Elementary Algebra and Geometry for Schools

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HOME : Mathematics - Geometry

Literal equations. Let us consider the following equations:

3x = 6
5x =-15
ax = b

All of these equations are in the same "form"; that is, all of them state that the product of x and some number is equal to a second number. The third equation, ax = b, is the general form for the first two equations. If we replace a by 3 and b by 6, we have the first equation, 3x = 6; if we replace a by 5 and b by -15, we have the second equation, 5x =-15. The equation ax = b is called a literal equation, since the numerical values of the coefficients are expressed by letters.

ILLUSTRATIVE EXAMPLES: Solving literal equations

Solving literal equations involves isolating a specific variable in an equation. The steps to solve a literal equation are similar to solving regular equations, but with multiple variables.

Here's a general approach to solving literal equations:

  1. Identify the variable you want to isolate. Let's say it's "x."

  2. Use algebraic operations to move terms containing "x" to one side of the equation and other terms to the other side. Perform the same operation on both sides to maintain equality.

  3. Simplify the equation by combining like terms and performing any necessary arithmetic operations.

  4. Continue solving for "x" until it is isolated on one side of the equation.

  5. If there are other variables present in the equation, express them in terms of the isolated variable "x" if needed.

Here's an example to illustrate the process:

Given the equation A = 2πr + h, solve for "r".

  1. We want to isolate the variable "r".

  2. Begin by moving the term "2πr" to the other side of the equation: A - 2πr = h

  3. If there are no other terms containing "r," we can proceed to the next step. Otherwise, simplify further.

  4. Now, isolate "r" by dividing both sides of the equation by 2π: (A - 2πr) / 2π = h / 2π r = (A - 2πr) / (2π)

  5. If necessary, simplify the expression involving other variables, such as "A" or "h," to express them solely in terms of "r."

Remember to carefully apply algebraic operations while solving literal equations and be mindful of maintaining equality throughout the process.

 

To solve a literal equation for a particular letter, we must express the value of that letter in terms of the other letters.

1. Solve the equation ax = b for x.

Solution. Following the method of solving a first degree equation, we divide both members of the equation by the coefficient of x. Dividing both members of the equation by a (the coefficient of x),

2. Solve the equation p +ax = b for x.

Solution. First, by transposing terms, we arrange the equation so that the term containing x appears on the left side of the equal sign, and every term that does not contain x appears on the right side of the equal sign:

ax = b -p

Dividing both members by a

EXERCISES: Solving literal equations

Solve each of the following equations for x in terms of the other letters:

Answers :

 

Answers :

ILLUSTRATIVE EXAMPLE: Solving formulas for one letter in terms of the other letters.

Solve the formula A = lw for l in terms of the other letters.

Solution. Following our method for solving first degree equations, we arrange the equation so that the term containing l appears on the left side of the equation and the term not containing l appears on the right side of the equation. Transposing the A term to the right and the lw term to the left:

-lw =-A

Multiplying both members by -1 lw = A

Dividing both members by w

In this particular problem, the work can be shortened if we simply leave the terms as they stand, and divide both members of the equation A = lw by w:

Since an equation may be read from right to left as well as from left to right, the form

are the same.

EXERCISES: Solving formulas for one letter in terms of the other letters

1. Given the formula A = lw, solve the equation for w in terms of the other letters.

2. The perimeter of a rectangle is given by the formula p = 2l + 2w, where p represents the perimeter of the rectangle, l represents its length, and w represents its width. Solve the formula for w in terms of the other letters.

3. The perimeter of a square is given by the formula p = 4s, where p represents the perimeter of the square and s represents a side of the square. Solve the formula for s in terms of p.

4. The distance d that is covered by traveling at a rate r for a time t is given by the formula d = rt. Solve this formula for r in terms of d and t; solve the formula for t in terms of d and r.

5. The interest i that is earned by investing a principal p at a rate r for a time t is given by the formula i= prt. Solve the formula for p in terms of the other letters.

6. The volume V of a rectangular solid whose length is l, width w, and height h is given by the formula V = lwh. Solve the formula for w in terms of the other letters.

Answers :

ILLUSTRATIVE EXAMPLES: Writing equations

As you learned that an important step in solving problems is to translate word statements into algebraic equations. In the simpler statements this translating could be done almost word for word.

1. Write an equation for the statement "3x, increased by 7, is equal to 70."

Solution. We can translate this statement into an equation as follows:

Frequently, however, a statement gives a comparison between several quantities, and the equality of the quantities is not specifically stated. In such problems, you must decide how to balance (equate) the given quantities.

2. Write an equation for the statement "5x exceeds (40 -x) by 14."

Solution. In this statement we are given a comparison between the quantity 5x and the quantity (40 -x). To help us write an equation, we ask ourselves the following questions:

1. Which of these two quantities is the larger?

2. By how much is it larger?

3. What can we do to one of these quantities so that the result is equal to the other quantity?

Answering these questions:

1. The quantity 5x is larger, since it exceeds the quantity (40 -x).

2. The quantity 5x is 14: larger than the quantity (40 -x).

3. To obtain equal quantities, we can add 14 to the smaller [(40 -x) + 14], or we can subtract 14 from the larger [5x -14].

Hence we write the equation as:

5x = (40 -x) + 14 (adding 14 to the smaller) or we write 5x -14 = (40 -x) (subtracting 14 from the larger)

Notice that either of these two equations can be obtained directly from the other by transposing the "14" from one member of the equation to the other.

In writing equations from statements such as these, it may be helpful to you to think of the quantities compared as being placed on opposite sides of a pair of scales; thus:

Since the quantities are not equal, the scales are not in balance. By asking yourself the questions listed above, you then decide what you must do to put the scales in balance. Since the quantity 5x is larger (heavier) by 14 units, you must either subtract 14 from 5x to make the scales balance, or you must add 14 to ( 40 -x), the smaller (lighter) side, to put the scales in balance.

EXERCISES: Writing equations

Form an equation for each of the following statements, and then solve your equation for x.

1. 5x is equal to 3x + 16.

2. 2x +50 is equal to 26-x.

3. 5x -21 is equal to 3x increased by 33.

4. 7x diminished by 31 is equal to 18.

5. 2x increased by 3 is equal to 5x diminished by 6.

6. 3x -5 is equal to 60 diminished by 2x.

7. 2x is 4 less than x + 7.

8. 5x -2 is 8 larger than 4x

9. 3x + 5 exceeds 2x by 7.

10. x + 4 exceeds 5 -2x by 20.

Answers:

1. 5x = 3x + 16, x = 8. 2. 2x + 50 = 26 -x, x =-8. 3. 5x -21 =3 x + 33, x = 27.

4. 7x -31 = 18, x = 7. 5. 2x + 3= 5x -6, x = 3. 6. 3x -5 = 60 -2x, x = 13. 7. 2x = (x + 7) -4, x = 3. 8. 5x -2= 8 + 4x, x = 10. 9. 3x + 5= 2x + 7, x = 2. 10. x + 4= (5 -2x) + 20, x = 7.

ILLUSTRATIVE EXAMPLES: Solving word problems

1. Three times a certain number, increased by 7, is equal to the number subtracted from 43. Find the number.

Solution. Let x represent the certain number; then 3x represents three times the number. We translate the statement into an algebraic equation:

3x + 7 = 43 -x

Transposing 3x + x = 43 -7

Collecting like terms 4x = 36

Dividing by 4   x=9

Answering the original question: the certain number is 9.

Check. Three times the certain number is 3 X 9, or 27; three times the number increased by 7 is 27 + 7, or 34; the number subtracted from 43 is 43 -9, or 34; thus, 34 = 34, and the original statement is satisfied.

2. Three times a certain number is 5 less than 29. Find the number.

Solution. Let x represent the number; then 3x represents three times the number. The statement gives us a comparison between the quantity 3x and the quantity 29. To write an equation, we write down the two quantities:

3x                       29

and then ask ourselves the questions: Which of these quantities is larger? By how much is it larger? What can we do to one quantity so that the result equals the other quantity?

3x                      29

29 is 5 larger than 3 x. We must therefore subtract 5 from this quantity to make the remainder equal to 3x.

Thus                          3x = 29 -5

Collecting                   3x = 24

Dividing by 3               x = 8

Answering the question: The certain number is 8.

Check. Three times the certain number is 3 X 8, or 24. Is 24 five less than 29? Yes, hence the original statement is satisfied.

3. The length of a rectangle is 6 ft. less than four times its width. If the perimeter is 38 ft., find the dimensions of the rectangle.

Solution. We draw a rectangle, and let x represent the number of feet in the width; then 4x -6 represents the number of feet in the length.

Our equation comes from the fact that the perimeter is 38 ft. To find the perimeter in terms of x, we add the four sides:

x + (4x -6) + x + (4x -6)

Then we write the equation

x + 4x -6 + x + 4x -6 = 38

Transposing x + 4x + x + 4x = 38 + 6 + 6

Collecting     10x = 50

Dividing by   10 x = 5

Answering the questions: The width of the rectangle, x, is 5 ft.; the length of the rectangle, 4x -6, is 4(5) -6, or 14ft.

Check. Is the perimeter 38? Does 5 + 14 + 5 + 14 equal 38?

EXERCISES: Problems

1. Five times a certain number, increased by 8, is equal to the number subtracted from 44. Find the number.

2. Find a number such that if 7 is added to four times the number the result is 31.

3. If a number is multiplied by 8, and then 21 is subtracted from this product, the result is 51. Find the number.

4. Four times a certain number, decreased by 5, equals 19 increased by the number. Find the number.

6. One number is three times another, and the smaller increased by 10 is equal to the larger diminished by 24. Find the numbers.

6. The larger of two numbers is 10 less than three times the smaller number, and the sum of the numbers is 78. Find the two numbers.

7. The larger of two numbers is twice the smaller. The smaller number, increased by 11, is the same as the larger number decreased by 17. Find the numbers.

8. Twice a certain number is 12 larger than 50. Find the number.

9. One number is seven times another. The larger, decreased by 60, is equal to three times the smaller increased by 16. Find the numbers.

10. One number is six times another. The larger number, diminished by 40 exceeds twice the smaller number by 8. Find the two numbers.

11. Twice a certain number is 12 less than the number increased by 28. Find the number.

12. The larger of two numbers is 8 more than three times the smaller number. If the sum of the two numbers is 44, find the numbers.

13. In a certain algebra class, the number of boys was 7 less than twice the number of girls. How many girls were there in the class, if the class contained 41 pupils?

14. How should 464 marbles be divided between two boys so that one boy receives 8 less than three times as many marbles as the other boy?

15. In a certain contest two prizes totaling $32 were awarded. If the first prize was $4 less than twice the second prize, find the amount of each prize.

16. A boy bought 28 stamps consisting of three-cent stamps and fivecent stamps at the post office. If the number of three-cent stamps that he bought was 4 more than twice the number of five-cent stamps, how many of each kind did he buy?

17. A rectangle is three times as long as it is wide, and its perimeter is 136 ft. Find the length and width of the rectangle.

18. One hundred eighty dollars was divided between a plumber and his helper so that the plumber received $15 less than twice what his helper received. How much did each receive?

19. The length of a rectangle is 8 ft. less than three times the width. If the perimeter of the rectangle is 80 ft., find the length and width.

20. The longest side of a triangle is 12 ft. less than twice the shortest side, and the third side is 3 ft. longer than the shortest side. If the perimeter of the triangle is 63 ft., find the lengths of the three sides.

Answers : 1. 6. 2. 6. 3. 9. 4. 8. 5. 17,51. 6. 22,56. 7. 28,56. 8. 31. 9. 19, 133. 10. 12, 72. 11. 16. 12. 9, 35. 13. 16. 14. 118, 346. 15. $20 first prize, $12 second prize. 16. 8 five-cent stamps, 20 three-cent stamps. 17. 17 ft. wide, 51 ft. long. 18. Helper received $65, plumber received $115. 19. 12 ft. width, 28 ft. length. 20. 18 ft., 21 ft., 24 ft.

CHAPTER REVIEW EXERCISES

In ex. 1-12 solve each equation for x and check your answer.

13. Solve for x in terms of a: 3x -a = 4x +a.

14. Solve the formula i= prt for t in terms of the other letters.

15. Write an equation for the statement, "3x is 9 larger than x -3."

16. Write an equation for the statement, "If 3x -6 is multiplied by 3, the result is 108."

17. Four times a certain number is 17 smaller than 89. Find the number.

18. Five times a certain number, decreased by 15, is equal to the number subtracted from 63. Find the number.

19. One number is 3 less than twice another number. If the sum of the two numbers is 21, find the numbers.

20. The length of a rectangle is 6 in. less than twice the width. If the perimeter of the rectangle is 90 in., find the dimensions of the rectangle.

CUMULATIVE REVIEW EXERCISES

1. Find the sum of 18x and 14x.

2. Find the sum of -5y and 7y.

3. Subtract 12a from -9a.

4. Subtract -11b from -4b.

5. Simplify by collecting like terms: 5x -7x2 +6x2 -8x +4x2

6. Subtract -a from a2.

7. Find the product (2ax)(-5ab).

8. If 3x =-27, find the value of x.

9. If a = 2, b = -6, and c = 10, find the value of:

ab; b -c; 2a +b; ac -bc.

10. How many terms are there in the expression 3x -5y +2?

11. Find the quotient: -6xy/3x

12. Express algebraically, "3 less than twice y."

13. Multiply -8a by 2a.

14. If y/2 =-3, find the value of y.

15. Write the following statement as an equation, "5x -8 is the same as 3x increased by 30."

16. What is the value of each of the following:

17. If x =-3, which is the larger: 6x or x3?

18. Solve and check the equation: 5x -28 -3x = 12+ 11x -16.

19. Solve and check the equation: .5y +1.8 = 2.6y -8.7.

20. In traveling a certain distance, a man rode 6 mi. more than four times the distance he walked. If the distance that he traveled was 41 mi., how far did he walk?