Direct and Inverse Proportion: Free Response

Direct variation means $y = kx$ for some constant $k$. Substitute the known pair, $y = 27$ when $x = 3$, and solve for $k$ by dividing.

$k = \frac{27}{3} = 9.$

The constant is 9 dollars per pound, so the completed equation is $y = 9x$.

Substitute $x = 7$ into the equation from part A.

$y = 9 \times 7 = 63.$

So 7 pounds costs 63 dollars.

Write the cost for a doubled purchase and factor.

$y_{\text{new}} = 9(2x) = 2(9x) = 2y.$

The constant $9$ is applied to whatever $x$ is fed into it, so multiplying the input by any factor multiplies the output by that same factor; doubling is just that factor being $2$.

Divide $y$ by $x$ in each pair.

$\frac{12}{2} = 6, \quad \frac{8}{3}, \quad \frac{6}{4} = 1.5, \quad \frac{4}{6} = \frac{2}{3}.$

The four values disagree, so the ratio does not stay fixed and the table is not direct variation.

Multiply $x$ and $y$ in each pair.

$2 \times 12 = 24, \quad 3 \times 8 = 24, \quad 4 \times 6 = 24, \quad 6 \times 4 = 24.$

The product is $24$ in every column, so the table is inverse variation with $k = 24$, and $y = 24/x$.

Substitute $x = 8$ into $y = 24/x$.

$y = \frac{24}{8} = 3.$

Since $xy = 24$ for every pair in this relationship, doubling $x$ while keeping the product fixed forces $y$ to be divided by the same factor of 2.

$16 \times y_{\text{new}} = 24 \implies y_{\text{new}} = 1.5.$

So $y$ falls from 3 to 1.5, exactly half, without ever recomputing the product from scratch: the fixed product does all the work.

Divide $y$ by $x$ in each of the first two rows.

$\frac{3}{2} = 1.5, \qquad \frac{6}{4} = 1.5.$

Both ratios agree at $1.5$, so these two rows by themselves look exactly like direct variation with $k = 1.5$.

Substitute $x = 6$ into the equation suggested by the first two rows.

$y = 1.5 \times 6 = 9.$

The table's actual value at $x = 6$ is $8$, not $9$, so the third row does not fit the equation the first two rows suggested. Two rows agreeing was consistent with direct variation, but it was not proof of it: the third row is the counterexample that disproves the claim.

The correct test checks $y/x$ (or $xy$, for inverse variation) across ALL of the rows, not a single chosen pair.

The reason a matching pair says nothing about the rows it did not touch is algebraic, not just bad luck: if $x$ doubles from $x_0$ to $2x_0$ and $y$ also doubles from $y_0$ to $2y_0$, then

$\frac{2y_0}{2x_0} = \frac{y_0}{x_0}$

automatically, for any starting pair with $x_0 \neq 0$ (and a row with $x_0 = 0$ has no ratio to test in the first place). Two rows related by doubling will always look consistent with direct variation, whatever the untested rows do, because that agreement is a fact about doubling itself, not about the table.

Note the exact limit of this. If those two rows are the only rows the table has, then checking both of them IS checking every row, and the matching ratios do settle it. The gap opens only when rows are left untested, which is why the test is stated as every row rather than as some fixed number of rows.

Joint variation means $m = k\ell w$. Substitute the known piece, $m = 30$ when $\ell = 20$ and $w = 15$, and solve for $k$.

$30 = k \times 20 \times 15 = 300k, \qquad k = \frac{30}{300} = 0.1.$

The completed equation is $m = 0.1\ell w$.

Substitute $\ell = 24$ and $w = 25$ into the equation from part A.

$m = 0.1 \times 24 \times 25 = 60.$

Substitute the doubled dimensions into $m = 0.1\ell w$.

$m = 0.1 \times 40 \times 30 = 120.$

That is four times the original 30 grams, not two. Joint variation multiplies by BOTH factors that change: doubling $\ell$ alone would give $0.1 \times 40 \times 15 = 60$, only double the original, because $w$ never changed. Doubling both at once compounds the two factors of 2 into one factor of 4.

Substitute the known triple into the equation and isolate $k$.

$y_0 = \frac{k x_0}{z_0} \implies k = \frac{y_0 z_0}{x_0}.$

Combined variation, like direct and inverse variation, has exactly one thing left open once the form of the equation is assumed: the constant $k$. A single equation in a single unknown pins down that unknown completely, so one complete triple of matching values is exactly enough, no more and no less, to determine $k$ and therefore the whole relationship for every other pair of $x$ and $z$.

Substitute into the formula from part A.

$k = \frac{y_0 z_0}{x_0} = \frac{8 \times 5}{4} = 10.$

Use that same constant at the new values.

$y = \frac{10 \times 6}{3} = 20.$

Use $y = 10x/z$ with $k = 10$ carried from part B.

Doubling both: $x = 12$, $z = 6$.

$y = \frac{10 \times 12}{6} = 20.$

Unchanged from the original 20, because the ratio $x/z = 6/3 = 2$ is the same as $12/6 = 2$.

Doubling only $x$: $x = 12$, $z$ stays at $3$.

$y = \frac{10 \times 12}{3} = 40.$

This time the ratio $x/z$ itself doubled, from $2$ to $4$, and $y$ doubled right along with it. The difference between the two outcomes comes entirely from whether the ratio $x/z$ changed or stayed put.

Suppose $x$ and $z$ are both scaled by the same factor $p \neq 0$, so $x \to px$ and $z \to pz$. Substitute into the combined-variation equation.

$y_{\text{new}} = \frac{k(px)}{pz} = \frac{p}{p} \cdot \frac{kx}{z} = \frac{kx}{z} = y.$

The factor $p$ appears in both the numerator and the denominator, so it cancels regardless of what $p$ is, and $y$ is left exactly as it was.

Adding the same number $n$ to both instead, $x \to x + n$ and $z \to z + n$, does not produce this cancellation:

$y_{\text{new}} = \frac{k(x + n)}{z + n},$

and there is no common factor to divide out of the top and bottom, so this new value generally differs from $y$. Multiplying by the same factor preserves the ratio $x/z$ that the formula actually depends on; adding the same number changes that ratio, which is why only the multiplicative version leaves $y$ untouched.