EXERCISES Perform the indicated operations and bring the results into their simplest Solution: 4 ab (2,4)2 — 1 2.24 3 + 3 + 3 1 + 3 + 24 10. + 26 ab ac bc +12 1-4. 2+素+景 a + b . 1 + + 12. a 26 3b + 13. b с с 2x a2 - 1 x CHAPTER IV EQUATIONS 52. Introduction. An equation is a statement of equality between two expressions. We assume the following AXIOM. If equals be added to, subtracted from, multiplied by, or divided by equals, the results are equal. As always, we exclude division by zero. In dividing an equation by an algebraic expression one must always note for what values of the letters the divisor vanishes and exclude those values from the discussion. 53. Identities and equations of condition. Equations are of two kinds : First. Equations that may be reduced to the equation 1 = 1 by performing the indicated operations are called identities. = are equations of this type. In identities the sign is often replaced by =. It should be noted that identities are true whatever numerical values the letters may have. = = 1, Second. Equations that cannot be reduced to the form 1 but which are true only when some of the letters have particular values, are called equations of condition or simply equations. Thus x 2 cannot further be simplified, and is true only when x has the value 2. Also x 2 a is true only when x has the value 2 a or a has the value If in 2 this equation x is replaced by 2 a, the equation of condition reduces to an identity. The number or expression which on being substituted for a letter in an equation reduces it to an identity is said to satisfy the equation. Thus the number 5 satisfies the equation x2 the equation (x − 3) (x + 4) = 0. 241. The number satisfies The process of finding values that satisfy an equation is called solving the equation. The development of methods for the solutions of the various forms of equations is the most important question that algebra considers. In an equation in which there are two letters it may be possible to find a value which substituted for either will satisfy the equation. Thus the equation x - 2 a = 0 is satisfied 2a = 0 is satisfied if x is replaced by х 2 2 a, or if a is replaced by In the former case we have solved for x, that is, have found a value that substituted for x satisfies the equation. In the latter case we have solved for a. In any equation it is necessary to know which letter we seek to replace by a value that will satisfy the equation, that is, with respect to which letter we shall solve the equation. The letter with respect to which we solve an equation is called the variable. Values which substituted for the variable satisfy the equation are called roots or solutions of the equation. When only one letter, i.e. the variable, occurs in an equation, the root is a number. When letters other than the variable occur, the root is expressed in terms of those letters. 54. Linear equations in one variable. An equation in which the variable occurs only to the first degree is called a linear equation. To solve a linear equation in one variable we apply the following RULE. Apply the axiom (§ 52) to obtain an equation in which the variable is alone on the left-hand side of the equation. The right-hand side is the desired solution. To test the accuracy of the work substitute the solution in the original equation and reduce to the identity 1 =1. Since the result of adding two numbers is a definite number, and the same is true for the other operations used in finding the solution of a linear equation, it appears that every linear equation in one variable has one and only one root. When both sides of an equation have a common denominator, the numerators are equal to each other. This appears from multiplying both sides of the equation by the common denominator and then canceling it from both fractions. 10. 7 5x + 10 + 8x −7+3x=x. 12. {[(x+5)-10]+3)-8=0. - 13. (1+6x)2+(2+8x)2=(1+10x)2. 14. 5=3x+(x+3)- (11x - 37). 15. 2(x+5) (x + 2) = (2 x + 7) (x + 3). 16. (7} 2 – 2}) – [4} – } (3} – 5x)]=18}. 18. (a - b) (x - c) + (a + b) (x + c) = 2 (bx + ad). 19. 2x3(5x) + (4x) (3x-16)= 0. 20. 5x-2= x + J x + F x + 1 % x + } } x + } }x. 22. 12.9x1.45 x 3.29.99 x 11 x + .32 = 0. |