* This should result in the following equations: K= ¾ Z, S = ⅔ Z, K = S 15Taking the equation K=S 15 students need to replace the K with K= ¾ Z and the S with S= ⅔ Z thus creating the equation ¾ Z= ⅔ Z 15Students then need to recognise and multiply each of the terms by the common denominator 12, and then solve the resulting equation 9Z= 8Z 15 ∴ Z = 15Students should first review the process of writing algebraic expressions and equations from a worded description or rule.The resource discusses and explains determining a formula to reinforce students' understanding.*

* This should result in the following equations: K= ¾ Z, S = ⅔ Z, K = S 15Taking the equation K=S 15 students need to replace the K with K= ¾ Z and the S with S= ⅔ Z thus creating the equation ¾ Z= ⅔ Z 15Students then need to recognise and multiply each of the terms by the common denominator 12, and then solve the resulting equation 9Z= 8Z 15 ∴ Z = 15Students should first review the process of writing algebraic expressions and equations from a worded description or rule.The resource discusses and explains determining a formula to reinforce students' understanding.*

Suppose you're told that Shelby earns "time and a half" for any hours she works over forty for a given week.

You would be expected to know that "time and a half" means dollars for every over-time hour.

You'll also be expected to know that "perimeter" indicates the length around the outside of a flat shape such as a rectangle (so you'll probably be adding lengths) and that "area" indicates the size of the insides of the flat shape (so you'll probably be multiplying length by width, or applying some other formula).

And "volume" is the insides of a three-dimensional shape, such as a cube or sphere (so you'll probably be multiplying).

When simplifying equations, one often drops expressions such as 7-7 that equal 0. You will need to read the problem through carefully to ensure you have some understanding of what you are being asked to solve.

Pay close attention to the problem to determine the key clues. Read the problem again to make sure you understand what you're being asked for. Does "" stand for "Shelby" or for "hours Shelby worked"?If the former, what does this mean, in practical terms?But figuring out the actual equation can seem nearly impossible. Be advised, however: To learn "how to do" word problems, you will need to practice, practice, practice.The first step to effectively translating and solving word problems is to read the problem entirely.When you’re solving algebra word problems, it’s smart to have a plan of attack ready to follow.Solving word problems may seem difficult, but when you read through the problem and can figure out what the specific equation is, it’s no harder than a regular algebra problem.Algebra word problems are very useful to solve real-life problems. Remember the famous words of Albert Einstein"Do not worry about your difficulties in mathematics, I assure you that mine are greater."When you take a real-world situation and translate it into math, you are actually 'expressing' it; hence the mathematical term 'expression'.Everything that is left of the equal sign is considered to be something you are expressing. For instance, suppose you're not sure if "half of (the unknown amount)" should be represented by multiplying by one-half, or by dividing by one-half. When solving this problem students will need to create generalised algebraic equations for each part of the question. Particular attention should be given to order of operation, the correct use of mathematical convention and potential problem areas with the use of language Students should also practice the use of algebraic substitution in a variety of situations including simple numeric substitution into formulas and cases where one algebraic expression is substituted for another.

## Comments How To Solve A Word Problem In Algebra

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