Chapter 7 of Algebra 1: Concepts and Skills, built for a visual learner. Systems solved by graphing, substitution and elimination, with the three outcomes shown as crossing, parallel and coincident lines; word problems with two unknowns drawn as bar models; and systems of inequalities as overlapping shaded regions.
Subject: Algebra 1 · 60 slides · symbolic lesson
Open the interactive version of this deck · Homework for this lesson
Title
Algebra 1 · Chapter 7
Two conditions at once: where lines cross, three ways to find it, and regions where every condition holds
Objectives
Until now every problem had one condition. Real problems usually have two, and two conditions is exactly what a system is.
Figure (svg): Two lines crossing at a single point, with that point circled and labelled as the solution of the system
Section
Section 7.1
Concept
A system is two equations that must both be true. Its solution is the point lying on both lines at once.
Figure (svg): Two lines crossing at a single point, with that point circled and labelled as the solution of the system
system of equations — Two or more equations considered together. A solution must satisfy every one of them, which is why one equation alone is never enough to pin down two unknowns.
Worked example
Solve the system below by graphing.
\[ y = x + 1 \qquad y = -x + 5 \]
Graph the first line from its slope and intercept
Why: It starts at 1 on the vertical axis and climbs one unit for every one across.
Graph the second the same way
Why: It starts at 5 and falls one unit for every one across.
Read the crossing point off the grid
Why: The lines meet where x is 2 and y is 3.
\[ (2, 3) \]
Figure (svg): Two lines crossing at a single point, with that point circled and labelled as the solution of the system
Verify: substitute the point into both equations
Why: The first gives 3 equals 2 plus 1, which is true. The second gives 3 equals negative 2 plus 5, also true. The point satisfies both, so it is genuinely the solution.
Prediction
Commit before checking.
\[ y = 2x - 1 \qquad y = x + 2 \]
Predict first
Where do these two lines cross?
Correct: (3, 5)
Why: Setting the two expressions for y equal gives 2x minus 1 equals x plus 2, so x is 3, and substituting back gives y equal to 5. Checking both: 2 times 3 minus 1 is 5, and 3 plus 2 is 5. The point (2, 3) is on the second line only, which is why it is tempting.
Socratic
One question, no computation.
Discussion prompt
Why does the solution of a system have to be a point that lies on both lines, rather than just one?
Hint: What does one line represent, all by itself?
Answer:
Because each line is the picture of every solution of its own equation. A point on the first line makes the first equation true, and a point on the second makes the second true.
The system demands both at once, so the solution has to belong to both pictures — which is precisely what a crossing point is. If the lines never cross, no such point exists and the system has no solution.
Error analysis
A student read a crossing point off a graph as (1.5, 2.5) and stopped. Diagnose the problem.
Annotate
On: \( y = 4x - 4 \qquad y = 2x - 1 \)
Use graphing to understand the situation and an algebraic method to get the number. That division of labour is the point of the next two sections.
Sorting
Compare slopes and intercepts. Do not graph anything.
Sort into buckets
Sort each system by its number of solutions.
Picture it
Before any algebra, learn what the three answers look like.
Figure (svg): Three planes showing lines crossing once, parallel lines never crossing, and two lines lying exactly on top of each other
This is the same classification you met in Chapter 3, now with the geometry attached. Two straight lines genuinely have no fourth option.
Tweak it
One line is fixed at y equals x. The other has an adjustable slope.
Parameter explorer
Drag the slope of the second line. What happens to the crossing point as the slope approaches 1?
\[ y = {m}x + 4 \]
Explain it
A classmate asks why you cannot just solve the two equations one after the other and be done.
Discussion prompt
Explain in two sentences why the two equations must be solved together rather than separately.
Hint: How many solutions does one equation with two unknowns have?
Answer:
Each equation on its own has infinitely many solutions — a whole line of them — so solving one alone tells you almost nothing about the answer.
The two equations have to be combined because the solution is the one pair of numbers that works for both at the same time, and only by using both together can you narrow a whole line down to a single point.
Section
Section 7.2
Concept
If one equation says what y equals, that expression can replace y in the other equation, leaving one equation in one letter.
Figure (svg): One equation solved for y being poured into the other equation, replacing the y and leaving only x
This is the same permission you used in Chapter 1: you may always swap a name for the thing it names.
Worked example
Solve the system below.
\[ y = 2x - 1 \qquad 3x + y = 9 \]
The first equation already says what y is, so substitute it into the second
Why: Replace y with the whole expression, brackets included, so the sign of every term is preserved.
\[ 3x + (2x - 1) = 9 \]
Solve the resulting one-variable equation
Why: Combine like terms, then undo the constant and the coefficient.
\[ 5x - 1 = 9 \;\Longrightarrow\; 5x = 10 \;\Longrightarrow\; x = 2 \]
Substitute back to find the other variable
Why: Use whichever equation is simpler — the first one is already solved for y.
\[ y = 2(2) - 1 = 3 \]
Figure (svg): The two lines of the system crossing at the point two comma three
Verify: substitute the pair into both original equations
Why: The first gives 3 equals 4 minus 1, which is true. The second gives 6 plus 3, which is 9, also true. Both hold, so (2, 3) is the solution.
Fill the middle
Fill each blank.
Fill in the blanks
x = y + 3 \;\texty + 3\; 2x + y = 12 \;\Longrightarrow\; 2(2) + y = 12 \;\Longrightarrow\; y = 5 \text___ x = ___
Why: Substituting y plus 3 for x gives 2y plus 6 plus y equals 12, so 3y equals 6 and y is 2. Then x equals 2 plus 3, which is 5. Forgetting the brackets around y plus 3 would give 2y plus 3, losing a 3 and producing the wrong answer.
Trap
Solve by substituting the first equation into the second.
\[ y = x - 4 \qquad 3y + 2x = 7 \]
Replace y with x minus 4 and multiply only the first term
Why: The three visibly touches the x, so it is easy to leave the minus 4 alone.
\[ 3x - 4 + 2x = 7 \;\to\; 5x = 11 \]
The 3 never reached the minus 4. The answer that comes out fails when substituted back.
Substitute the same expression, keeping it in brackets.
Write the substituted expression inside brackets, then distribute
Why: Brackets are what remind the 3 that it has two terms to reach.
\[ 3(x - 4) + 2x = 7 \;\Longrightarrow\; 5x - 12 = 7 \;\Longrightarrow\; x = \tfrac{19}{5} \]
Always write the brackets first and distribute second. It costs one extra symbol and prevents the commonest substitution error.
Discrimination
Substitution is easiest when one variable has a coefficient of 1. Sort each system.
Sort into buckets
In each system, which variable is easiest to isolate first?
Explain it to yourself
You replaced a letter with a whole expression. Say why that does not change the problem.
Discussion prompt
Why can you replace y with the expression 2x minus 1 in the second equation?
Hint: What exactly does the first equation claim about y?
Answer:
Because the first equation asserts that y and 2x minus 1 are the same number. At any point satisfying the system, they are two names for one value, so swapping one for the other changes nothing.
Notice the condition: this is only true at points satisfying the first equation. That is fine, because the solution of the system has to satisfy the first equation anyway — you are not losing any candidate you cared about.
Reverse engineer
Work backwards from a known solution.
Fill in the blanks
\text5 (4, 1) \text3 x + y = ___ \text___ x - y = ___
Why: Substituting the point into each equation tells you what the right-hand sides must be: 4 plus 1 is 5, and 4 minus 1 is 3. Building a system backwards like this is a useful way to make your own practice problems, and it is exactly how textbook questions are written.
Section
Section 7.3
Concept
Elimination adds the two equations together. If one variable has opposite coefficients, it disappears and leaves a single-variable problem.
Figure (svg): Two equations stacked and added, with the y terms cancelling to leave a single equation in x
When the coefficients are not already opposites, multiply one or both equations first to make them so.
Worked example
Solve the system below.
\[ 3x + 2y = 12 \qquad 5x - 2y = 4 \]
Notice the y terms are already opposites
Why: Positive 2y and negative 2y sum to nothing, so adding the equations will remove y entirely.
Add the two equations term by term
Why: Add the left sides together and the right sides together — this is legal because you are adding equal quantities to both sides.
\[ 8x = 16 \;\Longrightarrow\; x = 2 \]
Substitute back into either original equation
Why: The first gives 6 plus 2y equals 12, so 2y is 6 and y is 3.
\[ (2, 3) \]
Figure (svg): Two equations stacked and added, with the y terms cancelling to leave a single equation in x
Verify: substitute the pair into the second equation, the one not used
Why: Five times 2 is 10, minus 2 times 3 is 6, leaving 4, which matches. Checking against the equation you did not use for the back-substitution is the stronger test.
Worked example
Solve the system below, where nothing cancels yet.
\[ 2x + 3y = 13 \qquad 4x - y = 5 \]
Choose which variable to eliminate
Why: Multiplying the second equation by 3 will turn negative y into negative 3y, the opposite of the 3y in the first. That is the smallest multiplication available.
Multiply every term of the second equation by 3
Why: Every term, both sides. Scaling only part of an equation breaks it.
\[ 12x - 3y = 15 \]
Add the equations to eliminate y
Why: The 3y and negative 3y cancel, leaving a single-variable equation.
\[ 14x = 28 \;\Longrightarrow\; x = 2 \]
Substitute back
Why: The second original gives 8 minus y equals 5, so y is 3.
\[ (2, 3) \]
Figure (svg): The two equations with the second scaled by three, and the y column cancelling on addition
Verify: substitute into the first original equation
Why: Two times 2 is 4, plus 3 times 3 is 9, giving 13, which matches the right side exactly.
Ranking
For the system below, rank the moves.
\[ 5x + 2y = 11 \qquad 3x - 4y = 4 \]
Put in order
Why: Doubling the first equation turns 2y into 4y, the opposite of the negative 4y in the second. Adding then eliminates y and gives 13x equals 26, so x is 2. Substituting back into either original gives y equal to one half. Scaling must come first — adding before scaling eliminates nothing.
Comparison
Fill the blanks. All three give the same answer, but not at the same speed.
Comparison matrix
| method | best when | main risk |
|---|---|---|
| graphing | you want to see how many solutions there are | reading a non-integer crossing inaccurately |
| substitution | one variable has a coefficient of 1 | forgetting brackets around the substituted expression |
| elimination | the equations are in standard form | scaling only part of an equation |
Choosing the method before starting is worth more than being fast at any one of them.
Error analysis
This elimination goes wrong at the scaling step. Find it.
Annotate
On: \( \begin{aligned} 2x + 3y &= 13 \\ 4x - y &= 5 \;\overset{\times 3}{\longrightarrow}\; 12x - 3y &= 5 \end{aligned} \)
Say it as multiply the equation, not multiply the terms. The whole sentence gets scaled or none of it does.
Faded example
Fill each blank in this elimination.
Fill in the blanks
x + 2y = 7 \;\text4x\; 3x - 2y = 5 \;\Longrightarrow\; 3 = 12 \;\Longrightarrow\; x = 2 \text___ y = ___
Why: The y terms are already opposites, so adding gives 4x equals 12 and x is 3. Substituting into the first equation gives 3 plus 2y equals 7, so y is 2. Checking against the second equation: 9 minus 4 is 5, which is correct.
Prediction
Do not solve. Just choose the scaling.
\[ 2x + 5y = 1 \qquad 3x - 2y = 8 \]
Predict first
To eliminate x, what is the smallest pair of multipliers?
Correct: Multiply the first by 3 and the second by -2.
Why: That turns 2x into 6x and 3x into negative 6x, which cancel on addition. Multiplying by 3 and positive 2 makes both x terms positive, so adding doubles them instead of eliminating anything — you would have to subtract, which works but invites sign errors. Picking the multipliers so the column becomes exact opposites is the safer habit.
Pattern
One decision, then one of three procedures.
The last line matters: checking in the equation you already used cannot catch a back-substitution error, because that equation is what produced it.
Section
Section 7.4
Concept
Whenever a problem has two things you do not know, you need two independent pieces of information — and each one becomes an equation.
Figure (svg): Two unknowns shown as two empty boxes, with two separate facts each pinning down one relationship
Worked example
Tickets cost 8 dollars for adults and 5 dollars for children. 20 tickets were sold for 133 dollars. How many of each?
Name the two unknowns
Why: Let a be the number of adult tickets and c the number of child tickets. Naming first is what makes the equations writable.
Write the count equation
Why: Twenty tickets altogether, whatever the mix.
\[ a + c = 20 \]
Write the money equation
Why: Each adult ticket brings 8 dollars and each child ticket 5, and the total is 133.
\[ 8a + 5c = 133 \]
Solve by substitution
Why: The first equation gives c equals 20 minus a, which substitutes cleanly because its coefficient is 1.
\[ 8a + 5(20 - a) = 133 \;\Longrightarrow\; 3a = 33 \;\Longrightarrow\; a = 11 \]
\[ c = 20 - 11 = 9 \]
Figure (svg): A bar model showing twenty tickets split into eleven adult and nine child, with the money each group brings in
Verify: check both facts against the answer
Why: Eleven plus nine is 20 tickets, correct. Eight times 11 is 88, plus 5 times 9 is 45, giving 133 dollars, also correct. Both original facts hold.
Constraint
Both substitution and elimination have been taken away.
\[ a + c = 20 \qquad 8a + 5c = 133 \]
Discussion prompt
How would you find the answer using only reasoning about the numbers, and what does that tell you about why the algebra works?
Hint: What would the total be if every ticket were the cheaper kind?
Answer:
Start by assuming all 20 tickets were child tickets: that would give 100 dollars, which is 33 dollars short.
Every child ticket swapped for an adult one adds 3 dollars, so you need 11 swaps to make up the 33 — giving 11 adult and 9 child tickets.
This is elimination in disguise. Substituting c equals 20 minus a produces exactly the same arithmetic: a fixed baseline of 100 plus 3 for each adult ticket. The algebra is a compact way of writing the reasoning you just did.
Step zero
A boat travels 30 km downstream in 2 hours and the same 30 km back upstream in 3 hours.
Discussion prompt
Name the two unknowns and say what the two facts are, without solving anything.
Hint: What does the current do to the boat's speed in each direction?
Answer:
Unknowns: b, the speed of the boat in still water, and c, the speed of the current.
Fact one: going downstream the current helps, so the combined speed is 15 km per hour.
Fact two: going upstream the current hinders, so the combined speed is 10 km per hour.
\[ b + c = 15 \qquad b - c = 10 \]
Notice this system is already set up for elimination — adding the equations removes c immediately.
Missing information
A problem reads: two numbers add to 30. What are they?
Discussion prompt
Why can this not be answered, and what kind of extra fact would make it solvable?
Hint: How many pairs of numbers can you find that add to 30?
Answer:
One equation with two unknowns has infinitely many solutions — 1 and 29, 10 and 20, 15.5 and 14.5, and endlessly more. Graphed, they form a whole line.
You need a second independent fact: their difference, their ratio, or one of them being a multiple of the other. Any of those gives a second line, and the crossing point is the unique answer.
The word independent matters. Being told the numbers sum to 30 again in different words adds no line at all.
Real world
A coffee shop sells small cups for 3 dollars and large for 5. On one morning it sold 40 cups for 148 dollars.
Discussion prompt
Set up the system without solving it, then say which method you would use and why.
Hint: Which equation could you rearrange without creating a fraction?
Answer:
\[ s + l = 40 \qquad 3s + 5l = 148 \]
Method: substitution, because the first equation has coefficients of 1 and rearranging it introduces no fractions.
Solving gives 26 small and 14 large. Checking: 26 plus 14 is 40 cups, and 78 plus 70 is 148 dollars.
Analogy
Two-unknown problems all share one shape. Pair each situation with the pair of facts it gives you.
Match the pairs
Why: Every one of these is the same structure: one equation counting things and one equation weighting them. Once you recognise that pattern, setting up a two-unknown word problem becomes a matter of asking what is being counted and what is being weighted.
Estimation
Adult tickets cost 8 dollars, child tickets 5, and 20 tickets brought in 133 dollars.
Predict first
Before solving: roughly what proportion were adult tickets?
Correct: A bit more than half — the answer turns out to be 11 of 20.
\[ 100 \le \text{total} \le 160, \quad 133 \text{ is just past halfway} \]
Why: If all 20 were child tickets the total would be 100 dollars, and if all were adult it would be 160. The actual 133 sits a little past the middle of that range, so slightly more than half should be adult tickets. This estimate would immediately expose an answer like 3 adults as wrong.
Section
Section 7.5
Concept
Sometimes solving a system makes both variables disappear. What is left tells you which of the two special cases you are in.
Figure (svg): Three planes showing lines crossing once, parallel lines never crossing, and two lines lying exactly on top of each other
A false leftover statement means no solution; a true one means every point on the line is a solution.
Worked example
Solve the system below.
\[ 2x + y = 5 \qquad 4x + 2y = 3 \]
Scale the first equation to line up the coefficients
Why: Doubling the first gives 4x plus 2y equals 10, which now matches the second equation's left side exactly.
\[ 4x + 2y = 10 \qquad 4x + 2y = 3 \]
Subtract to eliminate
Why: Both variables cancel at once, leaving a statement about numbers alone.
\[ 0 = 7 \quad \text{false} \]
Interpret the leftover statement
Why: Zero is not seven, so no pair of numbers can satisfy both equations. The lines are parallel.
Figure (svg): Two parallel lines on a plane, never meeting, with the gap between them marked
Verify: compare the two slopes directly
Why: Rewriting both in slope-intercept form gives slopes of negative 2 in each case but different intercepts, which confirms parallel lines and no solution.
Anomaly
A student solved a system and every variable vanished.
\[ 3x - y = 4 \qquad 6x - 2y = 8 \]
Predict first
Subtracting twice the first from the second gives 0 equals 0. What does that mean?
Correct: Infinitely many solutions — the two equations describe the same line.
\[ 6x - 2y = 2(3x - y) = 2(4) = 8 \quad \text{the same equation} \]
Why: Doubling the first equation produces the second exactly, so they are two spellings of one line. Every point on that line satisfies both equations. A leftover statement that is TRUE means infinitely many solutions; one that is FALSE means none. The difference between the two cases is entirely in whether the leftover is true.
Comparison
Fill the blanks. This table turns a confusing moment into a decision.
Comparison matrix
| what is left after the variables cancel | how many solutions | what the lines look like |
|---|---|---|
| a false statement such as 0 = 7 | none | parallel, never meeting |
| a true statement such as 0 = 0 | infinitely many | the same line drawn twice |
| a value for one variable | exactly one | crossing at a single point |
Losing the variables is not a failure. It is the system telling you which of the three cases you are in.
Two truths and a lie
Three claims about special systems.
Eliminate the wrong options
Which statement is false?
Survives elimination: C
Why: Keep the false statement, which is C. Cancelling variables is a legitimate and informative outcome — it is precisely how the two special cases announce themselves. The only thing to do next is read whether the leftover number statement is true or false.
Error analysis
A student declared this system unsolvable. Diagnose the reasoning.
Annotate
On: \( \begin{aligned} 2x + 4y &= 10 \\ x + 2y &= 5 \end{aligned} \;\overset{?}{\Longrightarrow}\; 0 = 0 \;\text{ so no solution} \)
When the variables vanish, stop and read the number statement carefully. That one sentence is the entire answer.
Check
Solve it on paper before you click.
Check your understanding
Solve the system 3x + y = 10 and x - y = 2.
Answer: A
Why: The y terms are already opposites, so adding the equations gives 4x equals 12 and x is 3. Substituting into the second gives 3 minus y equals 2, so y is 1. Checking the first: 9 plus 1 is 10.
Section
Section 7.6
Concept
A system of inequalities asks for the points satisfying every condition. Shade each one and keep only the region shaded by all of them.
Figure (svg): Two shaded half-planes overlapping, with the overlap region highlighted as the solution of the system
This is the two-variable version of a compound and-inequality from Chapter 6.
Worked example
Graph the system and describe the solution region.
\[ y \le -x + 4 \qquad y \ge x \]
Graph the first boundary and shade below it
Why: The sign includes equality, so draw the line solid. Testing the origin gives 0 at most 4, which is true, so shade the side containing the origin.
Graph the second boundary and shade above it
Why: Also solid. Testing the point (1, 3) gives 3 at least 1, which is true, so shade that side.
Keep only the doubly shaded region
Why: A point must satisfy both conditions, so only the overlap counts. Everything shaded once and not twice is discarded.
Figure (svg): Two shaded half-planes overlapping, with the overlap region highlighted as the solution of the system
Verify: test one point inside the overlap and one just outside
Why: The point (1, 2) gives 2 at most 3 and 2 at least 1, so both hold. The point (3, 1) gives 1 at most 1, which holds, but 1 at least 3 fails — correctly outside.
Prediction
Substitute rather than guess.
\[ y > 2x - 1 \qquad y < -x + 6 \]
Predict first
Is the point (1, 3) a solution of this system?
Correct: Yes — it satisfies both conditions.
Why: The first gives 3 greater than 1, which is true. The second gives 3 less than 5, also true. A point only belongs to the solution region when every inequality holds, so both had to be checked — passing one is never enough.
Sorting
Test each point against both conditions.
\[ y \ge 0 \qquad y \le -x + 4 \]
Sort into buckets
Which points are in the solution region?
Real world
You can work at most 20 hours a week. A tutoring job pays 20 dollars an hour and a shop job pays 12, and you need at least 300 dollars.
Discussion prompt
Write the system, and say what the overlap region represents and why parts of the mathematical region are unusable.
Hint: Could you work minus three hours at the shop?
Answer:
\[ t + s \le 20 \qquad 20t + 12s \ge 300 \]
The overlap is every combination of hours that respects your time limit and still meets your income target.
Parts of the mathematical region are unusable because hours cannot be negative, so only the part in the first quadrant counts. Real constraints almost always add these invisible extra inequalities.
Edge cases
Push the system to its edge case.
Discussion prompt
What does it mean if the shaded regions of a system of inequalities have no overlap at all?
Hint: What was the equivalent situation for a compound and-inequality on a number line?
Answer:
It means the system has no solutions: there is no point in the plane satisfying every condition at once.
This is the two-variable version of the empty and-inequality from Chapter 6. In a real problem it usually means the constraints are contradictory — for example, needing to earn 800 dollars while working at most 20 hours at those rates.
A system with no overlap is not a failed calculation. It is a genuine and often useful answer: the demands cannot all be met.
Discrimination
For each inequality, decide the boundary style. Do not graph.
Sort into buckets
Sort each inequality by how its boundary should be drawn.
Trade off
You have at most 12 hours to study and want at least 8 topics covered. Maths takes 2 hours a topic and English 1. Fill the blanks.
Comparison matrix
| the time constraint | the coverage constraint | |
|---|---|---|
| inequality | 2m + e <= 12 | m + e >= 8 |
| which way it shades | below the line, towards the origin | above the line, away from the origin |
| what it rules out | plans that take too long | plans that cover too little |
The usable plans are the overlap, and the two constraints pull in opposite directions — which is exactly why the region is a wedge rather than the whole plane.
Commit first
Answer, then rate your confidence.
\[ y = 4x - 3 \qquad 8x - 2y = 6 \]
Predict first
How many solutions does this system have?
Correct: Infinitely many — the second equation is the first in disguise.
\[ 8x - 2y = 6 \;\Longrightarrow\; y = 4x - 3 \]
Why: Rearranging the second gives 2y equals 8x minus 6, so y equals 4x minus 3, which is exactly the first equation. The two describe the same line, so every point on it satisfies both. Substituting would produce the true statement 0 equals 0 rather than a value for x.
Exit ticket
Last commitment of the chapter.
Predict first
Which of these is shakiest right now?
Correct: Whatever you picked is the one to drill first.
Why: The third one is the most commonly under-practised, because it is the only step with no procedure to follow — you have to decide what the unknowns are before any algebra exists. Naming the two unknowns explicitly, in writing, fixes most of the difficulty.
Connect it up
One page, drawn by you.
Draw it
Put the three solution outcomes across the top: one, none, infinitely many. Under each, draw the picture and write what the algebra looks like when you reach it. Then, down the side, list the three methods and draw an arrow from each to the situation it suits best.
If your map does not connect parallel lines to a false leftover statement, add that arrow — it is the link that makes the special cases stop being surprising.
Recap
You can now handle problems with two unknowns, which is where algebra starts being genuinely useful.
| if you remember one thing | it should be |
|---|---|
| about solutions | the answer must satisfy every equation, not just one |
| about elimination | scale the whole equation, both sides, every term |
| about word problems | two unknowns need two independent facts |
| about regions | the answer is the overlap, and real constraints add hidden inequalities |
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