Inverse Variation

Inverse variation describes when two quantities always multiply to give a constant value, with one variable increasing as the other decreases.

y=kxy = \frac{k}{x}

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What each symbol means

What Inverse Variation takes
kk
xx
yy
Inverse Variation
SymbolMeaning
kkThe constant of proportionality, which equals the product xyxy no matter what values xx and yy have; it must be non-zero for the inverse relationship to exist.
xxThe independent variable, which can be any real number except zero (since division by zero is undefined).
yyThe dependent variable, the output of the inverse variation formula; it equals zero only if k=0k = 0, which breaks the inverse relationship.

When to use it

Use inverse variation when one variable increases as the other decreases, and their product stays the same.

Level

Usually taught in: Algebra I

Worked examples

1. Find y using a simple inverse variation

Problem

Given y=12xy = \frac{12}{x}, find yy when x=3x = 3.
  1. y=12xy = \frac{12}{x}

    Start with the inverse variation equation y=12xy = \frac{12}{x} where k=12k = 12.

  2. y=123y = \frac{12}{3}

    Substitute x=3x = 3 into the equation.

  3. y=4y = 4

    Divide: 12÷3=412 \div 3 = 4.

Answer: y=4y = 4

When you know the value of the independent variable xx, substitute it directly into y=kxy = \frac{k}{x} and simplify by dividing.

2. Solve for the independent variable using inverse variation

Problem

Given y=20xy = \frac{20}{x}, find xx when y=5y = -5.
  1. 5=20x-5 = \frac{20}{x}

    Substitute y=5y = -5 into the equation y=20xy = \frac{20}{x}.

  2. 5x=20-5x = 20

    Multiply both sides by xx to clear the fraction on the right side.

  3. x=205x = \frac{20}{-5}

    Divide both sides by 5-5 to isolate xx.

  4. x=4x = -4

    Compute: 20÷(5)=420 \div (-5) = -4.

Answer: x=4x = -4

When the variable is in the denominator, multiply both sides by that variable to move it to the numerator, then solve. Negative values can appear in inverse variation, so be careful with signs.

3. Apply inverse variation to a real-world scenario

Problem

A coffee shop discovers that the number of customers each day is inversely proportional to the outside temperature. When it is 40°F40°\text{F}, 80 customers visit. How many customers visit when the temperature is 50°F50°\text{F}?
  1. k=xy=4080=3200k = x \cdot y = 40 \cdot 80 = 3200

    In inverse variation, the product xyxy is constant. Use the first pair of values to find kk.

  2. y=3200xy = \frac{3200}{x}

    Write the inverse variation equation using the constant k=3200k = 3200.

  3. y=320050y = \frac{3200}{50}

    Substitute x=50x = 50 (the new temperature) into the equation.

  4. y=64y = 64

    Divide: 3200÷50=643200 \div 50 = 64 customers.

Answer: y=64y = 64

Word problems require you to first identify the constant kk using a known pair of values, then use that constant to find unknown values. This method works for any pair of inversely related quantities.

Common mistakes

Where Inverse Variation usually goes wrong
Answer came out wrong
Writing y=kxy = k \cdot x instead of y=kxy = \frac{k}{x}.
Remember: inverse means one variable is in the denominator. The formula must be y=kxy = \frac{k}{x} so that as xx grows, yy shrinks.
Forgetting that xx cannot be zero when you try to use x=0x = 0 in a problem.
Always note that x0x \neq 0 (and y0y \neq 0 if k0k \neq 0). If a problem seems to require x=0x = 0, re-read it—the scenario is not actually inverse variation.
Using the wrong operation to find the constant kk, such as computing x+yx + y or xyx - y instead of xyx \cdot y.
Always verify: k=xyk = x \cdot y for every pair of values. If one pair gives k=24k = 24 but another gives k=15k = 15, the relationship is not inverse variation.
The mistakeWhy it is wrongThe fix
Writing y=kxy = k \cdot x instead of y=kxy = \frac{k}{x}.Multiplying by xx creates direct variation, not inverse variation; the variables move in the same direction instead of opposite directions.Remember: inverse means one variable is in the denominator. The formula must be y=kxy = \frac{k}{x} so that as xx grows, yy shrinks.
Forgetting that xx cannot be zero when you try to use x=0x = 0 in a problem.Division by zero is undefined, so x=0x = 0 has no meaning in inverse variation; the relationship breaks down at that value.Always note that x0x \neq 0 (and y0y \neq 0 if k0k \neq 0). If a problem seems to require x=0x = 0, re-read it—the scenario is not actually inverse variation.
Using the wrong operation to find the constant kk, such as computing x+yx + y or xyx - y instead of xyx \cdot y.The constant in inverse variation is always the product xyxy, not the sum, difference, or any other combination; using a different operation gives the wrong formula.Always verify: k=xyk = x \cdot y for every pair of values. If one pair gives k=24k = 24 but another gives k=15k = 15, the relationship is not inverse variation.

Tips and when to use something else

  • Write the inverse variation formula y=kxy = \frac{k}{x} as soon as you identify the relationship, using the constant kk from the problem.
  • Check your constant by computing xyxy for at least two pairs; if you get different products, the relationship is not inverse variation—use direct variation (y=kxy = kx) instead.
  • Remember the graph of y=kxy = \frac{k}{x} is a hyperbola, not a line; it gets very steep near x=0x = 0 and flattens out far from the origin.
  • If a problem gives you one pair of values (x1,y1)(x_1, y_1), always find the constant first: k=x1y1k = x_1 \cdot y_1, then use y=kxy = \frac{k}{x} for any other pair.

Frequently asked questions

What is the difference between inverse and direct variation?
In direct variation, y=kxy = kx, and both variables change in the same direction: if xx doubles, yy doubles. In inverse variation, y=kxy = \frac{k}{x}, and the variables change in opposite directions: if xx doubles, yy is cut in half. Both have a constant, but direct variation multiplies and inverse variation divides.
How do I find the constant kk in an inverse variation?
Multiply the two known values together: k=xyk = x \cdot y. For example, if x=5x = 5 and y=8y = 8 satisfy the relationship, then k=58=40k = 5 \cdot 8 = 40, and the equation is y=40xy = \frac{40}{x}.
Can xx or yy be negative in inverse variation?
Yes. Negative values work just like positive values in the formula y=kxy = \frac{k}{x}. If kk is positive and xx is negative, then yy will be negative. If both xx and yy are negative, then kk is positive. Just be careful with signs when doing arithmetic.
Why does the graph of y=kxy = \frac{k}{x} never touch the axes?
The graph never crosses the xx-axis because yy can never equal zero (unless k=0k = 0, which is not inverse variation). Similarly, the graph never crosses the yy-axis because xx can never equal zero. The axes are called asymptotes—the curve approaches them but never reaches them.

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Reviewed 2026-09-18