Lecture: Systems of Linear Equations & Inequalities
Lecture script · about 15–20 minutes spoken · Week 5
Opening
Think about a moment when two conditions have to agree. A small-business owner wants to know how many sales cover the setup cost. A cook balancing two ingredients needs a mix that hits a total amount. A shopper comparing two phone plans wants to know where one plan becomes the cheaper choice. Each of those questions has the same shape: two lines, and the one point where they meet.
Here's where we are in this course. In Week 2 you solved an equation in one unknown and checked it by substitution. In Week 3 you read functions as input-output pairs. In Week 4 you read lines: the slope, the y-intercept, and the rule that parallel lines share a slope. In this lecture we put two lines on one graph and find where they agree. Next week, in Week 6, the course turns to polynomials and factoring.
I'll take eight ideas in order: what a solution of a system is, substitution, elimination, multiplying first, the three kinds of system, the break-even point, systems of inequalities, and the two-part check. Along the way we'll mix trail mix, follow a basketball season, spend a snack budget, and sell T-shirts.
1. See a solution: where two lines cross
Picture two people walking toward each other on a long straight path. One starts at mile marker 4, the other at marker 14. Wherever they meet, they stand at the same marker at the same moment. 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. A solution of a system is an ordered pair that makes every equation in the system true at once; on a graph, it's a point where the lines meet.
Here's our first pair. The first line is y equals 3x plus 4, the Week 4 line. The second is y equals negative 2x plus 14. Picture the graph: the first line climbs from a height of 4 on the y-axis, and the second falls from a height of 14. They cross at the point 2 comma 10.
Now check it in both. In the first equation, 3 times 2 plus 4 is 6 plus 4, which is 10. In the second, negative 2 times 2 plus 14 is negative 4 plus 14, which is 10 again. The pair works in both, so it's the solution.
You can find that point without a graph, too. Both equations say what y equals, so set the right sides equal. Then 3x plus 4 equals negative 2x plus 14. Add 2x to both sides, subtract 4, and you get 5x equals 10, so x is 2.
Pause and try this. The pair 1 comma 7 sits on the line y equals 3x plus 4. Is it a solution of the system?
Check the second equation before you decide.
Here's the answer. No: in the second equation, negative 2 times 1 plus 14 is 12, not 7, so the pair sits on only one line, and a solution has to sit on both.
2. Solve by substitution: replace one variable
Let's make trail mix. You want twice as many cups of oats as cups of nuts, and 9 cups in all. Call the cups of nuts x and the cups of oats y. Twice as many oats says y equals 2x. Nine cups in all says x plus y equals 9.
The first equation already tells you what y is. So wherever y shows up in the second equation, you may write 2x instead. That swap has a name. 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.
Watch it work. Replace y with 2x, and x plus y equals 9 becomes x plus 2x equals 9. That's 3x equals 9, so x is 3. Then y is 2 times 3, which is 6.
Here's the part I want you to notice. After the swap, the problem was an ordinary Week 2 equation in one unknown. Substitution doesn't invent new algebra. It turns two unknowns into one.
One trap: substitute into the other equation, not the one you started from. Putting 2x back into y equals 2x gives 2x equals 2x, which is true for every number and tells you nothing.
Pause and try this. Check the trail-mix pair, 3 comma 6, in both equations. Does it pass?
Run both checks before you read on.
Here's the answer. Yes: 6 equals 2 times 3, and 3 plus 6 is 9, so the mix is 3 cups of nuts and 6 cups of oats.
3. Solve by elimination: add to cancel a variable
A basketball team played 10 games and won 4 more than it lost. Call the wins x and the losses y. Ten games says x plus y equals 10. Four more wins than losses says x minus y equals 4.
Neither equation says what y equals. But look at the y terms: plus y in one equation, minus y in the other. They're opposites. Add the two equations, and the y's cancel. That's the 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.
Why is adding allowed? Each equation says two amounts are equal. Add equal amounts to equal amounts, and the totals stay equal. It's the Week 2 rule of doing the same thing to both sides.
So add. The left sides give 2x, because the y's cancel. The right sides give 10 plus 4, which is 14. Now 2x equals 14, so x is 7.
Then comes back-substitution, the step after you find one variable, in which you put its value into either original equation to find the other variable. Put 7 into x plus y equals 10, and 7 plus y equals 10 gives y equals 3.
Pause and try this. Someone wants to eliminate by adding x plus y equals 3 and x plus y equals 7. Will adding cancel a variable?
Look at the signs on the matching terms.
Here's the answer. No: adding gives 2x plus 2y equals 10, and nothing cancels, because terms with the same sign cancel when you subtract, not when you add.
4. Multiply first, then eliminate
Remember last week's snack budget: snacks at 2 dollars, drinks at 3 dollars, and 12 dollars to spend, which gave 2x plus 3y equals 12. Now add a second rule: buy one more snack than drinks. That's x minus y equals 1.
Look at the y terms: 3y and negative y. They aren't opposites yet. But multiply every term of x minus y equals 1 by 3, and you get 3x minus 3y equals 3. Now 3y and negative 3y are opposites.
Add the equations. The x terms give 5x, the y's cancel, and the right sides give 12 plus 3, which is 15. So 5x equals 15, and x is 3. Back-substitute into x minus y equals 1: 3 minus y equals 1, so y is 2. The group buys 3 snacks and 2 drinks.
Multiplying an equation doesn't move its line. The equation 3x minus 3y equals 3 says exactly what x minus y equals 1 says, only with every number three times as large.
The classic slip here is multiplying only the left side and writing 3x minus 3y equals 1. The 3 has to reach every term, including the number on the right.
Pause and try this. Check the snack-budget pair, 3 comma 2, in the budget equation, 2x plus 3y equals 12.
Do the two multiplications first.
Here's the answer. It checks: 2 times 3 is 6, and 3 times 2 is 6, and 6 plus 6 is 12, which spends the whole budget.
5. Classify a system: one, none, or infinitely many
Two lines in a plane can meet in only three ways. They cross once. They run side by side forever, like the two rails of a train track. Or they're the very same line, drawn twice.
Here are the names. A consistent system is a system that has at least one solution. An inconsistent system is a system with no solution; its lines are parallel, and solving it collapses to a false statement. An independent system is a system whose two equations describe different lines, and a dependent system is a system whose two equations describe the same line, so every point on that line is a solution.
Try x plus y equals 3 with x plus y equals 7. Subtract, and both variables vanish, leaving 0 equals 4. That's never true, so there's no solution. As lines, both have slope negative 1, with y-intercepts 3 and 7. Equal slopes and different intercepts make parallel lines.
Now try 2x plus y equals 5 with 4x plus 2y equals 10. Double the first equation and you get the second one exactly. Subtract, and you're left with 0 equals 0. That's always true, so every point on the line y equals negative 2x plus 5 is a solution, points like 0 comma 5, 1 comma 3, and 2 comma 1.
Students often swap these two. Here's the cure: when the variables vanish, read what's left. A true statement means every point works. A false statement means none does.
Pause and try this. You solve a system and the variables vanish, leaving 0 equals 0. How many solutions does the system have?
Read the leftover statement before you answer.
Here's the answer. Infinitely many: 0 equals 0 is always true, so the two equations are the same line, and the system is dependent.
6. Model a break-even point: where cost meets revenue
Your club orders T-shirts to sell at a fundraiser. The printer charges 60 dollars to set up and 5 dollars per shirt, so the cost is 60 plus 5x. Each shirt sells for 15 dollars, so the revenue is 15x.
Sell only a few, and you're behind. Sell a lot, and you're ahead. 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.
Set revenue equal to cost: 15x equals 60 plus 5x. Subtract 5x from both sides, and 10x equals 60, so x is 6. Check it. The revenue is 15 times 6, which is 90 dollars. The cost is 60 plus 5 times 6, which is 60 plus 30, also 90 dollars.
Read the point as a pair: 6 shirts and 90 dollars. Every shirt after the sixth earns 10 dollars of profit, because it sells for 15 and costs 5.
Here's an honest limit on a model like this. It assumes every shirt costs the same and every shirt sells. Real life wobbles, so treat the answer as a clear first estimate.
Pause and try this. The break-even point is 6 comma 90. What does each of those two numbers mean for the club?
Say each number with its unit.
Here's the answer. The 6 is the number of shirts the club must sell, and the 90 is the dollars that come in and go out at that point, so both parts carry meaning.
7. Shade a system of linear inequalities
One equation, like y equals 4, is a line. One inequality, like y is less than 4, is a whole region: every point below that line. Ask for two inequalities at once, and you want the place where the two regions overlap.
A linear inequality in two variables is an inequality in x and y whose solutions are all the ordered pairs on one side of a line, sometimes including the line itself. That side is a half-plane, all the points on one side of a line. The line itself is the boundary line. It's drawn dashed when the inequality is strict, less than or greater than, and solid when it also allows equality, at most or at least. To pick a side, use a test point, a point not on the boundary that you substitute in. Put two or more of these together and you have a system of linear inequalities, whose solutions sit in the overlap.
Take y is greater than x, with y less than 4. Both signs are strict, so both boundaries are dashed: the slanted line y equals x and the flat line y equals 4. Shade above the slanted line and below the flat one. The solutions live in the strip between them.
Test the point 1 comma 3. Is 3 greater than 1? Yes. Is 3 less than 4? Yes. So it's in the overlap.
Pause and try this. Is the point 2 comma 2 a solution of y greater than x and y less than 4?
Test the first inequality carefully.
Here's the answer. No: 2 is not greater than 2, so the point sits right on the dashed boundary y equals x, and a dashed boundary doesn't count.
8. Check every pair in both equations
The last idea protects all the others. A system has two equations, so your check has two parts. A pair that passes one equation and fails the other isn't a solution.
Here's a slip worth seeing once. Someone solves the trail-mix system and writes the answer as 6 comma 3. The numbers are right, but they're in the wrong order. Check the sum: 6 plus 3 is 9, so that check passes and fools you. Now check the other equation: y should equal 2x, and 2 times 6 is 12, not 3. That check fails.
The same habit protects you when a chatbot solves a system for you. A tidy answer isn't a checked answer. Substitute the pair into both original equations yourself.
Pause and try this. Which of the two trail-mix checks catches the swapped pair 6 comma 3?
Think about which equation cares about order.
Here's the answer. The check in y equals 2x catches it, because the sum 6 plus 3 equals 3 plus 6, so only the doubling rule can tell the order apart.
Closing
Let me pull it together. A solution of a system is one ordered pair that makes both equations true, the point where two lines cross. Substitution swaps a variable for an equal expression from the other equation. Elimination adds or subtracts the equations so one variable cancels. When nothing cancels yet, multiply one equation first, and reach every term. When the variables vanish, read what's left: 0 equals 4 means no solution, and 0 equals 0 means infinitely many. A break-even point is where cost meets revenue. For inequalities, strict signs get dashed boundaries, and a test point picks the side. And every pair gets checked in both equations.
Next in this module come the slide deck, the Readings and Resources page, the lecture tutorial, the practice set, the quiz, and the discussion. Next week, in Polynomials and Factoring, the distributive rule comes back as a method for multiplying and factoring.
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