CodalSearch this book — or all of Codal…⌘K
nydus/The First Steps in AlgebraPublic
Page 48 of 227
Table of Contents

Integral Compound Expressions.

&2a33a2+2a1a32a23a+5&=2a3a33a22a2+2a3a1+5&=a35a2a+4.

This process is more easily performed by writing the subtrahend below the minuend, mentally changing the sign of each term in the subtrahend, and adding. r*3cr2a3&&3a2&+&2a&&1a3&+&2a2&+&3a&&5\hlinea3&&5a2&&a&+&4

By changing the sign of each term in the subtrahend, the coefficient of a3 will be 21, or 1; the coefficient of a2 will be 32, or 5; the coefficient of a will be 23, or 1; the last term will be 1+5, or 4.

Again, suppose it is required to subtract x52ax43a2x3+4a3x2from4a3x22a2x35ax4. Here terms which are alike can be written in columns, as before: r*3cr&&5ax4&&2a2x3&+&4a3x2x5&&2ax4&&3a2x3&+&4a3x2\hlinex5&&3ax4&+&a2x3&&

There is no term of x5 in the minuend, hence the coefficient of x5 in the result will be 01, or 1; the coefficient of ax4 will be 5+2, or 3; the coefficient of a2x3 will be 2+3, or +1; the coefficient of a3x2 will be 4+4, or 0, and therefore the term a3x2 will not appear in the result.

Exercise 19.

Subtract:

  1. a2b+3c from 2a3b+4c.
  1. a3b5c from 3a5b+c.
  1. 2x4y+6z from 4xy2z.
  1. 5x11y3z from 6x7y+2z.
  1. abacbc+bd from ab+ac+bc+bd.
  1. 3ab+2ac3bc+bd from 5abac+bc+bd.
  1. 2x3x25x+3 from 3x3+2x23x5.
  1. 7x25x+1a from x3x+1a.
  1. 7b3+8c315abc from 9b3+3abc7c3.
  1. x4+x5x3+5 from 72x23x3+x4.
  1. a3+b3+c33abc from 3abc+a32b33c3.
  1. 2x45x2+7x3 from x4+22x3x2.
  1. 1x5x+x4x3 from x4+1+x+x2.
  1. a3b3+3a2b3ab2 from a3+b3a2bab2.
  1. a2bab23a3b3b4 from b45a3b32ab2+a2b.
48