Compute slope from two points, a graph, or a table.
Free, timed SHSAT practice sets on slope— every question includes a full worked explanation, and there's no account required.
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Slope is rise over run: the change in \(y\) divided by the change in \(x\), or \(m = \dfrac{y_2 - y_1}{x_2 - x_1}\).
Example: through \((1, 7)\) and \((3, -1)\), \(m = \dfrac{-1 - 7}{3 - 1} = \dfrac{-8}{2} = -4\).
Common traps:
Slope \(= \dfrac{y_2 - y_1}{x_2 - x_1} = \dfrac{9 - 5}{-6 - (-4)} = \dfrac{4}{-2} = -2\). (Dividing the change in \(x\) by the change in \(y\) gives \(-1/2\). Subtracting the \(y\)-values and the \(x\)-values in opposite orders gives \(2\). Adding the coordinates, \(\dfrac{5 + 9}{-4 + (-6)}\), gives \(-7/5\).)
Pick two points and apply \(\text{slope} = \dfrac{\Delta y}{\Delta x}\): \(\dfrac{-27}{9} = -3\). (Dividing \(\Delta x\) by \(\Delta y\) gives \(-1/3\). Subtracting the \(y\)-values and the \(x\)-values in opposite orders gives \(3\). The change in \(y\) from one point to the next is \(-9\), but \(x\) goes up by 3 each time, so you must still divide by 3.)
Solve for \(y\): \(-8y = -2 + 6x\), so \(y = \dfrac{6}{-8}x + \dfrac{-2}{-8}\). The slope is the coefficient of \(x\): \(-3/4\). (Forgetting the minus sign from moving \(-6x\) to the other side gives \(3/4\). Turning the fraction upside down gives \(-4/3\). Reading the slope straight off as the \(x\)-coefficient gives \(-6\), but that works only after you solve for \(y\).)