Here, if x be put for the smaller part, the greater will be 48 - x. Letters may be employed to express the known quantities in an equation, as well as the unknown. a = 720 d = 7392 b = 125 h = 462 Then by the conditions of the problem (x a)/b = d/h Therefore x = (bd - ah)/h Restoring the numbers,x = [(125.7392) - (720.462)]/462 = 1280.

By the conditions of the problem x/4 (48 - x)/6 = 9. A particular value is assigned to the numbers; when they are introduced into the calculation: and at the close, the numbers are restored. If to a certain number, 720 be added, and the sum be divided by 125 ; the quotient will be equal to 7392 divided by 462. When the resolution of an equation brings out a negative answer, it shows that the value of the unknown quantity is contrary to the quantities which, in the statement of the question,"are considered positive. A merchant gains or loses, in a bargain, a certain sum.

Having sold from one of these 39, and from the other 93, he finds twice as many remaining in the one as in the other.

A farmer had two flocks of sheep, each containing the same number. An express, travelling at the rate of 60 miles a day, had been dispatched 5 days, when a second was sent after him, travelling 75 miles a day.

Nothing then remains to be done, but to reduce the equation, and to find the aggregate value of the known quantities. 52.) As these are equal to the unknown quantity on the other side of the equation, the value of that also is determined, and therefore the problem is solved. A man being asked how much he gave for his watch, replied; If you multiply the price by 4, and to the product add 70, and from this sum subtract 50, the remainder will be equal to 220 dollars.

## Analysis On An Essay On Man By Alexander Pope - Solving Problems Involving Radicals

To solve this, we must first translate the conditions of the problem, into such algebraic expressions as will form an equation. A gentleman bought several gallons of wine for 94 dollars; and after using 7 gallons himself, sold 1/4 of the remainder for 20 dollars. Let the price of the watch be represented by x This price is to be multiplied by 4, which makes 4x To the product, 70 is to be added, making 4x 70 From this, 50 is to be subtracted, making 4x 70 - 50 Here we have a number of the conditions, expressed in algebraic terms; but have as yet no equation. We must observe then, that by the last condition of the problem, the preceding terms are said to be equal to 220. What number is that, to which 10 being added, 3/5 of the sum will be 66? Of the trees in an orchard, 3/4 are apple trees, 1/10 pear trees, and the remainder peach trees, which are 20 more than 1/8 of the whole. We have, therefore, this equation 4x 70 - 50 = 220 Which reduced gives x = 50. Here the value of x is found to be 50 dollars, which is the price of the watch. What number is that 1/4 of which is greater than 1/5 of it by 96? A post is in the earth, 3/7 in the water and 13 feet above the water. To prove whether we have obtained the true value of the letter wnich represents the unknown quantity, we have - only to substitute this value, for the letter itself, in the equation which contains the first statement of the conditions of the problem; and to see whether the sides are equal, after the substitution is made. For if the answer thus satisfies the conditions proposed, it is the quantity sought. A man spent one third of his life in England, one fourth of it in Scotland, and the remainder of it, which was 20 years, in the United States. It is one of the principal peculiarities of an algebraic solution, that the quantity sought is itself introduced into the operation. A person after spending 100 dollars more than 1/3 of his income, had remaining 35 dollars more than 1/2 of it. In the composition of a quantity of gunpowder The nitre was 10 lbs. This enables us to make a statement of the con ditions in the same form, as though the problem were already solved. What number is that, of which if 1/3, 1/4, and 2/7 be added together the sum will be 73? To avoid an unnecessary introduction of unknown quantities into an equation, it may be well to observe, in this place, that when the sum or difference of two quantities is given, both of them may be expressed by means of the same letter. Foi if one of the two quantities be subtracted from their sum, it is evident the remainder will be equal to the other.

• ###### Solutions of Word Problems Involving Equations -

Solutions of Word Problems Involving Equations. In the solution of problems, by means of equations, two things are necessary First, to translate the statement of the question from common to algebraic language, in such a manner as to form an equation Secondly, to reduce this equation to a state in which the unknown quantity will stand by itself, and its value be given in known terms, on the.…

• ###### Algebra - Equations with Radicals - Pauls Online Math Notes

In this section we will discuss how to solve equations with square roots in them. seems to imply that we're going to look at equations that involve any radicals.…

• ###### Solving problems involving rational exponents and radicals

الصفحة الرئيسية; لمحة عن الجامعة. الرسالة والرؤيا; مجمع الشيخ أحمد كفتارو…

• ###### Solving problems involving radicals -

Solving problems involving radicals. Jack Saturday the 31st. Essay statement of purpose national essay contest the giver persuasive essay topics. Psychology essays on memory critical thinking and nursing practice how to write a business plan proposal apa sample paper example of completed business plan.…

• ###### Solving problems involving radicals examples

Solving problems involving radicals examples. Aiden Saturday the 17th. Solving trig equations practice problems how to format a research paper sections essay about environment ielts. Persuasive essay lesson plan middle school Persuasive essay lesson plan middle school.…

• ###### IXL - Simplify radical expressions involving fractions.

Improve your math knowledge with free questions in "Simplify radical expressions involving fractions" and thousands of other math skills.…

• ###### Algebra - Equations with Radicals Practice Problems

Here is a set of practice problems to accompany the Equations with Radicals section of the Solving Equations and Inequalities chapter of the notes for Paul Dawkins Algebra course at Lamar University.…

• ###### Solving Radical Equations - ChiliMath

Learning how to solve radical equations requires a lot of practice and familiarity of the different types of problems. In this lesson, the goal is to show you detailed.…