Solving equations: balance and graph

Picture a market stall with an old-fashioned scale: a beam with a pan on each end. On the stall are bags that all weigh the same, but nobody knows how much. Call that weight x.

On the left pan are 4 bags. A balloon is tied to that pan too, and it pulls up by 5. On the right pan are 2 bags and a weight of 9. The scale hangs perfectly level. In maths you write that like this:

4x − 5 = 2x + 9

The = sign is the scale in balance. Solving the equation means finding out how much one bag weighs.

The balance method

The rule is short: whatever you do to one side, you do to the other side. Then the scale stays level. Take something off the left and not the right, and it tips.

Try it below. At each step, pick a button. One button keeps the scale level, the other does not.

Solve 4x − 5 = 2x + 9 on the scale

1. 4x − 5 = 2x + 9

Pick your next step

Move 2x to the left as +2x−2x on both sides

On paper it looks like this. After the bar is what you do to both sides.

4x − 5 = 2x + 9 | −2x
2x − 5 = 9 | +5
2x = 14 | ÷2
x = 7

Always check your answer. Put x = 7 into both sides:

4 × 7 − 5 = 28 − 5 = 23
2 × 7 + 9 = 14 + 9 = 23

Both sides are 23. The scale balances, so x = 7 is right. Each bag weighs 7.

Brackets and fractions

If the equation has brackets, get rid of them first. The number in front of the brackets multiplies every term inside. Watch the sign: in 3 − 2(x + 2), the minus belongs to the 2.

Expanding the brackets

• −2 × x = −2x
• −2 × 2 = −4

Now the whole equation, step by step:

3 − 2(x + 2) = 7
3 − 2x − 4 = 7
−2x − 1 = 7 | +1
−2x = 8 | ÷(−2)
x = −4

Check: 3 − 2 × (−4 + 2) = 3 − 2 × (−2) = 3 + 4 = 7. Correct.

Clear fractions first. Multiply both sides by the lowest common denominator. Take this equation:

x + 12 − x3 = 2

The denominators are 2 and 3. The smallest number both go into is 6. Multiply every term by 6:

6 × x + 12 = 3(x + 1)
6 × x3 = 2x
6 × 2 = 12

Keep the brackets around x + 1, because the whole top is multiplied by 3. After that it is an ordinary equation:

3(x + 1) − 2x = 12
3x + 3 − 2x = 12
x + 3 = 12 | −3
x = 9

Check: (9 + 1) ÷ 2 − 9 ÷ 3 = 5 − 3 = 2. Correct.

Two lines that cross

You can also draw each side of an equation as a line. For 4x − 5 = 2x + 9, those are y = 4x − 5 and y = 2x + 9. Where the lines cross, they have the same y. So that is where both sides are equal. The x of the crossing point is the solution.

Two lines, one equation

Line 1: y = ax + b

a = 4b = −5

Line 2: y = cx + d

c = 2d = 9

• Line 1: y = 4x − 5
• Line 2: y = 2x + 9

Equation

4x − 5 = 2x + 9

Solution

x = 7

Crossing point

(7, 23)

Show the numbers

x	Line 1	Line 2
−10	−45	−11
−8	−37	−7
−6	−29	−3
−4	−21	1
−2	−13	5
0	−5	9
2	3	13
4	11	17
6	19	21
8	27	25
10	35	29

The crossing point is (7, 23). So x = 7, and both sides are then 23. Those are the same numbers as in the check.

Now slide until a = c. The lines are then equally steep. Two things can happen:

• If b is not equal to d, the lines are parallel. They never cross, so there is no solution.
• If b is also equal to d, the lines lie on top of each other. Then every x is a solution.

This answers a classic test question: for which a and b does 5x + 6 = ax + b have no solution? Move everything with x to the left and the rest to the right:

5x − ax = b − 6
(5 − a)x = b − 6

• If a is not 5, divide by 5 − a. There is exactly one solution.
• If a = 5, you are left with 0 = b − 6. If b is not 6, that is never true: no solution.
• If a = 5 and b = 6, you are left with 0 = 0. That is true for every x.

The answer: a = 5, and b is any number except 6. To see it in the graph, set line 1 to a = 5 and b = 6. Set line 2 to c = 5 and slide d.

Equations with x²

If there is an x², you have a quadratic equation. It often has two solutions. Start with the simplest kind:

x² = 8

You are looking for a number that gives 8 when multiplied by itself. That is √8, but also −√8, because minus times minus is plus.

x = −√8 or x = √8

You can simplify √8: √8 = √(4 × 2) = 2√2 ≈ 2.83. So x = −2√2 or x = 2√2, in short x = ±2√2.

The graph shows why there are two. The line y = 8 crosses y = x² on the left and on the right, the same distance from 0.

x² = k has two, one or no solutions

Slide k: x² = 8

• y = x²
• y = 8

Two solutions

x = −√8 = −2√2 ≈ −2.83 or x = √8 = 2√2 ≈ 2.83

Show the numbers

x	x²
−6	36
−5	25
−4	16
−3	9
−2	4
−1	1
0	0
1	1
2	4
3	9
4	16
5	25
6	36

Slide k below 0. The line then misses the graph completely, because a square is never negative. At k = 0 there is exactly one solution: x = 0.

Factorising

Take x² − 4x − 21 = 0. Look for two numbers that multiply to −21 and add up to −4. Those are −7 and 3, because −7 × 3 = −21 and −7 + 3 = −4. Then:

x² − 4x − 21 = 0
(x − 7)(x + 3) = 0

Now one rule does the work: if a product is 0, one of the factors is 0. Two numbers multiplied together only give 0 when one of them is 0.

x − 7 = 0 or x + 3 = 0
x = 7 or x = −3

Check: 7² − 4 × 7 − 21 = 49 − 28 − 21 = 0 and (−3)² − 4 × (−3) − 21 = 9 + 12 − 21 = 0. Correct.

Make sure one side is 0 first. The rule only works with = 0.

The quadratic formula

Not every equation is easy to factorise. The quadratic formula always works. Write the equation as ax² + bx + c = 0, with a not 0. First work out the discriminant:

D = b² − 4ac

The discriminant decides how many solutions there are:

• D > 0: two solutions.
• D = 0: one solution.
• D < 0: no solution, because you cannot take the square root of a negative number.

If D is not negative, the solutions are:

x = −b − √D2a or x = −b + √D2a

For x² − 4x − 21 = 0, a = 1, b = −4 and c = −21:

D = (−4)² − 4 × 1 × (−21) = 16 + 84 = 100
x = 4 − 102 = −3 or x = 4 + 102 = 7

Those are the same solutions as with factorising. Slide a, b and c below and watch what D does.

The quadratic formula

a = 1b = −4c = −21

x² − 4x − 21 = 0

D = b² − 4ac

(−4)² − 4 × 1 × (−21) = 16 + 84 = 100

D > 0: two solutions

x = −3 or x = 7

Try it yourself