International projectionist (Nov-Dec 1933)

12 INTERNATIONAL PROJECTIONIST November 1933 will have 5 — 3 = 2 + 6, which when solved gives us 2 = 8 which is obviously not true. Equations are of use to us in solving for unknown quantities, and are most often set up from statements similar to the following: Mr. A has $5.30; Mr. B has $16.20; but Mr. C's holdings are not known. However, we do know that between them, the three men have a total of $53.00. To determine the amount of money held by Mr. C, we will set up an equation in which the letter "x" will designate this amount, and then solve for x. In this equation, we will express Mr. A's $5.30 by the letter "a", Mr. B's $16.20 by the letter "b", and the total $53.00 by the letter "d". From the statement of the problem given above, we know that a+b+x=d In order to put this equation in such a form that it may be easily solved, we will first (by rule b) subtract "a" from each side: a — a+b+x=d — a solving, b+x=d — a Similarly, we subtract b from each side of the equation, to put it in the form, x=d — a — b Now that we have the equation in the proper form for solution — that is, with the unknown on one side and all the known quantities on the other — we substitute the numerical values for their corresponding letters, and solve for x, as follows : x= 53.00— 5.30— 16.20 or x, Mr. C's holdings =$31.50. While in this case the algebraic solution of the problem may appear to be longer and more complicated than would a corresponding arithmetical solution, there are many examples which might be given wherein the use of algebra is absolutely necessary in order to arrive at a solution, and others wherein the algebraic solution will be very much shorter than would be the arithmetic. However, in order to prepare for the solution of these more complicated equations, it is necessary to become familiar with and adept at handling the simpler forms of equations. There are many types and kinds of algebraic equations, but before taking up a study of the more complicated types we will first consider the "simple" or "linear" equation. Linear Equations An equation, to be linear, must fulfill certain requirements. It may contain one or more unknowns, but must contain only unknowns of the first power (that is, no unknown may be squared, or cubed or raised to a higher power), which presupposes that no unknown can appear as an exponent, it must not have more than one unknown in the same term (that is, no unknown may be multiplied or divided by another unknown) ; and lastly, that it must be in such a form that by treating it according to one or more of the six fundamental rules given above, all of the unknowns may be removed from the denominator of any fraction appearing in the equation. In order that you may become familiar with the difference between a linear equation and those which are not of the linear type, the following examples are linear : 3x + 4y = 27. (Unknowns: x, y.) x —=15. y (Unknowns: x, y which can, by multiplying both sides of the equation by "y", be expressed as x=15y.) 5x + 3x — x — 12 = 2x. (Unknowns: x.) The following are not linear equations: 3xa+y=27 + x. (an unknown has the exponent 2.) xy — x = l. (x and y, both unknowns, occur in the same term.) ax+3b=l. (x, an unknown, occurs as an exponent.) 1 2x-{ =x+l. x— 3 (the equation cannot be legitimately changed so as to get rid of the unknown in the denominator. ) Within the classification of those equations which are linear, there are two sub-groups: (1) equations which have only one unknown and (2) equations which have more than one unknown. Considering first those equations which have only one unknown, let us solve a simple equation to determine the value of the unknown quantity. Let us take the equation: 5x=x+12 First (by Rule b) above, we may subtract x from each side of the equation, which will in effect change the equation so that all terms containing the unknown (x) will be on the same side of the equals sign: 5x — x = x — x + 12 or simplified, 4x=12 Then (by Rule d) above, dividing both sides of the equation by 4 in order to arrive at a value for x, we find that 4x=12 — — , or x = 3 4 4 In order to check the correctness of this answer (it is always advisable to check an algebraic computation when ever possible), we substitute 3 in the original equation wherever x appears, as follows : 5x=x+12 5 3=3+12, which upon solving becomes 15=15 The value 3 is known as a "root" of the equation, and mathematically is said to "satisfy the equation". Should this value of 3, which we have obtained for x have been incorrect, this would have been apparent upon checking the solution. Substituting an incorrect root in the original equation would result in an inequality. For instance, let us substitute the value x=4 in the equation and solve: 5x = x + 12, substituting 5 4=4+12, which upon completing the indicated multiplication and addition becomes 20 = 16, which is obviously incorrect Consequently, 4 is not a correct root of the equation, inasmuch as the equation is not satisfied when the value 4 is substituted for the unknown x. 'Transposing' You will notice that in preparing the example above for solving, we subtracted x from each side of the equation. In effect, this operation is the same as if we had moved the x from the right-hand side of the "equals sign" to the lefthand side, changing its sign at the same time. Had this been done, the second step of our solution would have been 5x=x+12 5x — x=12, which when simplified becomes 4x=12 Moving any term of an algebraic equation from one side of the "equals sign" to another is known as "transposing" and oftentimes results in considerable simplification of the algebraic operations. Any term may be transposed from one side of the equals sign to the other, provided its sign is changed. It should be borne in mind, however, that transposing a term is in effect actually adding or subtracting the same term to both sides of the equation. (To Be Continued) Problems, Article III: 10. In the following, solve for the unknown, showing each step in order to indicate the proper order of separation. (a) x=10 7— 72+1400 + 32— 49 + 20 20 (b) y=20 + 7 + (7 + 2)— (4— 2)' 11. Subtract: (—10 + 1—2 + 5) from (7 + 10—3+2) 12. (a) Multiply: —5 by 7. (b) Divide: —32 by —4. 13. Solve the following equation to obtain the value of "x": 4x + 7=2x— 3