Math 8 · Linear equations and graphs
Undo it in reverse, then draw the pattern
Grade 7 solved equations like 3x + 5 = 26. Grade 8 adds equations with a division, like x/5 + 8 = 3, and with brackets, like 3(x + 2) = 21, and lets every number in them be an integer. Then a rule like t = 5 − 3h becomes a table of values and a straight-line graph, which answers questions the table never listed.
- 1. The words, first
- 2. Solving linear equations
- 3. Writing equations for problems
- 4. Graphing linear relations
- 5. What costs marks
The words, first
The idea: The distributive property is the one new tool. It is what lets you open a bracket.
| Word | What it means |
|---|---|
| Equation, solution | An equation says two expressions are equal, like 4x = −28. Its solution is the value of the unknown that makes it true: x = −7. |
| Preservation of equality | Doing the same to both sides keeps an equation true, as long as you never divide by 0. |
| Distributive property | A number outside a bracket multiplies every term inside it: 3(x + 2) = 3x + 6. |
| Verify | Substitute the solution into the original equation and check that both sides come out equal. |
| Linear relation | A rule whose output changes by the same amount each time the input goes up by 1. Its graph is a straight line. |
| Table of values, ordered pair | A table of inputs and their outputs. Each pair is a point to plot: (2, −1) means 2 across, then 1 down. |
Solving linear equations
The idea: Undo what was done to the variable, last step first, doing the same to both sides. A bracket can be undone by dividing first, or multiplied out — both work.
| Equation | Undo, on both sides | Solution | Verify |
|---|---|---|---|
| 4x = −28 | divide by 4 | x = −7 | 4 × (−7) = −28 ✓ |
| x/(−3) = 5 | multiply by −3 | x = −15 | (−15) ÷ (−3) = 5 ✓ |
| −2x + 7 = 19 | subtract 7, then divide by −2 | x = −6 | (−2) × (−6) + 7 = 12 + 7 = 19 ✓ |
| x/4 − 3 = 2 | add 3, then multiply by 4 | x = 20 | 20 ÷ 4 − 3 = 5 − 3 = 2 ✓ |
| 3(x + 2) = 21 | divide by 3, then subtract 2 | x = 5 | 3 × (5 + 2) = 3 × 7 = 21 ✓ |
Worked example: −3(x − 4) = 18.
- Undo the multiplying by −3: divide both sides by −3, so x − 4 = −6.
- Undo the subtracting: add 4 to both sides, so x = −2.
- Verify: −3 × (−2 − 4) = −3 × (−6) = 18 ✓.
Or multiply out first. −3 times each term in the bracket gives −3x + 12, because −3 × (−4) = 12. Then −3x + 12 = 18, so −3x = 6 and x = −2, the same answer. Writing −3x − 12 by mistake leads to x = −10, and verifying catches it: −3 × (−10 − 4) = −3 × (−14) = 42, not 18.
Worked example: x/5 + 8 = 3. The last thing done to x was adding 8, so undo that first: x/5 = 3 − 8 = −5. Then multiply both sides by 5: x = −25. Verify: −25 ÷ 5 + 8 = −5 + 8 = 3 ✓. Multiplying only the x/5 by 5 gives x + 8 = 15 and x = 7, which fails the check.
Writing equations for problems
The idea: Say what the variable stands for, write one equation that says what the story says, solve it, and check the answer in the story rather than in your equation.
Worked example: brackets. Four friends each buy a movie ticket and a $3 popcorn. Together they spend $64. How much is one ticket?
- Name the variable: let t be the price of one ticket. Each friend spends t + 3.
- Write the equation: 4(t + 3) = 64.
- Solve: divide both sides by 4, so t + 3 = 16. Subtract 3: t = 13.
- Check in the story: a $13 ticket and a $3 popcorn is $16 each, and four people spend $64 ✓. A ticket costs $13.
4t + 3 = 64 is the tempting wrong equation. It pays for one popcorn instead of four.
With a division. 24 students share the cost of a bus equally, and each also pays $5 admission: $12 each in all. With b for the bus, b/24 + 5 = 12, so b/24 = 7 and the bus cost $168. Check: 168 ÷ 24 = 7, and 7 + 5 = 12 ✓.
With integers. Overnight the temperature fell steadily for 4 hours, from 3 °C to −13 °C. With c for the change each hour, 3 + 4c = −13, so 4c = −16 and c = −4: a drop of 4 degrees an hour. Check: 3 + 4 × (−4) = 3 − 16 = −13 ✓.
Graphing linear relations
The idea: Make a table of values from the rule, plot each pair as a point, and check that the points line up. Describe the pattern in words: where it starts, and how much it changes at each step.
Worked example. At 6 p.m. it is 5 °C, and the temperature falls 3 degrees every hour. After h hours it is t = 5 − 3h degrees.
| Hours after 6 p.m., h | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|
| Temperature, t (°C) | 5 | 2 | −1 | −4 | −7 |
- Plot it: hours across, temperature up and down. The points are (0, 5), (1, 2), (2, −1), (3, −4) and (4, −7); the last three are below the horizontal axis.
- Describe it: it starts at 5 and goes down 3 every hour. The change is the same at every step, so the points lie on a straight line, and since time runs on between the hours, they are joined.
- Between the points: at 7:30 p.m., h = 1.5, so t = 5 − 3 × 1.5 = 5 − 4.5 = 0.5 °C.
- Beyond the table: when will it be −13 °C? Solve 5 − 3h = −13: subtract 5 to get −3h = −18, then divide by −3, so h = 6, which is midnight. That assumes it keeps falling at the same rate, which real nights rarely do for long.
Missing values. For (4, ?), substitute: 5 − 3 × 4 = −7. For (?, −10), solve 5 − 3h = −10: h = 5.
What costs marks
The idea: Verifying in the original equation catches the first three.
- Undoing in the wrong order. In x/5 + 8 = 3, subtract the 8 before multiplying by 5.
- Multiplying only part of a bracket. −3(x − 4) is −3x + 12, not −3x − 4 or −3x − 12.
- Dropping the sign when dividing by a negative. −3x = 6 gives x = −2, not 2.
- Checking in your own equation. If the equation is wrong, the check agrees with it. Check in the story.
- Reading far beyond the table as if the pattern must go on forever.