Every variation problem is the same three steps: write it, find k, use it.
Find the value
y varies inversely as x. When x = 4, y = 15. What is y when x = 10?
Inverse variation means y = kx. From the pair given, k = xy = 4 × 15 = 60. Then y = 6010 = 6. A treats it as direct variation — but inverse means x going up sends y down, and 10 is bigger than 4, so the answer must be smaller than 15. That check alone rules out three options.
| The words | The equation | k is |
|---|---|---|
| y varies directly as x | y = kx | yx |
| y varies inversely as x | y = kx | xy |
| y varies jointly as x and z | y = kxz | yxz |
| y varies directly as x, inversely as z | y = kx/z | yz/x |
| y varies directly as the square of x | y = kx² | yx² |
The word directly puts the variable on top; inversely puts it underneath. Everything else is bookkeeping.
y varies directly as x; y = 12 when x = 3
find y when x = 7
y = kx, so 12 = 3k
k = 4
y = 4x
the full rule
y = 28
4 × 7
Direct variation is a straight line through the origin. Doubling x doubles y, every time — which is what "k is constant" means in practice.
x = 4, y = 15
from the opener
k = xy = 60
the product is the constant
y = 6010 = 6
and 10 × 6 = 60 ✓
Under inverse variation, a bigger x must give a smaller y. If your answer moves the same way as the input, you have used the direct formula by mistake.
w varies jointly as x and y
w = 36 when x = 3, y = 4
36 = k(3)(4) = 12k
k = 3
w = 3xy
now usable
x = 5, y = 2 → w = 30
substitute
"Jointly" means directly with both, so both go on top. Combined variation mixes the two: y = kx/z has x on top and z underneath.
Notice that k is often something with a name — π, an hourly rate, a distance. It is not an arbitrary letter; it is the fixed quantity the situation supplies.
Write the equation from the words first. Find k from the first set of values, then substitute EVERY new value.
y varies directly as x and inversely as z. y = 12 when x = 6 and z = 4. Find y when x = 10 and z = 5.
y = kx/z
x on top, z underneath
12 = k(6)4, so k = 486 = 8
the first set of values
y = 8(10)5 = 16
both new values
y varies inversely as the square of x. y = 5 when x = 4. Find y when x = 2.
y = kx²
the square goes underneath
5 = k16, so k = 80
square 4 before multiplying
y = 804 = 20
x halved, so y × 4 ✓
Using y = kx gives k = 20 and then y = 202 = 10 — only double the old y. With the square, halving x multiplies y by 2² = 4, so the answer is 20.
Three ways the same write-find-use routine gets disguised.
y varies directly as x; y = 10 when x = 4. Find y when x = 6.
k = 104 = 52
k need not be a whole number
y = (52)(6) = 15
check: 156 = 104 ✓
Keep k as a fraction; rounding it to 2 or 3 throws the answer off.
y varies jointly as x and the square of z. x is doubled and z is halved. What happens to y?
y = kxz²
no numbers needed
new y = k(2x)(z2)² = k(2x)(z²4)
put the changes in
new y = (12)kxz²
y is halved
Multiply the factors: × 2 from x, × 14 from the squared z, so × 12 overall.
y varies inversely as x; y = 4 when x = 9. For what x is y = 6?
k = xy = 36
the product is fixed
6x = 36, so x = 6
this time x is the unknown
y went up from 4 to 6, so x had to go down from 9 — the direction check still works.
Expect a sentence of variation words, a first set of values to find k, and a second set to substitute — often combined variation, a square, or a "what happens to y" question with no numbers at all. The fastest route is to translate the words into one equation, find k, then substitute every new value. The 10-second check is direction: under direct variation y moves with x, under inverse it moves the other way, and a square makes the change bigger.
Find the value
y varies directly as the square of x. When x = 2, y = 20. What is y when x = 5?
y = kx², so 20 = k(4) and k = 5. Then y = 5(25) = 125. A forgets the square and uses y = kx with k = 10. Notice x went up by a factor of 2.5, so y goes up by 2.5² = 6.25 — and 20 × 6.25 = 125 ✓
Hint: "as the square of x" puts x² in the equation, not x. Find k using the squared value before substituting the new one.
