Lecture: Linear Equations & Inequalities
Lecture script · about 15–20 minutes spoken · Week 2
Opening
Picture a phone plan that costs 20 dollars a month, plus 5 dollars for every gigabyte you use. Your budget is 45 dollars. How much data can you use? You could guess and check. You could try 3 gigabytes, then 4, then 5, and watch the bill climb. Or you could write one line of algebra and get every answer at once. That second way is what this lecture is about.
This kind of question shows up everywhere. A nursing student checks that a dose stays inside a safe range. A small-business owner works out how many items to sell to stay under a budget. A commuter decides which apartments sit within a few miles of campus. Each one is hunting for the numbers that make a statement true.
In Week 1, you learned the rules that never change: the order of operations, the properties of real numbers, and how to distribute and combine like terms. In this lecture we put those rules to work. Every solve starts by tidying each side, and tidying is exactly what Week 1 taught. Next week, in Week 3, Functions: Notation, Domain & Range, you'll use these same moves to decide which inputs a function allows.
We'll take five ideas in order. First, solving a linear equation. Second, the two equations that refuse to end in one number. Third, inequalities and a compact way to write a range of answers. Fourth, absolute-value equations. Fifth, absolute-value inequalities. Every answer we find, we'll check by putting it back in.
1. Solve a linear equation: one principle, four tactics
Think of an old two-pan kitchen scale with a bag of flour on each side, sitting perfectly level. If you add a cup to the left pan only, it tips. If you add a cup to both pans, it stays level. Solving an equation works on that one principle. You may do almost anything to an equation, as long as you do the same thing to both sides.
A linear equation is an equation in one variable that can be rewritten in the form a x plus b equals zero, with a not equal to zero. In plain words, the variable appears only to the first power and is never inside a denominator, a root, or an absolute value. A solution is a value of the variable that makes the statement true. For an equation, it's the number that makes the two sides equal.
Here are the four tactics, in the order you usually need them. Distribute to clear the parentheses. Clear any fractions. Collect the x terms on one side and the plain numbers on the other. Then divide by the number in front of x.
Let's run one. Take 3 times the quantity x minus 2, set equal to 2x plus 5. First, distribute. The 3 multiplies both terms inside, so 3 times x is 3x, and 3 times negative 2 is negative 6. Now the left side reads 3x minus 6. Next, collect. Subtract 2x from both sides, and you're left with x minus 6 on the left and 5 on the right. Add 6 to both sides, and x equals 11.
We're not done until we check. A check by substitution means you replace the variable in the original equation with your answer and evaluate each side. On the left, 11 minus 2 is 9, and 3 times 9 is 27. On the right, 2 times 11 is 22, and 22 plus 5 is 27. Both sides are 27, so 11 is the solution.
Fractions are the tactic people dread, so here's the shortcut. The least common denominator, or LCD, is the smallest number that every denominator divides into evenly. Multiply every term by it, and the fractions vanish in one step. Take x over 2, plus 1 third, equals 5 sixths. The denominators are 2, 3, and 6, and the LCD is 6. Six times x over 2 is 3x. Six times 1 third is 2. Six times 5 sixths is 5. Now you have 3x plus 2 equals 5, with no fractions anywhere.
Pause and try this. Finish that equation. Starting from 3x plus 2 equals 5, what is x, and how do you check it?
Take a moment and work it through.
Here's the answer. Subtract 2 from both sides, and 5 minus 2 leaves 3x equals 3. Divide both sides by 3, and x is 1. To check, go back to the original fractions: 1 half plus 1 third is 3 sixths plus 2 sixths, which makes 5 sixths, matching the right side.
One warning carries through this whole course. The number in front of the parentheses has to reach every term inside. If you write 3 times the quantity x minus 2 as 3x minus 2, the whole solve goes wrong from the first line.
2. Spot an identity or a contradiction
Most equations end in one number. Two kinds don't, and both give themselves away the same way: the x terms disappear.
Here's a picture for the first kind. Two friends describe the same lunch order in different words. Whatever the price turns out to be, their two descriptions always agree. That's an identity: an equation that is true for every real number. Solving it collapses to a true statement such as 0 equals 0, so its solution set is all real numbers.
Try 2 times the quantity x plus 3, set equal to 2x plus 6. Distribute on the left. 2 times x is 2x, and 2 times 3 is 6, so the left side is 2x plus 6. Now both sides say 2x plus 6. Subtract 2x plus 6 from both sides, and you get 0 equals 0. The x is gone, and what's left is always true. Any number you pick will balance it.
Now the second kind. Picture two receipts for the same order that print different totals. No price can make them agree. That's a contradiction: an equation that is true for no value of the variable. Solving it collapses to a false statement such as 3 equals 7, so it has no solution.
Try 2 times the quantity x plus 3, minus 3, set equal to 2x plus 7. Distribute, and the left side is 2x plus 6 minus 3, which is 2x plus 3. Now you have 2x plus 3 equals 2x plus 7. Subtract 2x from both sides, and you get 3 equals 7. The x is gone again, but this time what's left is never true.
Pause and try this. A classmate solves the identity, reaches 0 equals 0, and writes "x equals 0." What would you tell them?
Think about what x equals 0 actually claims.
Here's the answer. Writing x equals 0 claims that exactly one number, zero, works. But 0 equals 0 is true no matter what x was, so every real number works. The right way to write it is "all real numbers." And for the contradiction, where you land on 3 equals 7, you write "no solution," never x equals 0.
So when x vanishes, stop solving and read what's left. True means all real numbers. False means no solution.
3. Solve a linear inequality and write it in interval notation
Back to the phone plan. The bill is 20 dollars plus 5 dollars per gigabyte, and you can spend at most 45 dollars. "At most" doesn't mean exactly. Spending 30 is fine. Spending 45 is fine. Going over is not. So you're hunting for a whole range of amounts, and you need a tool built for ranges.
A linear inequality is a statement that compares two linear expressions with less than, greater than, less than or equal to, or greater than or equal to. Its solution is usually a whole range of numbers rather than a single value.
Let g stand for gigabytes. The bill is 20 plus 5g, and it must be at most 45. Subtract 20 from both sides, and 5g is at most 25. Divide by 5, and g is at most 5. You can use up to 5 gigabytes. Check the edge: 5 gigabytes costs 20 plus 25, which is exactly 45 dollars, and that's allowed.
So far an inequality behaves like an equation. There's one new rule, and here's why it exists. Start with a true statement: 2 is less than 5. Multiply both sides by negative 1, and you get negative 2 and negative 5. On a number line, negative 2 sits to the right of negative 5, so now negative 2 is the larger one. The order flipped. That's the flip rule: when both sides of an inequality are multiplied or divided by a negative number, the direction of the inequality sign reverses.
Let's use it. Solve negative 2x plus 1 is less than 9. Subtract 1 from both sides, and negative 2x is less than 8. Now divide both sides by negative 2. You're dividing by a negative, so the sign flips. 8 divided by negative 2 is negative 4, and x is greater than negative 4.
Check it. Try x equals 0, which should work. Negative 2 times 0, plus 1, is 1, and 1 is less than 9. Good. Now try the edge, x equals negative 4. Negative 2 times negative 4 is 8, plus 1 is 9, and 9 is not less than 9. So negative 4 itself is left out.
Now we need a compact way to write "every number greater than negative 4." Interval notation writes a set of real numbers by its two endpoints. A square bracket means the endpoint is included, and a parenthesis means it's excluded or infinite. Infinity marks an interval that keeps going without end; it isn't a number, so it always gets a parenthesis. So our answer is written: parenthesis, negative 4, comma, infinity, parenthesis.
Pause and try this. Solve negative 3x is greater than 9. Which way does the sign point when you're done?
Circle the number you're dividing by before you answer.
Here's the answer. You divide both sides by negative 3, and 9 divided by negative 3 is negative 3. Because you divided by a negative, the sign flips, so x is less than negative 3. Test negative 4: negative 3 times negative 4 is 12, and 12 is greater than 9, so negative 4 belongs, and it is indeed less than negative 3.
4. Solve an absolute-value equation with the split rule
Stand at a bus stop and walk 7 steps. You might end up 7 steps east, or 7 steps west. Either way, you're 7 steps away. The direction is lost, and only the distance is kept. That's what absolute value does.
Absolute value is the distance of a number from 0 on the number line. A distance is never negative, so the absolute value of 7 is 7, and the absolute value of negative 7 is also 7. We write it with a straight bar on each side of the number.
That gives us the split rule. For a positive number a, the absolute value of X equals a splits into two equations: X equals a, or X equals negative a. If a is negative, there's no solution at all.
Let's split one. Solve: the absolute value of 2x minus 1 equals 7. The inside, 2x minus 1, is 7 steps from zero, so it's either 7 or negative 7. Take the first case. 2x minus 1 equals 7, so 2x equals 8, and x is 4. Take the second case. 2x minus 1 equals negative 7, so 2x equals negative 6, and x is negative 3.
Check both. For x equals 4, 2 times 4 minus 1 is 7, and the absolute value of 7 is 7. For x equals negative 3, 2 times negative 3 minus 1 is negative 7, and the absolute value of negative 7 is 7. Both work. Each one puts the inside exactly 7 steps from zero, one on each side.
Pause and try this. Someone solves the absolute value of x equals negative 5 and writes "x is 5 or negative 5." Is that right?
Look at the right side before you think about splitting.
Here's the answer. It isn't right, and you can see why by checking. The absolute value of 5 is 5, not negative 5, and the absolute value of negative 5 is also 5. Neither candidate works. A distance is never negative, so this equation has no solution, and there's nothing to split.
5. Solve an absolute-value inequality: the middle or the two rays
Say you want an apartment within 2 miles of campus, and campus sits at mile marker 10 on a long straight road. Walk 2 miles one way, and you reach marker 8. Walk 2 miles the other way, and you reach marker 12. Every marker in between works. "Within" a distance always gives you one stretch in the middle.
Now flip the question. A greenhouse alarm sounds when the temperature is more than 4 degrees Celsius away from 20. That happens when the room is too cold, below 16, or too hot, above 24. "More than" a distance always gives you two pieces, one off each end.
An absolute-value inequality is an inequality with an absolute value on one side. For a positive a, the absolute value of X less than a is the bounded middle: X sits between negative a and a. The absolute value of X greater than a is the two outward rays: X is below negative a, or above a. The middle needs both conditions at once, so it's an AND. The two rays need only one condition or the other, so they're an OR. To write two rays as one answer, we use a union, a symbol that joins two sets into one and is read as "or."
Try the absolute value of x less than 5. That's every number within 5 of zero, so x sits between negative 5 and 5, with both ends left out because the sign is strict. Check zero: its absolute value is 0, and 0 is less than 5.
Now try the absolute value of x greater than or equal to 3. That's every number at least 3 away from zero. So x is at most negative 3, or x is at least 3. In interval notation, that's negative infinity up to negative 3 with a bracket, joined by a union to 3 up to infinity, again with a bracket on the 3. Check 4: its absolute value is 4, and 4 is at least 3, so it's in. Check zero: 0 is not at least 3, so zero is out.
The campus question works the same way. The distance from marker m to marker 10 must be at most 2. So m minus 10 sits between negative 2 and 2. Add 10 to every part, and m runs from 8 to 12, both ends included.
Pause and try this. A classmate writes the absolute value of x greater than 3 as "x is between negative 3 and 3." Test one number to show why that can't be right.
Say the inequality as a distance out loud before you test.
Here's the answer. Test zero. Its absolute value is 0, and 0 is not greater than 3, yet zero sits between negative 3 and 3. The middle describes numbers close to zero, but "greater than" asks for numbers far from zero, so the answer has to be the two outward rays.
Here's a sentence worth carrying: less than is the middle, and greater than is the two rays.
Closing
Let me pull it together. To solve a linear equation, do the same thing to both sides: distribute, clear fractions with the LCD, collect, divide, and always check by substitution. When the x terms vanish, read what's left. A true statement means all real numbers, and a false one means no solution.
A linear inequality solves the same way, with one new rule: flip the sign when you multiply or divide by a negative. Write the range in interval notation, with brackets for included endpoints and parentheses for excluded ones, and always a parenthesis beside infinity.
Absolute value is distance from zero. An absolute-value equation splits into two cases, unless the right side is negative, and then there's no solution. An absolute-value inequality is either the middle, an AND, or the two rays, an OR.
Next in this module are the slide deck, the Readings & Resources page, and the lecture tutorial, where you'll practice these moves with a chatbot and check its work. Next week, in Week 3, Functions: Notation, Domain & Range, these solving moves meet function notation, the rule that names one output for each input.
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