7. Adding to Both Sides of an Equation. If we wish to solve such an equation as 3 x1=14, it is necessary to add 1 to both sides. This can be done since the balance is not thereby disturbed. Adding 1 to both sides, we find 3 x= 15. Dividing both sides by 3, we have x=3. Ans. ORAL EXERCISES In order to solve certain of the following equations, it is necessary to subtract the same number from both sides; in other equations it is necessary to add the same number to both sides. State which should be done in each case. 1. x-3=10. 7. 2 st19=81. 13. x-69=25. 2. X+4=7. 8. 2x–7=13. 14. .3 x+.4=.9 3. x-8=16. 9. 2k+101=201. 15. .1 x+.65=.75 4. x+10=17. 10. 5c+39=61. 16. 4p-13=27. 5. r+32=50. 11. 12 k-12=12. 17. 7 x— 21 =42. 6. 9-5=3. 12. x+74= 12; 18. jm-2=8. WRITTEN EXERCISES Solve each of the following equations; that is, find the value of the unknown letter. Check each answer. 1. 2 2–1=5. 6. 1.2c+2.3=4.7 11. 8 1.23=.87 2. 2 x+1=5. 7. 5x-1.5=.75 12. 2–89=101. 3. 4 x–3= 17. 8. 5x+.03=.08 13. 3 g+.6=. 4. 5 x+3=18. 9. 11 s- . 14. 1{ d+7=22. 5. 2 x – 1}=1. 10. -15=23. 15. 3.7 x 7.4=0. 16. If I subtract 18 from a certain number, the result is 4. Write an equation expressing this fact and solve it to find the number. 17. If 16 is subtracted from three times a certain number the result is 110. Find the number. 18. Our team played ball with the seventh grade. Our score was 3 less than twice theirs. Our score was 11. What was theirs ? 19. In paying for 31 yards of cloth a lady gave the clerk a $2 bill and received 57 cents back in change. How much was the material a yard ? 20. A man said, “ For $2500 you can have a half interest in my business, and I will give you my automobile, which I value at $500, in the bargain.” At what figure did he value his business? For further exercises on this topic, see the review list, p. 20, and Appendix, p. 290. 8. Multiplying both sides of an Equation. We have already seen how we may divide both sides of an equation by the same number. Thus, if 2 x=10, we simply divide both sides by 2 to get x, obtaining x=5. Ans. Likewise, we may at any time multiply both sides of an equation by the same number, since the balance is not thereby disturbed. For example, to solve the equation, we multiply both sides by 3, thus obtaining x=15. Ans. ORAL EXERCISES Find the value of the letter in each of the following equations. WRITTEN EXERCISES Find the value of the letter in each of the following equations. 1. *=10. 4. 11=62. 7. SOLUTION. Multiplying both sides by 3, 2 x=12. · Dividing both sides by 2, x=6. Ans. Find the value of x in each of the following equations. 1.1 11. 32=9. 12. 59–3. 13. 28=4. 14. 38=12. 15. 24 = 8. 17. A certain number divided by 13 gives 50. Write an equation expressing this fact and solve it to find the number. 18. What number divided by .4 gives 4? (Work by algebra.) 19. If twice a certain number is divided by 3, the result is 12. What is the number? (Compare with Ex. 10.) 20. By what number must 7] be multiplied to give 425? 21. The diameter of an ear of corn is about one fourth of its length. What is the length if the diameter is 14 inches ? 9. Axioms. Our study of the equation has now brought out four different principles, which are called axioms. As we shall need these often, they should be committed to memory. Axiom I. Equal amounts added to equal amounts give equal amounts. AXIOM II. Equal amounts subtracted from equal amounts give equal amounts. AXIOM III. Equal amounts multiplied by equal amounts give equal amounts. Axiom IV. Equal amounts divided by equal amounts give equal amounts. 10. Common factor. In arithmetic a number which is a factor of two or more numbers is called a common factor of those numbers. Thus, 5 is a common factor of 10, 15, and 20. The same is true in algebra. Thus, x is a common factor of 3.x, 8 x, and 11 x. 11. Like Numbers. Literal numbers that have a common factor are called like numbers. Thus, 3 x, 8 x, and 11 x are like numbers, as also 2 a, 7 a, and 9 a. Like numbers are very easily added together. Thus, we evidently have 2 x+3 x=5 x; that is, all we need to do is to add the coefficients to obtain a new coefficient, then multiply the result by the common factor. Similarly, we have 5 x–3 x=2 x. Thus, we see that in order to subtract like numbers all we need to do is to subtract the coefficients to obtain a new coefficient, and then multiply the result by the common factor. ORAL EXERCISES State the simplest form for each of the following expressions. 1. 2 a+4 a. 7. 16 x–8 x+3 x– x. 2. 4a-3 a. 8. 25 x+4 x–3 x— 2 x. 3. 3 x+3 x+4 x. 9. 6 m–2 m – 3 m-m. 4. 10 y+2y+y. 10. *x- x. 5. 10 y+2 y-y. 11. 5.2 x+4.3 x. 6. 10 y–2 y-y. 12. 5.2 y+4.3 y. 12. Further Study of the Equation. Sometimes we have to solve such an equation as .: 5x=16–3 x. This is different from any we have considered before because the x appears on both sides. If we remember the axioms of $ 9, however, we can easily solve it. The steps are as follows: 5 x=16–3 x. Adding 3 x to both sides, 8 x=16. (Axiom I) Dividing both sides by 8, x=2. Ans. (Axiom IV) · Again, suppose we wish to solve the equation 6+6 x=3 x+12. |