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This page sets out the method for direct & inverse proportion (algebraic) in the order you would use it, then walks through the full lesson explanation. For three complete worked examples with the slips that cost marks, see the Direct & inverse proportion (algebraic) topic page.
When a question says 'is proportional to' or 'varies inversely as', follow the same three-stage routine every time.
When you see 'inversely proportional' or 'as one doubles, the other halves'.
If y is proportional to the square of x, then y = kx².
Substitute the known pair, then use k for a new x.
If y is inversely proportional to x squared, then y = k / x² and k = yx².
Once k is known, square the new x before you multiply or divide.
If y = kx², then x is the square root of y/k. Take the positive root here.
A square law is not y = kx. The graph bends.
The sign propto means “is proportional to”. The sentence y propto x^2 becomes y = kx^2 as soon as you introduce k. There is no extra added term. When x is 0, y is 0. The square is the whole of the variable part.
A known pair fixes k. If y = 20 when x = 2, then 20 = k × 2^2 = 4k, so k = 5. The equation is now y = 5x^2 for every later x. You do not find a new k for each question. One pair determines the constant.
Method: 1. Write y = kx² from the proportion statement. 2. Square the known x. 3. Divide the known y by that square to find k. 4. Keep k and use it for every new x.
Once k is known, a new x is squared before it is multiplied. For k = 5 and x = 5, the square is 25 and y = 5 × 25 = 125. Multiplying 5 by 5 and forgetting the square gives 25, which is x^2 itself, not y.
The four-times rule is a useful check when the new x is double the old one. Doubling x multiplies x^2 by 4, so it multiplies y by 4. From x = 2 to x = 4, y should be four times the original y, if k has not changed.
Method: 1. Square the new x. 2. Multiply that square by k. 3. If x has doubled, check that y has become four times larger. 4. Substitute the original pair and check k still fits.
Inverse square means the square is in the denominator. If y = 12 when x = 3, then k = 12 × 3^2 = 12 × 9 = 108. The equation is y = dfrac108x^2. A new x is squared, then 108 is divided by that square.
Do not divide 12 by 3^2 and call that k. That division belongs to the direct-square model. Here y gets smaller when x gets larger, so k has to be a product. When x changes from 3 to 2, the denominator gets smaller and y rises. That direction is the check.
Method: 1. Square the known x. 2. Multiply that square by the known y to find k. 3. Write y = k/x². 4. Square the new x and divide k by it.
Sometimes y is given and x is not. From y = kx^2, divide by k to get x^2, then take the positive square root. If k = 5 and y = 80, then x^2 = 80 ÷ 5 = 16, so x = 4. The negative root is set aside here because the question asks for the positive value.
Square your answer as a check. 4^2 = 16, and 5 × 16 = 80. Also check the original pair: 5 × 3^2 = 45. Both pairs share the same k. If the square root is not a whole number, look again at whether you squared before you divided.
Method: 1. Find k from the known pair, using the square. 2. Divide the new y by k to get x². 3. Square-root, and keep the positive result. 4. Square the result and multiply by k to check the new y.
Every question starts with a sentence you must turn into an equation with a constant k. "Directly proportional" puts the variable on top, as in y = kx, y = kx^2, y = kx^3 or y = k√(x). "Inversely proportional" puts the variable underneath, as in y = k/x, y = dfrackx^2 or y = dfrack√(x). Write the equation before you use any number from the question.
The words "the square of" or "the cube of" go with the variable that follows, not with y. "y is directly proportional to the cube of x" is y = kx^3. A pair of values then fixes k: if y = 54 when x = 3 then 54 = k × 27 and k = 2. Rewrite the full equation with the value of k in it, such as y = 2x^3, before you substitute anything else.
Method: 1. Underline the type of proportion and which variable has a power or a root. 2. Write the equation with k in the right place. 3. Substitute the given pair of values and solve for k. 4. Write the equation again with the value of k, then use it.
Tip: If your value of k is a simple number, you have probably set up the equation correctly.
Some questions give two statements: "A is directly proportional to the square of B" and "B is inversely proportional to C". Work out each constant separately, one statement at a time. If A = 12 when B = 2, then k = 3 and A = 3B^2. If B = 2 when C = 9, then B = 18/C. Each statement uses its own pair of values.
To go from A to C, work backwards through the chain. If A = 27, then B^2 = 9 and B = 3 because B is positive. Then 3 = 18/C so C = 6. To find a formula for one variable in terms of another, substitute one equation into the other. Do not mix the constants. Each equation has a different k.
Method: 1. Write both equations with separate constants such as k and m. 2. Find each constant using its own pair of values. 3. Either work through the chain step by step, or substitute one equation into the other. 4. Check your final pair of values in both equations.
In a "show that" question the values may be letters. If y is inversely proportional to x^2 and y = 18 when x = a, then 18 = dfracka^2 so k = 18a^2. When x = 3a, y = dfrac18a^29a^2 = 2. The a^2 cancels. Write each line out in full so the examiner can follow the algebra.
Percentage change questions need no value of k. If x is directly proportional to the square of y and y increases by 20%, then y is multiplied by 1.2, and x is multiplied by 1.2^2 = 1.44. That is a 44% increase in x. If y is decreased by 20% and x is inversely proportional to y^2, then x is multiplied by dfrac10.8^2 = 1.5625, an increase of 56.25%.
Method: 1. Write the multiplier for the change, such as 1.2 for a 20% increase. 2. Apply the power or the inverse to the multiplier. 3. Turn the new multiplier back into a percentage change. 4. For letter values, substitute and cancel the common factor.
Read the last line of the question for the form of answer. "Work out an equation connecting" needs y = ldots with the number for k. "Find the positive value" or "find the negative value" asks for one root of x^2 = number. "Give your answer correct to 3 significant figures" asks for a rounded final answer. Do not round k early unless it is exact.
Use the shape of the relationship as a check. In y = kx^2, doubling x multiplies y by 4. In y = dfrackx^2, doubling x divides y by 4. In y = k√(x), quadrupling x doubles y. A graph of y = kx is a straight line through the origin and y = k/x is a curve that never touches the axes. Put your answer back into the equation to be sure.
Method: 1. Underline the form of the answer asked for. 2. Keep k exact until the end. 3. Take the correct root: positive or negative as asked. 4. Check the direction of the change: larger or smaller than before.
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