Direct and Inverse Proportion - Forming Equations

Substitute the known pair of values into the proportionality equation.

1. Substitution gives 18=6k.
2. Divide by 6 to get k=3.

Start with the general equation for direct proportion.

1. Write y=kx.
2. Using the given values, 14=4k, so k=\dfrac72.
3. Therefore y=\dfrac{7x}{2}.

For inverse proportion, the product of the two variables is constant.

1. Write y=\dfrac kx.
2. Then 9=\dfrac k4, so k=36.
3. Therefore y=\dfrac{36}{x}.

Use the square of the input in the proportionality equation.

1. Write y=kx^2.
2. Then 45=9k, so k=5.
3. Therefore y=5x^2.

Place the square of the input in the denominator.

1. Write y=\dfrac{k}{x^2}.
2. Then 6=\dfrac k4, so k=24.
3. Therefore y=\dfrac{24}{x^2}.

Find the constant first, using 2^3=8.

1. Direct proportion to the cube of r means V=kr^3 for a constant k.
2. Substitute r=2, V=56: 56=k\times2^3=8k, so k=7 and V=7r^3.
3. When r=5: V=7\times5^3=7\times125=875.

Take the square root of the given input when finding the constant.

1. Write s=k\sqrt t.
2. Then 12=3k, so k=4.
3. Therefore s=4\sqrt t.

Think about what happens to y when x doubles, and whether y can ever be 0.

1. Inverse proportion means y=\frac{k}{x}. When x doubles, y halves.
2. So y is very large when x is small, and gets closer and closer to 0 as x grows, but never reaches 0.
3. Graph C has this shape.
4. Graph D also goes down, but it is a straight line that reaches the x-axis, where y=0; \frac{k}{x} is never 0.
5. Graph A shows y directly proportional to x, and Graph B shows y directly proportional to x^2.

The square root belongs in the denominator for this inverse relationship.

1. Write y=\dfrac{k}{\sqrt m}.
2. Then 8=\dfrac k3, so k=24.
3. Therefore y=\dfrac{24}{\sqrt m}.

Square the negative input before dividing to find the constant.

1. Write y=kx^2.
2. Since (-2)^2=4, the given pair gives -12=4k.
3. Thus k=-3 and y=-3x^2.

Find the cost per kilogram first.

Part (a)

1. Write C=km.
2. The given values give 7.60=4k, so k=1.9.
3. Therefore C=1.9m.

Part (b)

1. The cost per kilogram is 7.60\div4=1.9 pounds.
2. For 13 kg, C=1.9\times13=24.70, so the cost is £24.70.

Find the constant product before solving for the new area.

1. Write P=\dfrac kA. The given values give k=18\times2.5=45.
2. For the new area, 12=\dfrac{45}{A}.
3. Therefore A=45\div12=3.75 m^2.

Consider how the square of the input changes.

1. Multiplying x by 3 multiplies x^2 by 9.
2. Since y=\dfrac{k}{x^2}, the output is divided by 9.
3. The multiplying factor is \dfrac19.

Find the constant, then square the value obtained for the square root of the new input.

1. Write T=k\sqrt l. Using l=81 gives 1.8=9k, so k=0.2.
2. When T=2.4, \sqrt l=2.4\div0.2=12.
3. Therefore l=12^2=144.

Find the constant using the squared radius, then take the positive root at the end.

1. Write R=\dfrac{k}{r^2}. The given values give k=36\times0.5^2=9.
2. For the new radius, 16=\dfrac9{r^2}, so r^2=\dfrac9{16}.
3. The radius is positive, so r=\dfrac34=0.75 mm.

Find a constant for each relationship before combining them.

1. Write y=au^2. Using y=18 and u=3 gives a=2.
2. Write u=\dfrac bx. Using u=4 and x=5 gives b=20.
3. Substitute u=\dfrac{20}{x} into y=2u^2.
4. Therefore y=2\left(\dfrac{20}{x}\right)^2=\dfrac{800}{x^2}.

It can be simpler to square the square-root relationship before combining it with the other equation.

1. Write A=ar^2. Then 72=36a, so A=2r^2.
2. Write r=b\sqrt m. Squaring gives r^2=b^2m.
3. The pair r=4, m=8 gives 16=8b^2, so r^2=2m.
4. Therefore A=2\times2m=4m.

Compare the multiplying factor for the input with the multiplying factor for the output.

1. The input is multiplied by 3 and the output is multiplied by \dfrac19.
2. An inverse-square relationship divides the output by the square of the input factor.
3. The constant also agrees: 45\times2^2=5\times6^2=180.
4. Therefore y is inversely proportional to x^2.

Relate the multiplying factors using the inverse-square law before calculating a percentage change.

1. y=\frac{k}{x^2}, so x^2=\frac{k}{y} and x=\sqrt{\frac{k}{y}}.
2. If y is multiplied by 1.3, then x is multiplied by \frac{1}{\sqrt{1.3}}=0.877058\ldots
3. The decrease in x is 1-0.877058\ldots=0.122941\ldots of its old value.
4. That is 12.294\ldots\%, which is 12.3\% to 3 significant figures.

Find both constants, then work backwards from the new value of the third variable.

1. The inverse relationship gives y=\dfrac{12}{x} because xy=4\times3=12.
2. Write z=by^2. The given pair gives 18=9b, so z=2y^2.
3. When z=8, y^2=4. Positivity gives y=2.
4. Then x=12\div2=6.