# Always show substitutions whether whole equations

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*Always show substitutions, whether whole equations, constants, zeros, or any numbers. Example: Solve for ? as a function of ? and t with ? 0 = 0 given: ? = ?? and ? = ? 0 ? + 1 2 𝑎? 2 and ? = ? 0 + 𝑎? Choose starting equation that contains desired variable: ? = ?? Need to find a way to get x into the equation. The second equation listed has x in it, but does not have v, so substitute in the third equation to eliminate v: Since ? = ? 0 + 𝑎? Then ? = ?(? 0 + 𝑎?) As defined in the problem, ? 0 = 0 , so substitute in to simplify: ? = ?(0 + 𝑎?) = ?𝑎? Still need to get x involved, so use second equation: Let ? = ? 0 ? + 1 2 𝑎? 2 As defined in the problem, ? 0 = 0 , so substitute in to simplify: ? = (0)? + 1 2 𝑎? 2 = 1 2 𝑎? 2 In many cases, t is an undefined variable because it is difficult to measure with precision. While not required for working the example problems, I will choose to eliminate t from the equation. You will learn how to identify from a problem which variables are or are not allowed in a final answer. Solve the x equation for t: Since ? = 1 2 𝑎? 2 Then ? = √ 2𝑥 𝑎 Substitute in to the equation for p: Therefore ? = ?𝑎? = ?𝑎√ 2𝑥 𝑎 At this point we could rewrite and box this solution, since it meets the requirements of the problem, or we could simplify by combining the a terms. Simplification is not always necessary unless you are trying to match your algebra to the solutions in a multiple choice question. However, to practice some simplification skills, we will simplify this result below. Recall that 𝑎 = 𝑎 1 = √ 𝑎 2 1 2 = √ 𝑎 2 1 therefore we can simplify by: ? = ?𝑎√ 2𝑥 𝑎 = ?√ 𝑎 2 1 2𝑥 𝑎 Recall that √? √? = √?? therefore we can further simplify by: ? = ?√ 𝑎 2 1 2𝑥 𝑎 = ?√2𝑎? Always rewrite and box the final answer: ? = ?√2𝑎?