Solve an equation or formula for one variable in terms of the others (e.g., express x in terms of y, or isolate a variable in a given formula). Includes equations arising from stated geometric facts, such as two expressions that are both radii of the same circle.
Free, timed SHSAT practice sets on literal equations & rearranging formulas— every question includes a full worked explanation, and there's no account required.
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A literal equation has more than one letter. To solve for one letter, treat the other letters like numbers and undo the operations in the usual order.
Example: solve \(5w - 9 = P\) for \(w\). Add \(9\) to both sides: \(5w = P + 9\). Divide both sides by \(5\): \(w = \dfrac{P + 9}{5}\).
Divide the whole side, not one term. From \(6w = 5P - 15\), you get \(w = \dfrac{5P - 15}{6}\). The answer \(\dfrac{5P}{6} - 15\) is wrong, because it divides only \(5P\).
Equal lengths give an equation. Two radii of the same circle are equal. If the radii are \(4x + 2\) and \(x + 14\), then \(4x + 2 = x + 14\), so \(3x = 12\) and \(x = 4\). Then answer the question that was asked: the radius is \(4 + 14 = 18\), not \(4\).
Subtract 3 from both sides: \(-5r = C - 3\). Divide by -5: \(r = (C - 3)/(-5)\). (Adding 3 instead gives \((C + 3)/(-5)\). Dividing only \(C\) by -5 gives \(C/(-5) - 3\). Multiplying by -5 instead of dividing gives \((C - 3) × (-5)\).)
Divide both sides by 5: \(r = (-9C - 12) \div 5\), which is \((-9C - 12)/5\). (Multiplying by 5 instead of dividing gives \(5(-9C - 12)\). Dividing in the wrong order gives \(5/(-9C - 12)\). Dividing only the \(-9C\) term by 5, and not the -12, gives \(-9C/5 - 12\).)
Set the expressions equal and solve: \(2x = 24\), so \(x = 12\). The radius length is \(x + 35 = 12 + 35 = 47\). (Stopping at \(x\) gives \(12\). Doubling the radius gives the diameter, \(94\). Adding the \(x\)-terms, \(4x = 24\), gives \(x = 6\) and a radius of \(41\).)