First Course in Algebra, Βιβλίο 1Macmillan, 1919 - 334 σελίδες |
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Σελίδα 7
... sides by 2. This gives ( see Fig . 3 ) the new equation x = 12 . Therefore , the number sought is 12. Ans ... sides , we find 2 x = 24 . Dividing both sides of last equation by 2 , gives x = 12. Ans . We shall now solve another example ...
... sides by 2. This gives ( see Fig . 3 ) the new equation x = 12 . Therefore , the number sought is 12. Ans ... sides , we find 2 x = 24 . Dividing both sides of last equation by 2 , gives x = 12. Ans . We shall now solve another example ...
Σελίδα 8
... sides , we find Dividing both sides by 3 , we have 3x = 5 x = 13 , the number of ounces each ball weighs . Ans . CHECK . Whenever an answer does what the problem demands of it , it is said to check . In this problem , the answer 13 ...
... sides , we find Dividing both sides by 3 , we have 3x = 5 x = 13 , the number of ounces each ball weighs . Ans . CHECK . Whenever an answer does what the problem demands of it , it is said to check . In this problem , the answer 13 ...
Σελίδα 9
... One team won three games and the other two teams divided the remaining victories equally between them . How many games did each of the other teams win ? 7. Adding to Both Sides of an Equation . If I , § 6 ] 9 LITERAL NUMBERS.
... One team won three games and the other two teams divided the remaining victories equally between them . How many games did each of the other teams win ? 7. Adding to Both Sides of an Equation . If I , § 6 ] 9 LITERAL NUMBERS.
Σελίδα 10
... sides . This can be done since the balance is not thereby disturbed . Adding 1 to both sides , we find 3x = 15 . Dividing both sides by 3 , we have x = 3 . Ans . ORAL EXERCISES In order to solve certain of the following equations , it ...
... sides . This can be done since the balance is not thereby disturbed . Adding 1 to both sides , we find 3x = 15 . Dividing both sides by 3 , we have x = 3 . Ans . ORAL EXERCISES In order to solve certain of the following equations , it ...
Σελίδα 11
... Sides of an Equation . We have already seen how we may divide both sides of an equation by the same number . Thus , if 2x = 10 , we simply divide both sides by 2 to get x , obtaining x = 5 . Ans . Likewise , we may at any time multiply both ...
... Sides of an Equation . We have already seen how we may divide both sides of an equation by the same number . Thus , if 2x = 10 , we simply divide both sides by 2 to get x , obtaining x = 5 . Ans . Likewise , we may at any time multiply both ...
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a²+2ab+b² ab+b² added algebra Appendix arithmetic Axiom called cents CHAPTER check your answer circle coefficient common denominator common factor completing the square cube diameter divisor Draw the graph equal EXAMPLE EXERCISES Solve figure Find the H. C. F. Find the numbers Find the value following equations following exercises following expressions fraction further exercises given equation gives highest common factor HINT illustrated inches length lowest common means miles an hour minuend monomial multiply negative numbers numerator and denominator ORAL EXERCISES parentheses perfect square polynomial quadratic quadratic equation quotient radical radicand radius ratio represents result Similarly Simplify simultaneous equations SOLUTION Solve the equation square root subtract subtrahend SUPPLEMENTARY EXERCISES surd Transposing triangle trinomial weight wheel write WRITTEN EXERCISES x²y² yards