is a square, are called quadratic equations, or equations of the second degree. Those in which the highest power is a cube, are called cubic equations, or equations of the third degree. So, also, we have biquadratic equations, or equations of the fourth degree; equations of the fifth, sixth, degree. = Thus, x"+px”―r, is an equation of the nth degree. nth In general, the degree of an equation is determined by the highest of the exponents with which the unknown quantity is affected. (97.) Numerical equations are those which contain only particular numbers, with the exception of the unknown quantity, which is always denoted by a letter. Thus, x+4x=3x+12 is a numerical equation. Literal equations are those in which the known quantities are represented by letters, or by letters and numbers. Thus, x+px'+qx=r x*—3px3+5qx2=5 are literal equations. To solve an equation is to find the value of the unknown quantity, or to find a number which, substituted for the unknown quantity in the equation, renders the first member identical with the second. The difficulty of solving equations depends upon their degree, and the number of unknown quantities. We will begin with the most simple case. SIMPLE EQUATIONS CONTAINING BUT ONE UNKNOWN QUAN TITY. (98.) The various operations which we perform upon equations in order to deduce the value of the unknown quantities, are founded upon the following principles: 1. If to two equal quantities the same quantity be added, the sums will be equal. 2. If from two equal quantities the same quantity be subtracted, the remainders will be equal. 3. If two equal quantities be multiplied by the same quantity, the products will be equal. 4. If two equal quantities be divided by the same quantity, the quotients will be equal. (99.) The unknown quantity may be combined with the known quantities in the given equation by the operations of addition, subtraction, multiplication, or division. We shall consider these different cases in succession. I. The unknown quantity may be combined with knowu quantities by addition. Let it be required to solve the equation x+6=24. If from the two equal quantities, x+6 and 24, we subtract the same quantity 6, the remainders will be equal, according to the last Article, and we shall have x+6-6=24-6, So, also, in the equation =18, the value of x required x+a=b, subtracting a from each of the equal quantities, x+a and b, the result is x=b-a, the value of a required. (100.) II. The unknown quantity may be combined with known quantities by subtraction. Let the equation be x-6=24. If to the two equal quantities, x-6 and 24, the same quantity 6 be added, the sums will be equal, according to Art. 98 and we have x-6+6=24+6, or x=30, the value of x required. So, also, in the equation x-a=b, adding a to each of these equal quantities, the result is x=b+a, the value of x required. From the preceding examples, it follows that We may transpose any term of an equation from one member to the other by changing its sign. We may change the sign of every term of an equation with out destroying the equality. This is, in fact, the same thing as transposing every term in each member of the equation. If the same quantity appear in each member of the equation affected with the same sign, it may be suppressed. (101.) III. The unknown quantity may be combined with known quantities by multiplication. Let the equation be 6x=24. If we divide each of the equal quantities, 6x and 24, by the same quantity 6, the quotients will be equal, and we shall have So, also, in the equation 24 x= 6' =4, the value of x required. ax=b, dividing each of these equals by a, the result is When the unknown quantity is multiplied by a known quantity, the equation is solved by dividing both members by this known quantity. (102.) IV. The unknown quantity may be combined with known quantities by division. Let the equation be x =24. 6 x If we multiply each of the equal quantities, and 24, by the 6 same quantity 6, the products will be equal, and we shall have x=144, the value of x required. multiplying each of these equals by a, the result is x=ab, the value of a required. From this it follows, that When the unknown quantity is divided by a known quantity, the equation is solved by multiplying both members by this known quantity. (103.) V. Several terms of an equation may be fractional. Let the equation be x 2 4 2 3 5 Multiplying each of these equals by 2, the result is 4-8 Multiplying each of these last equals by 3, we obtain We might have obtained the same result by multiplying the original equation at once by the product of all the denominators. Thus, multiplying by 2×3×5, we have multiplying successively by all the denominators, or by a ce at once, we obtain Canceling from each term the letter which is common to its numerator and denominator, we have cex abe+acd, an equation clear of fractions. Hence it appears that An equation may be cleared of fractions by multiplying each member into all the denominators. (104.) From the preceding remarks, we deduce the folTowing RULE FOR THE SOLUTION OF A SIMPLE EQUATION CONTAINING ONE UNKNOWN QUANTITY. 1. Clear the equation of fractions, and perform in both members all the algebraic operations indicated. 2. Transpose all the terms containing the unknown quantity to one side, and all the remaining terms to the other side of the equation, and reduce each member to its most simple form. 3. Divide each member by the coefficient of the unknown quantity. EXAMPLES. 1. Given 5x+8=4x+10, to find the value of x. Transposing 4x to the first member of the equation, and 8 to the second member, taking care to change their signs (Art. 100), we have 5x-4x-10-8.. Uniting similar terms, x=2. х In order to verify this result, put 2 in the place of x wherever it occurs in the original equation, and we shall obtain That is, 5x2+8=4x2+10. 18=18, or an identical equation, which proves that we have found the correct value of x. х 2. Given x-7=+, to find the value of x. 5 3' Multiplying every term of the equation by 5 and also by 3, in order to clear it of fractions (Art. 103), we obtain To verify this result, put 15 in the place of x in the original equation, and we have |