Of the solution of equations of the first degree, having only one unknown quantity. Now, the unknown quantities are found mixed with the known, in three ways: 1st.-By Addition and Subtraction, as in the Equations, - 2ndly. By Addition, Subtraction, and Multiplication, as in the Equations, 3rdly. By Addition, Subtraction, Multiplication, and Division, as in the Equations, 11 mx + cx d= + P In taking the first equation above, for an Example, to solve, (or to find the value of x) it will be necessary to make terms pass from one side of the equation to the other; and, in order to perform this, the terms changing sides must also have their signs changed, the reason of which will appear evident, because changing the signs of a term, when removing it to the other side, is equivalent to subtracting or adding that term to both sides. 1. Thus + 5 = 9 becomes x + x = 9 and it becomes 2x = then 4, - 5, by transposition, by Addition and Subtraction, 2, by taking 2 from both sides. Therefore, a term can be changed from one side of an equation to the other, by also changing its sign, which we call transposition. K It must be remembered, that taking the co-efficient of the unknown quantity from both sides, is the same as dividing the whole equation by it, therefore, we shall use the word division for the future. The next example contains fractions; and, in order to bring the whole equation to a form that each term shall be a whole number, we must clear it of fractions. This can be performed, either by multiplying the entire equation by each denominator or by finding a common multiple, and then multiplying the whole equation by it, at the same time dividing the product of the numerator of each fraction and the common multiple, by its own denominator. Next, 60x33x= 324-288, (by multiplying by next denominator, Then we have 27x= 36, by Subtraction, 36 And x = by dividing by the co-efficient of x. 27' anqx + bcnqx — bmqx = bdnq + bnp, by × by q, Then it becomes, (ang + bcnq — bmq)x = bdnq + bnp, by using the (), -- Fractions can be also cleared, by multiplying by a common multiple, and then dividing each fraction by its denominator, as before remarked, thus, 70x+420=84x + 1575—75x, by multiplying by 105, next, 70x-84x+75x-1575-420, by transposition, then, 61x 1155, by incorporation, 1155 and finally, x = by dividing by 61. " 61 8. Find the value of x, in the following example, It becomes, 5 X by a com40y5y+5=10y+4y-4+240 mon multiple, 40y5y-10y-4y= 240—4—5, by transposition, then, 21y 231, by Addition and Subtraction, and y 11, by dividing by the co-efficient of y. In this example, the sign minus stands before the fraction, therefore the sign or signs must be changed, when the quantity is brought down; the reason is evident, because it means that 4y is less by y 1, and therefore, according to the Rule for Subtraction, the sign must be changed. The same Rule is to be observed in all cases, when stands before a fraction. 9. What number, when added to of itself, will be equal to 20? 1 Here we take x for the unknown quantity, therefore we have, x + x = 20, which becomes, 3x + x = 60, by clearing fractions, then, 4x = 60, by Addition, and, x = 15, by dividing by the co-efficient of x. Wherefore, we have 15, added to of itself, equal to 20. 10. What sum of money, being added to of itself, will be equal to £30 ? Here we have x + Zx = £30, it becomes, 9 + 2x = £270, by clearing fractions, and 11x = £270, by Addition, Therefore, a = £24 10s. 101d., the Answer. 11.-In the following example, the terms are proportional, therefore the product of the means shall be equal to the product of the extremes. 126, by Multiplication, 56x+7x= 70 126, by transposition, next, 70+7x=56x and, and, 49x= 196, by Addition and Subtraction, therefore, x = 4, by Division. Of Questions containing two or more unknown quantities of the first degree. 1.—In the example, 12x + 7y = 46, We have 4x X =16, by Subtraction. |