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First Degree Equations
A first degree equation is one which, when reduced to its simplest form, contains the unknown letter or letters raised to the first power only. Thus the equations 5x 7=18 and 3x + 5x 2 = 34 x are first degree equations. The equation 2x^{2} + 7 x 3x 2x^{2} = 28, as it is written, does not look like a first degree equation, since it contains the unknown raised to the second power. However, when it is written in simplest form by collecting like terms, the two x^{2} terms disappear, and the equation reduces to 4x = 28. Thus this equation is a first degree equation.
We have already learned to solve first degree equations by adding, subtracting, multiplying, or dividing both members of the equation by the same number. In this pages we will continue to apply these methods to solve equations; however, we will now solve equations which may contain negative as well as positive numbers. Also, we will learn a few "short cuts" that will make our work easier.
Let us state once more the four principles that we have applied in solving equations. These principles apply to equations which contain negative numbers as well as positive numbers. Trese principles are called axioms. An axiom is a statement that is accepted without proof.
Axioms
1. The same number may be added to both members of an equation
( we call this the" addition axiom").
2. The same number may be subtracted from both members of an equation (subtraction axiom).
3. Both members of an equation may be multiplied by the same number (multiplication axiom).
4. Both members of an equation may be divided by the same number, providing that the number by which they are divided is not zero (division axiom).

ILLUSTRATIVE EXAMPLES: Applying axioms
1. Solve the equation 6x =18.
Solution. We are given the value of 6x. To solve
the equation, we must find the value of x (that is,
the value of 1x); hence we divide both members
of the equation by 6:
2. Solve the equation 4x = 22.
Solution. We are given the value of 4x. To solve the equation, we must find the value of x (that is, the value of 1x); hence "ve divide both members of the equation by 4:
3. Solve the equation 6x 7x = 14 8.
Solution. Collecting like terms, we have x = 6.
We now have the value of x (that is, the value of 1x); to find the value of x (that is, the value of
+ 1x), we must multiply (or divide) both members of the equation by 1:
EXERCISES: Equations
Solve each of the following equations:
Answers : 1.7. 2.9. 3.6. 4. 4. 5. 6. 6.6½. 7. 11. 8. 15. 9. 5. 10.  4/3 ;. 11.8. 12. 2. 13. 5/2. 14. 12. 15. 1/5. 16. 3/4.
17. 8. 18.6. 19. 48. 20. 9. 21.4. 22. 5. 23. 3. 24. 10. 25. 8. 26. 11. 27.3. 28.13. 29.18. 30.14/3. 31. 13. 32. 27. 33. 1/3,.
34. 1. 35. 11.
Transposing terms. We have already learned how to solve equations of the type: 3x 4 = 8 by adding the same number to both members of the equation. Let us study this method again:
3x 4= 8
Adding 4 to both members 3x 4 +4 = 8 +4
Collecting on the left 3x = 8 +4
Let us pause here to compare what we now have with the original equation. We notice that the only change in the two equations is that the term 4 appeared in the original equation in the left member, and that it has now been changed in the new equation to a positive term, + 4, of the right member. In other words, we can go directly from the form 3x  4 = 8 to the form 3x = 8 +4 by moving (transposing) the term 4 from the left member to the right member providing we change the sign of the term. When we transpose (move) terms from one member of an equation to the other member of the equation, we are merely applying to the equation the axioms of addition and subtraction which were stated above in this pages.
A term may be transposed (moved) from one member of an equation to the other member of the equation provided
the sign of the term is changed. 
ILLUSTRATIVE EXAMPLES: Transposing terms
1. Solve the equation 5x 3 =18.
2. Solve the equation 27x = 3x 60
EXERCISES: Equations
Solve each of the following equations by transposing the terms containing x to the left member and the terms not containing x to the right member, and by making the correct changes of signs.
Answers : 1. 6. 2. 2. 3. 11. 4. 4. 5. 3. 6. 2. 7. 3. 8. 7. 9. 9. 10.4. 11. 2. 12. 13. 13. 13. 14. 7. 15. 8. 16. 13. 17. 8. 18. 15. 19. 11. 20. 3. 21. 5. 22. 14. 23. 12. 24. 6. 25. 4. 26. 8. 27. 10. 28. 15. 29. 9. 30. 4. 31. 10. 32. 7. 33. 2. 34. 3. 35. 12. 36. 17. 37. 6 38. 3. 39./15. 40. 9. 41.0. 42. ½. 43. 1/3 . 44. 2/3. 45.1/5. 46.7/8. 47. 3/5. 48. 2/3. 49. 17/2. 50. 17/3. 51.5/4. 52. 4½. 53.  58/5. 54. 4½. 55. 35/3. 56. 63/8. 57.1/10. 58. 8/3. 59. 23/6. 60. 7/2.
ILLUSTRATIVE EXAMPLES: Solving equations containing decimals
1. Solve the equation .3x =1.2.
Solution. We can solve this equation directly by dividing both members by .3, obtaining the solution x =4. An alternate method of solution is frequently used, however, in which the decimals are first eliminated by multiplying both sides of the equation by 10 (or by a power of 10):
Given equation .3x = 1.2
Multiplying by 10 (multiplication axiom) 3x =12
Dividing by 3 (division axiom) x = 4
2. Solve the equation 1.5x 16.8 = 2 .85x.
Solution. To eliminate the decimals, we must multiply
.85x by 100; hence we multiply both members by 100. This is the same as multiplying every term by 100:
Given equation 1.5x 1.8 = 2 .85x
Multiplying by 100 (multiplication axiom) 150x 1680 = 200 85x
Transposing (addition axiom) 150x + 85x = 200 + 1680
Collecting 235x = 1880
Dividing by 235 (division axiom)
x = 8.
EXERCISES: Solving equations containing decimals
Solve each of the following equations:
Answers : 1. 270. 2. 6. 3. 5. 4. 90. 5. 19. 6. 10. 7. 20,000. 8. 6400.
9. 8. 10.3/4 11. 6. 12. 760.
Summary. The following methods are applied in solving first degree equations:
1.
Write the equation.
2.
Collect like terms in each member of the equation.
3. Using the rule for transposing terms, rewrite the equation so that every term containing the unknown letter appears on one side of the equality sign and every term that does not contain the unknown letter appears on the other side of the equality sign. (Usually we transpose the terms containing the unknown letter to the left side of the equation, but this is not necessary. We can transpose the terms containing the unknown letter to the right and the other terms to the left.)
4. Collect like terms.
5.
Divide both members of the equation by the coefficient of the unknown letter.
ILLUSTRATIVE EXAMPLES: Solving first degree equations
1. Solve the equation 4x 18 + 5x = 22 + 15x 10,
Solution. Given equation 4x 18 + 5x = 22 + 15x 10
Collecting like terms in each member 9x 18 = 15x + 12
Transposing terms containing x to the left and terms not containing x to the right 9x 15x = 12 + 18
Collecting like terms 6x = 30
Dividing both members by 6 x = 5 Answer
Check. Given equation 4x 18 + 5x = 22 + 15x 10
Substituting 5 for x 4(5) 18 +5(5) = 22 + 15(5) 10
Simplifying 20 18 25 = 22 75 10
63 =63 Check
2. Solve the equation 6x + 17 3x = 25 x 8.
Solution. Given equation 6x + 17 3x = 25 x 8
Collecting like terms in each member 9x + 17 = 17 x
Transposing terms containing x to the left and terms not containing x to the right 9x + x = 1717
Collecting like terms 8x =0
Dividing both members by 8 x = 0 (Zero divided y 8 is zero.)
Check Given equation 6x + 17 3x = 25 x 8
Substituting 0 for x 6(0) + 17 3(0) = 25 0 8
Simplifying 0 +17 0 = 25 8
17 = 17 Check
EXERCISES: Solving flrst degree equations
Solve each of the following equations and check your answer:
Answers: 1.12. 2. 3. 3. 20. 4. 3. 5. 8. 6. 7. 7. 3. 8. 1 . 9. 1. 10. 2. 11. 5. 12. 7. 13. 5. 14. 2. 15. 5. 16. 64/7. 17. 2. 18. 1. 19. 9/2 20. 45/7. 21. 20/3 . 22. 42. 23. 1. 24. 11/5. 25. 3/2 . 26. 4. 27. 9000. 28. 3.
Equations containing parentheses. We have learned to remove parentheses in algebraic expressions such as 2(x +4), 3(x 5), etc. by multiplying each term within the parentheses by the monomial outside; that is
2(x +4) is equal to 2x +8
and 3(x 5) is equal to 3x + 15
In solving equations containing parentheses, we first remove the parenheses and then we solve the equation by the methods we have been using.
ILLUSTRATIVE EXAMPLE: Solving an equation containing parentheses
Solve the equation 2(2x + 3) 6(x 2) = 24.
Solution. Given equation 2(2x +3) 6(x 2) = 24
Removing parentheses 4x + 6 6x + 12 = 24
Collecting like terms in the left member 2x +18 = 24
Transposing the term 18 to the
right member 2x = 24 18
Collecting terms in the right member 2x = 6
Dividing both members by 2 x = 3 Answer
Check. Given equation 2(2x +3) 6(x 2) = 24
Substituting 3 for x 2(6 +3) 6(3 2) = 24
Simplifying 2(3) 6(5) = 24
6 + 30 = 24
24 = 24 Check
EXERCISES: Solving equations containing parentheses
Solve each of the following equations and check your answer:
Answers : 1.1. 2.3. 3. 7. 4. 3. 5.0. 6.  4/3. 7. 8/5 . 8. 3/2. 9. 3. 10. 11. 11. 1. 12. 2. 13.0. 14. 13. 15. 2. 16. 10. 17.0. 18. 6. 19. 1. 20. 1/10
Solve x: (2 + x) (3 x) = 7
(2 + x) (3 x) = 7
2 +x 3 +x = 7
2x 1 =7
2x = 6
x = 3; Answer
