Demonstration 1 of 4
Test a candidate
For 3x + 5 = 20, which inputs make the promise true?
The top bar is the left side, built from 3x and 5. The bottom bar is 20. The equation is true only where the two bars end at the same place.
\[3x + 5 = 20\]
x may be any value on the number line. The left side 3x + 5 changes with x; the right side stays 20. A solution is a value that makes both sides match.
Predict first. Before switching to x = 7, guess: will the left side overshoot 20, and by how much?
Choose an example
Constructed example: the chapter's equation 3x + 5 = 20 and its candidates 2, 5 and 7.
Calculated values
- Candidate
- x = 5
- Left side 3x + 5
- 3(5) + 5 = 20
- Right side
- 20
- Equation true?
- yes
Substitute x = 5: 3(5) + 5 = 15 + 5 = 20, against 20 on the right. Both sides have value 20, so 5 belongs in the solution set.
Use the idea
When someone hands you an answer to an equation, substitute it and evaluate each side separately. A yes or no is the whole verdict.
Where the conclusion applies
Three candidates cannot show that no other solution exists. Solving with equal changes to both sides does that work, and it gives the single solution 5.
Check your understanding: What does the left side give at x = 6, and is the equation true there?
Chapter 14 source: section "A statement with a condition". Demonstration C14-D01.
Demonstration 2 of 4
Keep the pans balanced
Why must you do the same thing to both sides?
Subtracting 5 from both pans keeps them equal, leaving 3x = 15. Splitting both into 3 equal groups leaves x = 5. Changing one pan alone writes a new equation with a different answer.
\[3x + 5 - 5 = 20 - 5\]
\[(3x) / 3 = 15 / 3\]
Each tall block is one x. Each small square is one unit. The equals sign between the pans says their totals match.
Predict first. Remove the 5 units from the left pan only. Will the last line still say x = 5?
Choose an example
Constructed example: the chapter's balance for 3x + 5 = 20, plus a one-sided mistake.
Calculated values
- Current line
- x = 5
- Left pan
- 1 x
- Right pan
- 5 units
- Solution of this line
- 5
- Same solution as 3x + 5 = 20?
- yes
Split both pans into 3 equal groups: (3x) / 3 = 15 / 3, so x = 5. Check in the original: 3(5) + 5 = 15 + 5 = 20, and the right side is 20.
Use the idea
Write each operation on both sides before you shorten the line. Nothing jumps across the equals sign by itself.
Where the conclusion applies
The divisor 3 is not zero, so every step can be undone. A step on one side only is not reversible in this way, and it loses the original solution.
Check your understanding: Starting from 3x + 5 = 20, what happens if you divide only the left side by 3 at the start?
Chapter 14 source: section "Equal treatment preserves equality". Demonstration C14-D02.
Demonstration 3 of 4
Unknowns on both sides
In 5x - 7 = 2x + 8, why is it fair to subtract 2x before you know x?
Subtracting 2x removes the same amount from both bars, whatever x is. Both bars drop, but the gap between them stays put, so the solution cannot change.
\[5x-7=2x+8\]
\[3x-7=8\]
x is the unknown. Each bar is one side of the equation evaluated at the chosen x. Left minus right is the gap between them.
Predict first. At x = 3, the original sides give 8 and 14. After subtracting 2x from both sides, will the gap still be -6?
Choose an example
Constructed example: the chapter's equation 5x - 7 = 2x + 8 at chosen inputs.
Calculated values
- x
- 5
- Left side 5x - 7
- 18
- Right side 2x + 8
- 18
- Left minus right
- 0
- Equation true?
- yes
At x = 5: 5(5) - 7 = 25 - 7 = 18 and 2(5) + 8 = 10 + 8 = 18. Left minus right is 18 - 18 = 0. Subtracting 2x took the same 10 from both sides, so the gap is the same in both forms: it is 3(5) - 15 = 0. The gap is zero only at x = 5.
Use the idea
Gather the unknown on one side by subtracting the same variable term from both sides, then finish with the moves from 3x + 5 = 20.
Where the conclusion applies
x may be any number. The gap is 3x - 15 in both forms, which is zero only at x = 5. Subtracting 2x is reversible by adding 2x back.
Check your understanding: At x = 5, what are the two sides of the original equation?
Chapter 14 source: section "Worked example: variables on both sides". Demonstration C14-D03.
Demonstration 4 of 4
The one reversal
Solving -2x + 3 ≤ 11, why does the sign flip when you divide by -2?
Multiplying or dividing by a negative number reflects every point across zero, so left and right trade places. The comparison sign must turn around to describe the new order.
\[-2x+3\leq11\]
\[x\geq-4\]
≤ means less than or equal to, and ≥ means greater than or equal to. The shaded part of the line is the claimed solution set; the marker is one value to test.
Predict first. Keep the sign instead of reversing it, then test x = 0. Does the shaded answer agree with the original inequality?
Choose an example
Constructed example: the chapter's inequality -2x + 3 ≤ 11 and its test values.
Calculated values
- Rule used
- divide by -2 and reverse
- Shaded answer
- x ≥ -4
- Original at the test value
- -2(-4) + 3 = 11
- Is 11 ≤ 11?
- yes
- Shaded answer agrees?
- yes
Subtract 3 from both sides: -2x ≤ 8. Dividing by -2 gives x ≥ -4 because a negative divisor reverses the order. Test x = -4 in the original: -2(-4) + 3 = 8 + 3 = 11, and 11 ≤ 11 is true. The shaded answer includes -4, which agrees.
Use the idea
After solving an inequality, test the boundary and one value on each side in the original. Any disagreement exposes a missed reversal.
Where the conclusion applies
x may be any real number. Subtracting 3 does not reverse anything; only the division by -2 does. Testing points checks the answer but is not the reason it works.
Check your understanding: Test x = -5 in -2x + 3 ≤ 11. Should it be in the solution set?
Chapter 14 source: section "Inequalities preserve order, with one reversal". Demonstration C14-D04.