Week 5 · Chapter

Chapter 5 — Systems of Linear Equations & Inequalities

College Algebra Free sample edition

The module's primary reading · ~35–45 minutes · Objective 4 (analyze linear functions (slope, intercepts, and the forms of a line, including parallel and perpendicular lines) and solve systems of linear equations and inequalities in two variables), Objective 2 (solve linear equations and linear inequalities in one variable, including absolute-value equations and inequalities, and express solution sets in interval notation) · Read it first this week; the lecture, the tutorial, the quiz, and the Model/Build Studio all build on it.

In plain language

Last week every line stood on its own. This week two lines share one graph, and you look for the place where they agree. That place is the solution of a system (the one pair of numbers that makes both equations true at the same time). You will learn two ways to find it with algebra, substitution (swapping one variable for an equal expression) and elimination (adding the equations so one variable cancels). You will also spot the two odd cases, lines that never meet and two copies of one line, and you will shade the region where two inequalities are true together.

Why you already care about this

Say your club orders T-shirts to sell at a fundraiser. The printer charges 60 dollars to set up the design and 5 dollars for every shirt. You sell each shirt for 15 dollars. Sell 2 shirts and you are still behind. Sell 20 and you are well ahead. Somewhere in between, the money coming in catches the money going out. That moment has a name, the break-even point, and it is the crossing point of two lines.

Now say you are mixing trail mix for a hiking trip. You want twice as many cups of oats as cups of nuts, and 9 cups in all. How much of each? Two facts, two unknowns, one answer that has to fit both facts. That is a system, and you can find its answer with one substitution.

You already balance two conditions every day, like a plan that fits both your budget and your data use. This chapter writes the balancing down so you can find the answer and prove it.

By the end of this chapter you will be able to:

check whether an ordered pair is a solution of a system, in both equations;
solve a system by substitution;
solve a system by elimination, multiplying one equation first when you need to;
classify a system as having one solution, no solution, or infinitely many;
set up and solve a system from a scenario, such as a break-even question;
shade a system of linear inequalities and test a point in it.

Carry this sentence with you: one pair, two equations — check it in both.

Where we are

Already covered: Week 2, Linear Equations & Inequalities, taught you to solve for one unknown and to check by substitution. Week 3, Functions: Notation, Domain & Range, taught you to read input-output pairs. Week 4, Linear Functions & Their Graphs, taught you slope, intercepts, and the forms of a line.
This week: systems of two linear equations, solved by graphing, substitution, and elimination; classifying a system; and systems of linear inequalities.
How they connect: every system you solve this week ends with a one-variable equation from Week 2, and every graph is a Week 4 line.
Where we're going next: Week 6, Polynomials & Factoring, starts a new unit, where expressions get squared terms and you learn to multiply and factor them.

Week 4 is the week this chapter leans on hardest. Recall slope-intercept form, y = mx + b, where the slope m multiplies x and the y-intercept b stands alone. Recall also that parallel lines have equal slopes and different y-intercepts, so they never meet. Week 2 supplies the other half: do the same thing to both sides, and check your answer in the original equation. You will use both moves on nearly every page below.

1. See a solution: where two lines cross

Picture two friends walking toward each other on a long straight path. One starts at mile marker 4 and walks one way; the other starts at marker 14 and walks the other way. Wherever they meet, they stand at the same marker at the same time. Two lines on a graph meet the same way: at one shared point.

A system of linear equations is two or more linear equations that share the same variables and are meant to hold at the same time; in this course a system has two equations in x and y.

A solution of a system is an ordered pair (x, y) that makes every equation in the system true at once; on a graph it is a point where the lines meet.

The key word is every. A pair that fits one equation is a point on one line. A solution has to sit on both lines.

Worked example — the crossing at (2, 10). Take the two lines y = 3x + 4 and y = −2x + 14. The first is the Week 4 line with slope 3 and y-intercept 4. The second falls 2 for every step to the right and starts at 14.

Graph them. The first line rises from (0, 4). The second falls from (0, 14). They cross at the point (2, 10).
Check the first equation. Put x = 2 into y = 3x + 4: 3(2) + 4 = 6 + 4 = 10. The y-value is 10, so the pair fits.
Check the second equation. Put x = 2 into y = −2x + 14: −2(2) + 14 = −4 + 14 = 10. Again the y-value is 10, so the pair fits.
Conclude. The pair works in both, so (2, 10) is the solution of the system.

In words: at x = 2, both lines have the same height, 10, and that shared point is the answer.

Worked example — a pair that fits only one line. Is (1, 7) a solution of the same system?

Check the first equation. 3(1) + 4 = 7. That matches the 7, so (1, 7) is on the first line.
Check the second equation. −2(1) + 14 = 12. But the pair says y = 7, and 12 ≠ 7.
Conclude. The pair fails the second equation, so (1, 7) is not a solution of the system.

You can also find the crossing without a graph. Both equations say what y equals, so the two right sides must be equal where the lines meet. Set 3x + 4 = −2x + 14. Add 2x to both sides to get 5x + 4 = 14. Subtract 4 to get 5x = 10. Divide by 5 to get x = 2. Then y = 3(2) + 4 = 10, and the pair (2, 10) is the same point the graph showed.

Misconception check — one equation is enough. Students often check a pair in the first equation, see it work, and stop. It is tempting because the first check feels like proof. But (1, 7) passes the first equation and fails the second. The cure: a system has two equations, so a check has two parts. Write "first equation" and "second equation" and fill in both.

In practice (your tools). In Desmos, type y = 3x + 4 on one line and y = -2x + 14 on the next. Click the point where the two lines cross. Desmos labels it (2, 10). Then click any other point on the first line, such as (1, 7), and see that it sits off the second line.

2. Solve by substitution: replace one variable

Back to the trail mix. You want twice as many cups of oats as cups of nuts, and 9 cups in all. Let x be the cups of nuts and y be the cups of oats. "Twice as many oats as nuts" says y = 2x. "Nine cups in all" says x + y = 9.

The first equation already tells you what y is: it is 2x. So wherever you see y in the second equation, you may write 2x instead. That swap is the whole idea.

The substitution method is a way to solve a system by solving one equation for a variable and putting that expression in place of the variable in the other equation, which leaves one equation in one variable.

Worked example — the trail mix. Solve y = 2x and x + y = 9.

Substitute. Replace y with 2x in x + y = 9. That gives x + 2x = 9.
Combine like terms. x + 2x is 3x, so 3x = 9.
Divide by 3. x = 3.
Find y. Put x = 3 into y = 2x: y = 2 · 3 = 6.
Check the first equation. 6 = 2 · 3 is true.
Check the second equation. 3 + 6 = 9 is true.

In words: the mix uses 3 cups of nuts and 6 cups of oats, and the pair (3, 6) fits both facts.

Notice what happened in step 2. Once you substituted, the system became an ordinary Week 2 equation in one variable. Substitution does not invent a new kind of algebra. It turns a two-variable problem into a one-variable problem you already know how to solve.

Substitution works best when one equation already has a variable alone, like y = 2x. If neither does, solve one equation for a variable first; x + y = 9 can become y = 9 − x.

Misconception check — substituting into the same equation. Students sometimes solve y = 2x for y and then put 2x back into y = 2x. The result is 2x = 2x, which is true for every x and tells you nothing. It is tempting because that equation is right in front of you. The cure: substitute into the other equation, the one you have not used yet.

In practice (your tools). In Desmos, type y = 2x and x + y = 9. Desmos graphs both directly. Click the crossing point, and Desmos labels it (3, 6), the same pair your substitution found.

3. Solve by elimination: add to cancel a variable

A basketball team played 10 games this season and won 4 more games than it lost. Let x be the wins and y be the losses. "Ten games" says x + y = 10. "Four more wins than losses" says x − y = 4.

This time neither equation says "y equals." But look at the y terms: +y in the first equation and −y in the second. They are opposites. If you add the two equations, the y terms cancel, and only x is left.

The elimination method is a way to solve a system by adding or subtracting the equations, after multiplying one or both by a number if needed, so that one variable cancels out.

Why is adding allowed? Each equation says two amounts are equal. If you add equal amounts to equal amounts, the totals are still equal. That is the Week 2 "same thing to both sides" principle, used twice at once.

Worked example — the season record. Solve x + y = 10 and x − y = 4.

Add the left sides. (x + y) + (x − y) = 2x, because y and −y cancel.
Add the right sides. 10 + 4 = 14.
Write the new equation. 2x = 14.
Divide by 2. x = 7.
Back-substitute. Put x = 7 into x + y = 10: 7 + y = 10, so y = 3.
Check both equations. 7 + 3 = 10 is true, and 7 − 3 = 4 is true.

In words: the team won 7 games and lost 3, and the pair (7, 3) fits both facts.

Step 5 has a name. Back-substitution is the step after you find one variable, in which you put its value into either original equation to find the other variable. Either equation works. Try it with x − y = 4: 7 − y = 4, so y = 3, the same answer.

Misconception check — adding when you should subtract. Students often add two equations out of habit. Take x + y = 3 and x + y = 7. Adding gives 2x + 2y = 10, and nothing cancels. It is tempting because adding worked last time. The cure: look at the pair of terms you want gone. Opposite signs, such as y and −y, cancel when you add. The same sign, such as y and y, cancel when you subtract. You will finish this example in Section 5.

4. Multiply first, then eliminate

Last week's study group had a 12-dollar budget for snacks at 2 dollars each and drinks at 3 dollars each, which gave the standard-form line 2x + 3y = 12. This week the group adds a second condition: buy one more snack than drinks. With x snacks and y drinks, that says x − y = 1.

Now look at the y terms: 3y and −y. They are not opposites yet, so adding the equations would not cancel anything. But you can make them opposites. Multiply every term of x − y = 1 by 3, and −y becomes −3y.

Worked example — the snack budget. Solve 2x + 3y = 12 and x − y = 1.

Multiply the second equation by 3. 3 times x is 3x, 3 times −y is −3y, and 3 times 1 is 3. So x − y = 1 becomes 3x − 3y = 3.
Add the equations. Add 2x + 3y = 12 and 3x − 3y = 3. The left sides give 2x + 3x = 5x, and the 3y and −3y cancel. The right sides give 12 + 3 = 15. So 5x = 15.
Divide by 5. x = 3.
Back-substitute. Put x = 3 into x − y = 1: 3 − y = 1, so y = 2.
Check the first equation. 2(3) + 3(2) = 6 + 6 = 12 is true.
Check the second equation. 3 − 2 = 1 is true.

In words: the group buys 3 snacks and 2 drinks, which spends exactly 12 dollars and keeps one more snack than drinks.

Multiplying an equation does not change its line. The equation 3x − 3y = 3 says the same thing as x − y = 1, only with every number three times as large. Every pair that fits one fits the other.

Which variable should you eliminate? Either one works here. Multiplying x − y = 1 by 3 makes the y terms opposites, so adding cancels y. Multiplying it by 2 instead makes the x terms match, so subtracting would cancel x. Pick the route that feels safer to you; this example adds, because adding opposites is the move this section practises.

Misconception check — multiplying only one side. Students often multiply x − y = 1 by 3 and write 3x − 3y = 1. It is tempting because the left side is where the action is. But the 3 has to reach every term on both sides, the same as the distributive property from Week 1. The right side becomes 3 · 1 = 3. The cure: draw an arrow from the 3 to every term, including the number on the right, and count the arrows.

In practice (your tools). In Desmos, type 2x + 3y = 12 and x - y = 1. Click the crossing point, and Desmos labels it (3, 2). Now add a third line, 3x - 3y = 3. It lands exactly on top of x - y = 1, which shows that multiplying by 3 did not move the line.

5. Classify a system: one, none, or infinitely many

Two lines in a plane can meet in only three ways. They can cross once. They can run side by side forever, like the two rails of a train track. Or they can be the very same line drawn twice. Each picture has its own algebra.

A consistent system is a system that has at least one solution; its lines meet at one point or are the same line.
An inconsistent system is a system with no solution; its lines are parallel, and solving it collapses to a false statement such as 0 = 4.
An independent system is a system whose two equations describe different lines; when the lines cross, it has exactly one solution.
A dependent system is a system whose two equations describe the same line, so every point on that line is a solution; solving it collapses to a true statement such as 0 = 0.

Every system you solved in Sections 1 through 4 was consistent and independent: one crossing, one solution.

Worked example — an inconsistent system. Solve x + y = 3 and x + y = 7.

Choose the operation. The x terms have the same sign, and so do the y terms, so you subtract.
Subtract the first equation from the second. (x + y) − (x + y) = 7 − 3, which gives 0 = 4.
Read the leftover statement. 0 = 4 is never true. Both variables vanished, and what remains is false.
Conclude. There is no solution. The system is inconsistent.

The graph tells the same story. Solve each equation for y: y = −x + 3 and y = −x + 7. Both lines have slope −1, and their y-intercepts are 3 and 7. Equal slopes and different intercepts make parallel lines, and parallel lines never meet.

Worked example — a dependent system. Solve 2x + y = 5 and 4x + 2y = 10.

Multiply the first equation by 2. 2 times 2x is 4x, 2 times y is 2y, and 2 times 5 is 10. So you get 4x + 2y = 10.
Subtract it from the second equation. (4x + 2y) − (4x + 2y) = 10 − 10, which gives 0 = 0.
Read the leftover statement. 0 = 0 is always true.
Conclude. There are infinitely many solutions: every point on the line y = −2x + 5.

You can name some of them. The points (0, 5), (1, 3), and (2, 1) all work. Check (1, 3): 2(1) + 3 = 5 is true, and 4(1) + 2(3) = 4 + 6 = 10 is true. The line y = −2x + 5 may look familiar. In Week 4 it was the line where students swap the slope and the intercept.

The three kinds of linear system, with their algebra and their graphs
What solving gives	Number of solutions	What the graph shows	Name
one pair, such as (2, 10)	exactly one	two lines crossing once	consistent and independent
a false statement, such as 0 = 4	none	two parallel lines	inconsistent
a true statement, such as 0 = 0	infinitely many	one line drawn twice	consistent and dependent

You can even classify before you solve. Put both equations in slope-intercept form. Different slopes mean one crossing. Equal slopes with different intercepts mean parallel lines and no solution. Equal slopes with equal intercepts mean the same line.

Misconception check — swapping the two odd cases. Students often call 0 = 0 "no solution" and 0 = 4 "infinitely many." It is tempting because both variables vanished, and a vanished variable feels like a dead end. The cure is the Week 2 habit: when the variables disappear, read the leftover statement. A true statement, 0 = 0, means every point on the line works. A false statement, 0 = 4, means no point works.

In practice (your tools). In Desmos, type x + y = 3 and x + y = 7. You see two parallel lines and no crossing point to click. Now replace them with 2x + y = 5 and 4x + 2y = 10. Only one line appears, because the second equation is drawn exactly on top of the first.

6. Model a break-even point: where cost meets revenue

Here is the club's T-shirt sale from the start of the chapter. Every business question of the form "how many until we stop losing money?" is a system in disguise. One equation tracks the money going out; the other tracks the money coming in.

The cost has a fixed part and a per-item part, the same shape as a Week 4 line. The printer charges 60 dollars once and 5 dollars per shirt, so the cost of x shirts is C = 60 + 5x dollars. Each shirt sells for 15 dollars, so the revenue from x shirts is R = 15x dollars.

The break-even point is the point where a cost line and a revenue line meet, so the money taken in equals the money spent; selling more than that number of items earns a profit.

Worked example — the T-shirt break-even point. Find where C = 60 + 5x meets R = 15x.

Set revenue equal to cost. 15x = 60 + 5x.
Collect the x terms. Subtract 5x from both sides: 10x = 60.
Divide by 10. x = 6.
Check the revenue. 15 · 6 = 90 dollars.
Check the cost. 60 + 5 · 6 = 60 + 30 = 90 dollars.
Interpret. At 6 shirts, the money in equals the money out, so the break-even point is (6, 90).

In words: the club must sell 6 shirts to cover the 60-dollar setup fee. The seventh shirt and every one after it earns 10 dollars of profit, because each sells for 15 dollars and costs only 5.

Here is an honesty note about models like this one. A break-even line assumes every shirt costs the same and every shirt sells. Real printers give bulk discounts, and real sales stall. The model is a clear first answer, not a promise.

Misconception check — reading only half the point. Students often report the break-even point as "90" or as "6" alone. It is tempting because the question sounds like it wants one number. But the point has two parts, and each answers a different question: 6 is how many shirts, and 90 is how many dollars. The cure: write the pair (6, 90) and then say each number in words with its unit.

In practice (your tools). In Desmos, type y = 60 + 5x and y = 15x. Click the crossing point, and Desmos labels it (6, 90). To the right of that point, the revenue line sits above the cost line; that gap is the profit.

7. Shade a system of linear inequalities

A single equation such as y = 4 is a line. A single inequality such as y < 4 is a whole region: every point below that line. Now ask for two conditions at once, and you want the region where both shaded areas overlap.

A linear inequality in two variables is an inequality in x and y, such as 2x + 3y ≤ 12, whose solutions are all the ordered pairs on one side of a line, sometimes including the line itself. That one side is a half-plane: all the points on one side of a line in the coordinate plane; the solutions of a linear inequality in two variables form a half-plane.

The line itself has a name. The boundary line is the line that separates the half-plane of an inequality from the rest of the plane. It is drawn dashed for a strict inequality (less than or greater than) and solid when equality is allowed (at most or at least).

To choose a side, use a test point: a point that is not on a boundary line, substituted into an inequality to decide which side to shade: if the inequality is true there, that side holds the solutions.

A system of linear inequalities is two or more linear inequalities in the same variables that must all hold at once; its solutions are the points in the overlap of their half-planes.

Worked example — dashed boundaries. Graph the system y > x and y < 4.

Draw the boundaries. The boundary lines are y = x and y = 4. Both signs are strict, so you draw both lines dashed.
Shade the first inequality. y > x means the y-value is bigger than the x-value, so you shade above the line y = x.
Shade the second inequality. y < 4 means the y-value is less than 4, so you shade below the line y = 4.
Find the overlap. The solutions sit above y = x and below y = 4 at the same time.
Test the point (1, 3). 3 > 1 is true, and 3 < 4 is true. Both are true, so (1, 3) is in the overlap.
Test the point (2, 2). 2 > 2 is false. The point sits on the dashed boundary y = x, so it is not a solution.
Test the point (0, 5). 5 > 0 is true, but 5 < 4 is false. The point fails one inequality, so it is outside.

In words: the solution region is the strip between the two dashed lines, and a point on either dashed line does not count.

Worked example — solid boundaries. Return to the snack budget. Now the group may spend at most 12 dollars and wants at least one more snack than drinks. With x snacks and y drinks, that is 2x + 3y ≤ 12 and x − y ≥ 1.

Draw the boundaries. The boundary lines are 2x + 3y = 12 and x − y = 1, the two lines from Section 4. Both signs allow equality, so you draw both lines solid.
Test the point (4, 1). 2(4) + 3(1) = 8 + 3 = 11, and 11 ≤ 12 is true. Also 4 − 1 = 3, and 3 ≥ 1 is true. So (4, 1) is in the region.
Look at the corner. The two solid boundaries cross at (3, 2), the solution from Section 4. That corner sits on both solid lines, so it counts.

In words: buying 4 snacks and 1 drink fits both rules, and so does the exact-budget choice of 3 snacks and 2 drinks.

Misconception check — a solid line for a strict sign. Students often draw every boundary solid. It is tempting because a solid line is what you drew all of last week. But in y < 4, the points on the line y = 4 make the statement 4 < 4, which is false. The cure: before you draw, read the sign. Strict signs, less than and greater than, get dashed lines. Signs with "or equal to" get solid lines.

In practice (your tools). In Desmos, type y > x on one line and y < 4 on the next. Desmos shades each half-plane and draws strict boundaries dashed for you. The darker region where the two shadings overlap is the solution region. Type y <= 4 instead, and the top boundary turns solid.

8. Check every pair in both equations

The last move of the week is the one that protects all the others. A system has two equations, so your check has two parts. A pair that works in one equation only is not a solution. The check is cheap, and it catches nearly every slip.

Here is a slip worth seeing once. Someone solves the trail mix system, y = 2x and x + y = 9, and reports the answer as (6, 3). The numbers 3 and 6 are right, but they are in the wrong order.

Worked example — catching a swapped pair. Is (6, 3) a solution of y = 2x and x + y = 9?

Read the pair. In (6, 3), x = 6 and y = 3.
Check the second equation. 6 + 3 = 9 is true. This check alone would fool you.
Check the first equation. 2 · 6 = 12, but the pair says y = 3, and 3 ≠ 12. False.
Conclude. (6, 3) is not a solution. The correct pair is (3, 6): 3 cups of nuts and 6 cups of oats.

The sum check could not catch the swap, because 6 + 3 and 3 + 6 are both 9. Only the second check could. That is why both checks matter, every time.

The same habit protects you when a chatbot solves a system for you. A chatbot can write a confident pair that fits one equation and not the other. Here is a caveat worth repeating: a tidy answer is not a checked answer. Substitute the pair into both original equations yourself before you trust it.

Misconception check — trusting the pair you found. Students often skip the check because the arithmetic "felt right." It is tempting at the end of a long problem. But a dropped sign or a swapped order shows up only when you substitute. The cure: make the two-part check the last line of every system, written out, with each side's value.

In practice (your tools). In Desmos, type both equations of any system you solved by hand, then click the crossing point. If Desmos labels the same pair you found, your algebra agrees with the picture. If it labels a different pair, recheck your steps.

A reflection to choose from. Pick one option, (a) or (b), and write two or three sentences.

(a) Think of a choice in your own life that has two conditions at once, such as a budget and a time limit. Name the two unknowns and say what each condition says about them.
(b) Think of a time two plans or prices crossed over, so one was cheaper at first and the other was cheaper later. Describe what the crossing point meant for your choice.

Key terms

System of linear equations — Two or more linear equations that share the same variables and are meant to hold at the same time; in this course a system has two equations in x and y. (first used: 1. See a solution: where two lines cross)
Solution of a system — An ordered pair (x, y) that makes every equation in the system true at once; on a graph it is a point where the lines meet. (first used: 1. See a solution: where two lines cross)
Substitution method — A way to solve a system by solving one equation for a variable and putting that expression in place of the variable in the other equation, which leaves one equation in one variable. (first used: 2. Solve by substitution: replace one variable)
Elimination method — A way to solve a system by adding or subtracting the equations, after multiplying one or both by a number if needed, so that one variable cancels out. (first used: 3. Solve by elimination: add to cancel a variable)
Back-substitution — The step after you find one variable, in which you put its value into either original equation to find the other variable. (first used: 3. Solve by elimination: add to cancel a variable)
Consistent system — A system that has at least one solution; its lines meet at one point or are the same line. (first used: 5. Classify a system: one, none, or infinitely many)
Inconsistent system — A system with no solution; its lines are parallel, and solving it collapses to a false statement such as 0 = 4. (first used: 5. Classify a system: one, none, or infinitely many)
Independent system — A system whose two equations describe different lines; when the lines cross, it has exactly one solution. (first used: 5. Classify a system: one, none, or infinitely many)
Dependent system — A system whose two equations describe the same line, so every point on that line is a solution; solving it collapses to a true statement such as 0 = 0. (first used: 5. Classify a system: one, none, or infinitely many)
Break-even point — The point where a cost line and a revenue line meet, so the money taken in equals the money spent; selling more than that number of items earns a profit. (first used: 6. Model a break-even point: where cost meets revenue)
Linear inequality in two variables — An inequality in x and y, such as 2x + 3y ≤ 12, whose solutions are all the ordered pairs on one side of a line, sometimes including the line itself. (first used: 7. Shade a system of linear inequalities)
Half-plane — All the points on one side of a line in the coordinate plane; the solutions of a linear inequality in two variables form a half-plane. (first used: 7. Shade a system of linear inequalities)
Boundary line — The line that separates the half-plane of an inequality from the rest of the plane. It is drawn dashed for a strict inequality (less than or greater than) and solid when equality is allowed (at most or at least). (first used: 7. Shade a system of linear inequalities)
Test point — A point that is not on a boundary line, substituted into an inequality to decide which side to shade: if the inequality is true there, that side holds the solutions. (first used: 7. Shade a system of linear inequalities)
System of linear inequalities — Two or more linear inequalities in the same variables that must all hold at once; its solutions are the points in the overlap of their half-planes. (first used: 7. Shade a system of linear inequalities)

Summary

A solution of a system is one ordered pair that makes both equations true.
On a graph, that pair is the point where the two lines cross.
Substitution: replace a variable with an equal expression from the other equation.
Elimination: add or subtract the equations so one variable cancels.
Multiply an equation first when no pair of terms cancels yet, and multiply every term.
A false leftover such as 0 = 4 means no solution; a true one such as 0 = 0 means infinitely many.
A break-even point is where the cost line meets the revenue line.
For inequalities, strict signs get dashed boundaries, and a test point picks the side.
Check every pair in both equations, never only one.

Check your understanding

Work these before looking at the answers — they are ungraded, and the answers (with reasoning) are at the bottom of the page.

Solve by substitution and check: y = 4x − 1 and 2x + y = 11.
A community theater sold 30 tickets for a total of 280 dollars. Student tickets cost 8 dollars and general tickets cost 12 dollars. Write a system and solve it by elimination to find how many of each were sold.
Classify each system without finding a pair: (a) 3x − y = 2 and 6x − 2y = 4; (b) 3x − y = 2 and 6x − 2y = 5.
For the system y ≥ 2x and x + y < 6, say which boundary is dashed and which is solid. Then decide whether (1, 4) and (2, 4) are solutions.

Answers (worked reasoning — read after attempting)

The first equation says y is 4x − 1, so substitute it into the second: 2x + (4x − 1) = 11. Combine like terms: 6x − 1 = 11. Add 1 to both sides: 6x = 12. Divide by 6: x = 2. Then y = 4(2) − 1 = 8 − 1 = 7. Check the first equation: 4 · 2 − 1 = 7 is true. Check the second: 2 · 2 + 7 = 4 + 7 = 11 is true. The pair (2, 7) works in both equations.
Let x be the student tickets and y the general tickets. Then x + y = 30 and 8x + 12y = 280. Multiply the first equation by 8: 8x + 8y = 240. Subtract it from the second: (8x + 12y) − (8x + 8y) = 280 − 240, so 4y = 40 and y = 10. Back-substitute: x + 10 = 30, so x = 20. Check: 20 + 10 = 30 is true, and 8 · 20 + 12 · 10 = 160 + 120 = 280 is true. The theater sold 20 student tickets and 10 general tickets.
In (a), multiply 3x − y = 2 by 2 to get 6x − 2y = 4. Subtracting it from the second equation gives 0 = 0, which is always true, so the system is dependent: one line, infinitely many solutions. In (b), the same multiplication gives 6x − 2y = 4, and subtracting it from 6x − 2y = 5 gives 0 = 1, which is never true, so the system is inconsistent: parallel lines, no solution. Calling 0 = 0 "no solution" would be the classic swap.
The boundary y = 2x is solid, because ≥ allows equality. The boundary x + y = 6 is dashed, because < is strict. For (1, 4): 4 ≥ 2 · 1 = 2 is true, and 1 + 4 = 5 < 6 is true, so (1, 4) is a solution. For (2, 4): 4 ≥ 2 · 2 = 4 is true, because the solid boundary counts. But 2 + 4 = 6, and 6 < 6 is false, so (2, 4) sits on the dashed boundary and is not a solution.

Next up in this module: the lecture, the slide deck, the Readings & Resources page, the lecture tutorial, the practice set, the quiz, and the discussion — and next week the distributive rule returns as a method for multiplying and factoring polynomials.

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