# Solving For X Word Problems

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Using the rate equation in the form distance = speed How much of a 10% vinegar solution should be added to 2 cups of a 30% vinegar solution to make a 20% solution?

Having difficulty turning a word problem into an algebra equation? With this tutorial, you'll learn how to break down word problems and translate them into mathematical equations.

They may involve a single person, comparing his/her age in the past, present or future.

They may also compare the ages involving more than one person.When the value that a particular variable will represent is determined, it is defined by writing a statement like, By using simple arithmetic, this problem probably could have been solved faster without setting up an algebra equation.But, knowing how to use an equation for this problem builds awareness of concepts which are useful, and sometimes critical to solving much harder problems.Many students find solving algebra word problems difficult.The best way to approach word problems is to “divide and conquer”.If the perimeter of a rectangle is 10 inches, and one side is one inch longer than the other, how long are the sides? Then the given condition may be expressed as x x (x 1) (x 1) = 10 Solving: 4x 2 = 10 4x = 8 x = 2 so the sides have length 2 and 3.A fast employee can assemble 7 radios in an hour, and another slower employee can only assemble 5 radios per hour.Go back to the top of the paper you used to solve this problem.It should contain the following work: Keep going to learn about what to do when you encounter more than two consecutive numbers, negative consecutives, and even or odd consecutives.A word problem in algebra is the equivalent of a story problem in math.When you solved story problems in your math class you had to decide what information you had and what you needed to find out. Addition was used to find a totals and subtraction was used to find changes in values.

## Comments Solving For X Word Problems

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