To beginning algebra students, literal equations, those involving more than one letter, often seem inherently difficult to solve than simpler, univariate ones.
But this isn’t so.
The trick is to treat the extra letters in literal equations like simple numerals, using the same steps you would ordinarily, just as if you had only a single letter to deal with.
Solving equations is a matter of “undoing” what’s done to the unknown by using inverse operations, starting as far from the variable as possible.
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Example 1
To solve 3(x+4)–8 = 19, we first undo subtracting 8 by adding 8, then divide by 3 to undo multiplying by 3, and finally subtract 4 to undo adding 4. The result is x = 5. [Note: we could first simplify by distributing 3 across (x+4) and adding like terms, but this would take four steps, not three.]
If the letters a, b, and c were to replace 3, 4, and 19 in the same equation, we’d carry out exactly the same series of steps. This time we’d start with a(x+b)–8 = c. We’d then add 8, divide by a, then subtract b to get x = (c+8)/a – b.
When solving univariate equations, it’s important to add like variable terms as soon as possible. Sometimes this isn’t possible when solving literal equations. In such a case, factoring out the variable leads to a simple solution.
Example 2
To solve (nx–mn)/q = x+p for x, we first multiply by q and then collect and isolate variable terms on the same side by subtracting qx and adding mn to yield nx–qx = qp+mn. Since nx and x are not like terms, we factor out the variable x to produce x(n–q) = qp+mn, and divide by n–q. The result is x = (qp+mn)/(n–q).
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Solving equations is a matter of “undoing” what’s done to the unknown by using inverse operations, starting as far from the variable as possible.
-
Example 1
To solve 3(x+4)–8 = 19, we first undo subtracting 8 by adding 8, then divide by 3 to undo multiplying by 3, and finally subtract 4 to undo adding 4. The result is x = 5. [Note: we could first simplify by distributing 3 across (x+4) and adding like terms, but this would take four steps, not three.]
If the letters a, b, and c were to replace 3, 4, and 19 in the same equation, we’d carry out exactly the same series of steps. This time we’d start with a(x+b)–8 = c. We’d then add 8, divide by a, then subtract b to get x = (c+8)/a – b.
When solving univariate equations, it’s important to add like variable terms as soon as possible. Sometimes this isn’t possible when solving literal equations. In such a case, factoring out the variable leads to a simple solution.
Example 2
To solve (nx–mn)/q = x+p for x, we first multiply by q and then collect and isolate variable terms on the same side by subtracting qx and adding mn to yield nx–qx = qp+mn. Since nx and x are not like terms, we factor out the variable x to produce x(n–q) = qp+mn, and divide by n–q. The result is x = (qp+mn)/(n–q).
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For practice, search Google for “ solving literal equations worksheets,” pick a worksheet that provides answers, complete the worksheet, analyze any mistakes, and redo it until you can complete that worksheet with no errors. Then repeat, with additional worksheets, as needed, until you’ve mastered this important topic.
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