a variable there |
| The answer is fractional or negative | Ask whether the situation allows it |
Lesson 7.5 returns to the systems themselves and asks what happens when the two lines do not cross once. Parallel lines and coincident lines both produce recognisable algebraic signals, and knowing them completes the picture.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 7 Systems of Linear Equations and Inequalities — Lesson 7.4 Linear Systems and Problem Solving §7.4, pp. 409-414 — everything on these slides traces back here
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