Notes of Lessons: Algebra, Class IV

[We have thought that it might be of use to our readers (in their own families) to publish from month to month during the current year, Notes of Lessons prepared by students of the House of Education for the pupils of the Practising School. We should like to say, however, that such a Lesson is never given as a tour de force, but is always an illustration or an expansion of some part of the children’s regular studies (in the Parents’ Review School), some passage in one or other of their school books.—Ed.]

Group: Mathematics • Class IV • Time: 30 minutes

By J. H. Morris
The Parents’ Review, 1906, pp. 143-146

Objects

I. To introduce simple equations.

II. To stimulate interest in algebra by showing how easily many problems may be solved.

III. To encourage accuracy.

IV. To increase the power of reasoning.

Lesson

Step I.—Explain that an “equation” asserts that two expressions are equal, but that we do not usually employ the word equation in so wide a sense. Thus the statement x + 3 + x + 4 = 2x + 7, which is always true, whatever value x may have, is called an “identical equation” or an “identity.”

Step II.—But certain equations are only true for particular values of the symbols employed. Thus 3x = 6 is only true when x = 2, and is called an “equation of condition” or simply an “equation.”

Step III.—Introduce the solution of an equation in a simple concrete instance. A father is four times as old as his son: in 24 years he will only be twice as old. Find their ages.

Let x years be the son’s age.

Then 4x years = the father’s age.

In 24 years the son will be x + 24 and the father 4x + 24 years old. Therefore by first supposition:—

\(4x+24=2(x+24)\)

\(4x+24=2x+48\ \ \ \ \ \ (1)\)

By this equation (1), taken in the abstract, is meant that if x be replaced by a certain number (in this case the age in years of the son) then the left-hand side of (1) can be transformed into the right by means of the laws of arithmetic or algebra. The object, therefore, is to find the value of x which will make (1) an identity. This is called “solving the equation.”

Although the value of x, which makes (1) an identity, is not known, we proceed to transform the equation on the hypothesis that x has such a value.

For, 4x + 24 = 2x + 48, and if equals be taken from equals the remainders are equal.

∴\(4x+24-(2x+48)=4x+24-(2x+48)\)

\(2x=24\)

But if equals be divided by equals the quotients are equal, therefore dividing by 2 : x = 12

∴ the “solution” is x = 12, i.e., the boy was 12 years old.

By substituting x = 12 in (1) we may show that 4 x 12 + 24 = 2 x 12 + 48, which is described as verifying the solution.

Step IV.—Explain what conclusions may be deduced from this problem.

(1) That the process of solving an equation consists in finding a value for the unknown quantity, such as will make the equation an actual identity.

(2) That in every transformation of the equation we suppose the unknown quantity to have values such that the equation is an identity.

(3) That in each step of the process of solution we deduce from a previous equation (A) another (B) which has all the solution or solutions of (A).

Step V.—Consider the equation

\(5x=10\)

Dividing both sides by 5 we get

\(x=2\)

Similarly if

\(\frac{x}{2}=-8\)

Multiplying both sides by 2 we get

\(x=-16.\)

Step VI.—To solve 3x + 15 = x + 25. Here the unknown quantity occurs on both sides of the equation, but show that any term can be transposed from one side to the other by simply by changing its sign. For, subtract x from both sides of the equation and

\(3x+15-x=25\)

Subtract 15 from each side

\(3x-x=25-15\)

Thus it is seen that + x has been removed from one side and appears as – x on the other, and that + 15 has been removed from one side and appears as – 15 on the other. Therefore, we have the rule that any term may be transposed from one side of the equation to the other by changing its sign.

Step VII.—From the last step it follows that we may change the sign of every term in an equation, since this is equivalent to transposing all the terms, and then making the right and left hand members change places. For:—

\(-5x+10=-3x-4\)

\(Transposing\ \ \ \ \ 3x+4=5x-10\)

\(or\ \ \ \ \ 5x-10=3x+4.\)

Step VIII.—Let the girls work out on the blackboard:—

\((1)\ \ \ \ 2x+3=16-(2x-3)\)

\((2)\ \ \ \ 8(x-1)+17(x-3)=4(4x-9)+4\)

Help them to solve

\((3)\ \ \ \ \frac{x}{2}-3=\frac{x}{4}+\frac{x}{5}\)

by showing them that it is convenient to clear the equation of fractions by multiplying both sides by the L. C. M. of the denominators. Thus, multiplying by 20

\(10x-60=5x+4x.\)

Step IX.—Draw from the girls the rules for the solution of an equation. First, if necessary, clear of fractions, secondly, transpose all the terms containing the unknown quantity to one side of the equation and the known quantities to the other. Then collect the terms on each side and finally divide both sides by the co-efficient of the unknown quantity, and the value required is obtained.

Step X.—If time let them work two or three easy problems involving simple equations.

(1) One number exceeds another by 5 and their sum is 29; find them.

(2) What two numbers are those whose sum is 58 and difference 28.