# Solving Problems With Equations

This can be done by first multiplying the entire problem by the common denominator and then solving the resulting equation.

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In this case, one pipe is filling the pool and the other is emptying the pool so we get the equation: Step 2: Solve the equation created in the first step.

This can be done by first multiplying the entire problem by the common denominator and then solving the resulting equation. Click Here for Practice Problems Example 4 – One roofer can put a new roof on a house three times faster than another. How long would it take the faster roofer working alone?

How long would it take to fill the pool if both pipes were accidentally left open?

Step 1: A problem involving work can be solved using the formula , where T = time working together, A = the time for person A working alone, and B = the time for person B working alone.

Worked-out word problems on linear equations with solutions explained step-by-step in different types of examples. Solution: Then the other number = x 9Let the number be x. Therefore, x 4 = 2(x - 5 4) ⇒ x 4 = 2(x - 1) ⇒ x 4 = 2x - 2⇒ x 4 = 2x - 2⇒ x - 2x = -2 - 4⇒ -x = -6⇒ x = 6Therefore, Aaron’s present age = x - 5 = 6 - 5 = 1Therefore, present age of Ron = 6 years and present age of Aaron = 1 year.5. Then the other multiple of 5 will be x 5 and their sum = 55Therefore, x x 5 = 55⇒ 2x 5 = 55⇒ 2x = 55 - 5⇒ 2x = 50⇒ x = 50/2 ⇒ x = 25 Therefore, the multiples of 5, i.e., x 5 = 25 5 = 30Therefore, the two consecutive multiples of 5 whose sum is 55 are 25 and 30. The difference in the measures of two complementary angles is 12°. ⇒ 3x/5 - x/2 = 4⇒ (6x - 5x)/10 = 4⇒ x/10 = 4⇒ x = 40The required number is 40.

There are several problems which involve relations among known and unknown numbers and can be put in the form of equations. Sum of two numbers = 25According to question, x x 9 = 25⇒ 2x 9 = 25⇒ 2x = 25 - 9 (transposing 9 to the R. S changes to -9) ⇒ 2x = 16⇒ 2x/2 = 16/2 (divide by 2 on both the sides) ⇒ x = 8Therefore, x 9 = 8 9 = 17Therefore, the two numbers are 8 and 17.2. A number is divided into two parts, such that one part is 10 more than the other. Try to follow the methods of solving word problems on linear equations and then observe the detailed instruction on the application of equations to solve the problems. According to the question; Ron will be twice as old as Aaron. Complement of x = 90 - x Given their difference = 12°Therefore, (90 - x) - x = 12°⇒ 90 - 2x = 12⇒ -2x = 12 - 90⇒ -2x = -78⇒ 2x/2 = 78/2⇒ x = 39Therefore, 90 - x = 90 - 39 = 51 Therefore, the two complementary angles are 39° and 51°9. If the table costs more than the chair, find the cost of the table and the chair. Solution: Let the number be x, then 3/5 ᵗʰ of the number = 3x/5Also, 1/2 of the number = x/2 According to the question, 3/5 ᵗʰ of the number is 4 more than 1/2 of the number. Solution: Let the breadth of the rectangle be x, Then the length of the rectangle = 2x Perimeter of the rectangle = 72Therefore, according to the question2(x 2x) = 72⇒ 2 × 3x = 72⇒ 6x = 72 ⇒ x = 72/6⇒ x = 12We know, length of the rectangle = 2x = 2 × 12 = 24Therefore, length of the rectangle is 24 m and breadth of the rectangle is 12 m. Then Aaron’s present age = x - 5After 4 years Ron’s age = x 4, Aaron’s age x - 5 4. Then the cost of the table = \$ 40 x The cost of 3 chairs = 3 × x = 3x and the cost of 2 tables 2(40 x) Total cost of 2 tables and 3 chairs = 5Therefore, 2(40 x) 3x = 70580 2x 3x = 70580 5x = 7055x = 705 - 805x = 625/5x = 125 and 40 x = 40 125 = 165Therefore, the cost of each chair is 5 and that of each table is 5. If 3/5 ᵗʰ of a number is 4 more than 1/2 the number, then what is the number?How long would it take the triplets to complete the project if they work together?In this case, there are three people so the equation becomes: Step 2: Solve the equation created in the first step. Two step algebraic equations are relatively quick and easy -- after all, they should only take two steps.Example 1 – Walter and Helen are asked to paint a house.The equations are generally stated in words and it is for this reason we refer to these problems as word problems. If the two parts are in the ratio 5 : 3, find the number and the two parts. With the help of equations in one variable, we have already practiced equations to solve some real life problems. Solution: Let one part of the number be x Then the other part of the number = x 10The ratio of the two numbers is 5 : 3Therefore, (x 10)/x = 5/3⇒ 3(x 10) = 5x ⇒ 3x 30 = 5x⇒ 30 = 5x - 3x⇒ 30 = 2x ⇒ x = 30/2 ⇒ x = 15Therefore, x 10 = 15 10 = 25Therefore, the number = 25 15 = 40 The two parts are 15 and 25. Then Robert’s father’s age = 4x After 5 years, Robert’s age = x 5Father’s age = 4x 5According to the question, 4x 5 = 3(x 5) ⇒ 4x 5 = 3x 15 ⇒ 4x - 3x = 15 - 5 ⇒ x = 10⇒ 4x = 4 × 10 = 40 Robert’s present age is 10 years and that of his father’s age = 40 years. Their difference = 48According to the question, 7x - 3x = 48 ⇒ 4x = 48 ⇒ x = 48/4 ⇒ x = 12Therefore, 7x = 7 × 12 = 84 3x = 3 × 12 = 36 Therefore, the two numbers are 84 and 36.3. If the perimeter is 72 metre, find the length and breadth of the rectangle.More solved examples with detailed explanation on the word problems on linear equations.6. After 5 years, father will be three times as old as Robert.

## Comments Solving Problems With Equations

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When you enter an equation into the calculator, the calculator will begin by expanding simplifying the problem. Then it will attempt to solve the equation by using one or more of the following addition, subtraction, division, taking the square root of each side, factoring, and completing the square.…

• ###### Ways to Solve Two Step Algebraic Equations - wikiHow

Method 1 Solving Equations with One Variable 1. Write the problem. The first step to solving a two step algebraic equation is just to write. 2. Decide whether to use addition or subtraction to isolate the variable term. 3. Add or subtract the constant on both sides of the equation. 4. Eliminate.…

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• ###### Translating Word Problems into Equations -

Site Navigation. Translating Word Problems into Equations. Most of the time when someone says “word problems” there is automatic panic. But word problems do not have to be the worst part of a math class. By setting up a system and following it, you can be successful with word problems.…