Direct · inverse · finding k.
The problem
y varies inversely as x. If y = 4 when x = 3, what is the value of y when x = 6?
Read the first three words carefully.
Inverse variation holds the product constant: k = 4 × 3 = 12. So at x = 6, y = 12 ÷ 6 = 2. D doubles y as though this were direct variation — doubling x has to halve y.
One word changes the whole formula.
y = kx
x goes up, y goes up too
y = kx
x goes up, y comes down
Our problem says INVERSELY, so k sits on the bottom.
“Varies as” means direct. “Varies inversely as” means the x moves under the fraction bar. Nothing else in the method changes.
y = kx
k = x · y
multiply both sides by x
| Pair | x | y | x · y |
|---|---|---|---|
| given pair | 3 | 4 | 12 |
| new pair | 6 | ? | 12 |
The last column must stay the same all the way down.
Inverse fixes the product — direct fixes y ÷ x.
You are given y = 4 when x = 3, so put those in.
4 = k3
4 · 3 = k
· 3 both sides
12 = k
so k is 12
k is the one thing shared by both pairs. Without it the two pairs are unrelated numbers.
y = 12x
this now works for every pair
y = 126
swap x for the new value 6
x doubled from 3 to 6, so y had to halve from 4 to 2. Inverse means they pull in opposite directions.
Miss the word “inversely” and you write y = kx: k = 4 ÷ 3 = 43, then y = (43)(6) = 8. That is the 8 among the choices — y went up when x went up, which inverse variation never allows.
Because x · y never changes, you can jump straight to it.
x1 · y1 = x2 · y2
3 · 4 = 6 · y
12 = 6y
the left side is just 12
2 = y
÷ 6 both sides
Same answer, no k written down.
Translate the sentence into a formula with k, find k from the first set of values, then use it. Only the formula changes.
| The sentence says | Formula |
|---|---|
| y varies directly as x | y = kx |
| y varies inversely as x | y = kx |
| y varies directly as the square of x | y = kx² |
| y varies inversely as the square of x | y = kx² |
| z varies jointly as x and y | z = kxy |
| y varies directly as x and inversely as z | y = kx/z |
y varies directly as x and inversely as the square of z. If y = 10 when x = 5 and z = 2, find y when x = 4 and z = 4.
y = kxz²
directly → top, inversely → bottom
10 = k(5)2² → 10 = 5k4 → k = 8
first set of values
y = 8(4)4² = 3216 = 2
second set
Dropping the square — writing y = kx/z — gives k = 4 and y = 4, a wrong answer that will be offered. Underline “square” or “cube” in the sentence before writing the formula.
y varies directly as the square of x, and y = 12 when x = 2. Find x when y = 75.
12 = k(2)² → k = 3
y = kx²
75 = 3x² → x² = 25
÷ 3
x = ±5
only 5 if x is a length
Different: the unknown is the squared letter, so the last move is a square root — and both signs fit unless the context rules one out.
y varies directly as the square root of x, and y = 6 when x = 4. Find y when x = 25.
6 = k√4 = 2k → k = 3
y = k√x
y = 3√25 = 15
x grew 6.25 times, y only 2.5 times
Different: “the square root of x” puts √x in the formula, so y grows much more slowly than x.
A spring’s stretch varies directly as the mass hung on it. A 2 kg mass stretches it 4 cm. How far does a 750 g mass stretch it?
4 = k(2) → k = 2 cm per kg
s = km
750 g = 0.75 kg
match k’s units first
s = 2(0.75) = 1.5 cm
less mass, less stretch ✓
Different: k carries units, so the second value has to be in the same units as the first.
Using 750 with a k found in kilograms gives s = 1,500 cm — fifteen metres of stretch for less than a kilogram. Convert before substituting.
y varies directly as the square of x and inversely as z. If x is tripled and z is doubled, y is multiplied by what? Put the multipliers into the formula in place of the letters.
y = kx²z
the formula
x → 3x gives x² → 9x²
a square multiplies by 3² = 9
z → 2z divides by 2
it is on the bottom
y is multiplied by 92 = 4.5
k never needed
Percent changes work the same way. If A = kr² and r grows by 10%, r becomes 1.1r, so A becomes (1.1)² = 1.21 times as large: a 21% increase, not 10%.
Variation items are worded problems — light and distance, travel time and speed, area and radius — or “if x is doubled…” scaling questions. Translate word by word into y = k…, find k from the first set, then substitute the second. For scaling, skip k entirely and multiply the factors.
Same product, so the pairs belong to the same rule.
x doubled · y halved
Find x
y varies inversely as x. If y = 5 when x = 8, for what value of x will y be 10?
k = 5 × 8 = 40, so when y = 10, x = 40 ÷ 10 = 4. A moves the wrong way — doubling y from 5 to 10 must halve x from 8 to 4, because their product cannot change.
Hint: this time the missing letter is x, not y. Find k the same way, then solve for x at the end.
