Distinguish y = kx from y = mx + b, interpret slope and intercept, and identify a proportional relationship from a table, graph, or equation.
Free, timed SHSAT practice sets on proportional vs. linear relationships— every question includes a full worked explanation, and there's no account required.
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A proportional relationship has the form \(y = kx\). Its graph is a straight line through the origin \((0, 0)\), and \(\dfrac{y}{x}\) is the same number \(k\) for every pair.
A linear but not proportional relationship has the form \(y = mx + b\) with \(b \ne 0\). The line does not pass through the origin. For \(y = 4x - 3\), \(y = -3\) when \(x = 0\).
Test a table: divide \(y\) by \(x\) in every row. For \((1, 2), (2, 5), (3, 8)\), the ratios are \(2\), \(2.5\) and about \(2.67\). They change, so the table is not proportional.
Use \(k\) to find a new value: if \(y = -15\) when \(x = 5\), then \(k = \dfrac{-15}{5} = -3\), and when \(x = 7\), \(y = -3 \times 7 = -21\).
Common traps: using \(\dfrac{x}{y}\) instead of \(\dfrac{y}{x}\), and stopping after you find \(k\).
When \(x = 0\): \(y = -8(0) + 9 = 9\). This nonzero value at \(x=0\) is exactly why the relationship is NOT proportional. (Giving the slope gives \(-8\). Answering \(0\) treats the line as proportional, but only a proportional line passes through \((0, 0)\). Putting in \(x = 1\) instead of \(x = 0\) gives \(1\).)
Find \(k = y/x = 6/1 = 6\). Then \(y = k \times 10 = 60\). (Stopping after finding \(k\) gives \(6\). Using \(x/y\) as the constant gives \(5/3\). Adding the change in \(x\), 9, to \(y\) gives \(15\), but a proportional relationship multiplies, it does not add.)
A relationship is proportional when \(y / x\) is the same constant for every point (so it passes through the origin). \((1, 4), (2, 8), (3, 12)\) gives \(k = 4\) and \((2, 6), (4, 12), (5, 15)\) gives \(k = 3\); each of the others has a changing ratio.