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#M-2 Origin of Algebra


   

(M–1a) Algebra Proficiency Drill

Section (M–1) gave you the principles of simple algebra. These excercises will give you practice in applying them.

Do not expect anything profound or interesting--this is just a drill, like the finger excercises you do if you wish to master a musical instrument, or the parallel parking practice before your driver's test. Do them all--do not leave any out!

(1) Isolate x in each equation and find its value, following the
        rule that "when equal operations are performed on both
        sides of an equation, the result is still equal."

To the right of each problem are instructions for solving it, a list of the required operations. The sequence in which they are applied reads from left to right. Write down a new equation for each step.

The notation of the instructions is as follows. For operations applied to both sides:

      (+2)     add 2
      (–6)     subtract 6
      (*3)     multiply by 3
      (/5)     divide by 5
      For other operations:
      (+/–)     add or subtract terms wherever you can
      (*)     multiply terms wherever you can
Note: Not every equation has a unique solution. Some can be identities, e.g. 2(x+1) = 2x + 2, which hold for any value of x. Some can be mis-identities, e.g. 2(x+1) = 2x + 3, which hold for no value, because they reduce the the impossible requirement 2 = 3.

5+x = 7 (–5)
x/2 = 3 (*2)
x/3 + 4 = 8 (–4)(*3)
4x – 5 = 15 (+5)(/4)
3x + 6 = 5x (–3x)(/2)
6x + 4 = 1.5x + 13 (–1.5x)( –4)(/4.5)
15x – 2 = 6x + 16 (–6x)(+2)(/9)
21x – 3 = (7x+9)/2 (*2)(–7x)(/5)(/3)
Note multiplication by (–1) reverses all signs on both sides!

10 – 3x = –2 (–10)(*( –1))(/3)
1/(x+1) = 2/(x+3) (*(x+1))(*(x+3)(*)( –x)( –2)
(x+2)(x+1) = (x+7)(x–1) (*)(+/–)( –x2)( –2)( –6x)(*( –1)(/3)

                (2) Same types of equations, but now without instructions:

7 + 2x = 13
15 + 7x = 1
4x – 3 = 2x
5x – 3 = 1 – 2x

(x/2)+5 = (x/3)+6
5x – 20 = x+8
(x+6)/2 = 2x – 21
(2x–3)/(4x–3) = 1

2/(3–x) + 1/(2+x) = 0
(x+10)/(3x+5) = 2
(11x+1)/(6x–2) = 2
(x+2)(x+3) = (x+1)(x+7)

                (3) Is anything wrong with these equations? And if so, what?

(15x–5)/(3x–1) = 5
4(3x–5) = 2(6x+7)
5(x–3) = 7x – 15

                (4) The relations below all contain both x and y. :
                      Express y in terms of x, for instance

                                x + y = 7                 Answer: y = 7 – x

                All operations are indicated as before, but watch out:
                      the problems include one badly-posed example.

2x + 3y = 7 (–2x)(/3)
(3y+1)/(x+2) = –2 (*(x+2))( –1)(/3)
(4x – 5y –2) = 13 (+2)( –4x)(* –1)(/5)
(3y + x + 6)/(y–x+2) = 2 (*(y–x+2)) ( –2y)( –x)( –6)
(y–4x)/(y+x+6) = 1 (*(y+x+6))( –y)( –x)(*( –1))(/5)
(15x–2y+6) = (y–6) (–y)( –15x)( –6)(*(–1))(/3)

                (5) Below are pairs of equations involving two unknown numbers, x and y.

Solve each of the sets of equations twice. Solve once by
    (a) expressing y of one equation in terms of x, then
    (b) substituting the expression for y in the other equation, then
    (c) deriving x, and finally
    (d) putting that value in the substituted expression and getting y.
Then solve again by exchanging the roles--express x in one equation, substitute that expression in place of x in the other, derive y, then derive x too.

(a) x+3y = 5 2x – y = 3
(b) x+y = –1 3x+4y = 2
(c) x+34 = 15 3x+y = 5

                (6) Given two equations, marked here I and II, you may also

multiply or divide each equation by any number. You may furthermore add one equation to the other, or subtract it: because the quantities you add or subtract to both sides are equal, what is left is also a valid equality.

Here are some examples--the first is worked out, for the rest just the steps are given. In this notation, II always means the 2nd equation at this stage of the calculation--it need not be the original 2nd equation but could have been (say) multiplied by 6. If the instructions just name an operation, it is to be applied to the equation obtained in the preceding step.

                5x – 12y = 2 (I)
                –3x + 2y = 4 (II)

(II*6)
                –18x + 12y = 24
(I+II)
                5x – 18x = 26                 (12y and –12y cancel)
(+)
                –13x = 26
(*(–1))
                13x = –26
(/13)
                x = –2
To find y, put this in (I)

–10 – 12y = 2
–12y = 12
12y = –12
y = –1
To check your result, see if (II) also holds

(–3)( –2) + 2(–1) = 4?     (4 = 4, result OK)

In what follows, only the steps for obtaining one variable are given. On your own, derive also the other variable and check the result.

(a) 3x+4y = 19         (I) 5x + 2y = 13         (II) (II*2)(II – I)(/7)
(b) 2x+3y = 5         (I) 3x+2y = 0         (II) (I*3)(II*2)(I–II)(/5)
(c) 4x+3y = 16         (I) 3x+5y = 12         (II) (I*3)(II*4)(II–1)(/11)
(d) 2x+6y = 34         (I) 5x+2y = 46         (II) (II*3)(II–1)(/13)
(e) 3x+5y = 31         (I) 2x–3y = 11         (II) (I*2)(II*3)(I–II)(/19)

                (7) Now solve on your own:

(a) 2x–3y = 1 (I) 3x+2y = 21 (II)
(b) 5x–2y = 20 (I) 10x + 3y = 5 (II)
(c) 6x + 2y = 8 (I) 5x + 4y = 16 (II)
(d) 3x – 4y = 1 (I) 2x + 3y = –5 (II)


(8)     The two most widely used scales for measuring temperature are those introduced by Farenheit (used in the US) and by Celsius (the centigrade scale, used in the rest of the world and by scientists).
        (a)     If the temperature F degrees Farenheit corresponds to C degrees centigrade, then

    F = x C + y

    Find x and y, given that 100 degrees centigrade (boiling point of water) corresponds to 212 degrees Farenheit, and 0 degrees centigrade (freezing point of water) corresponds to 32 degrees Farenheit.

        (b)     Using the solution of (a)--at what temperature C = F?


Next Stop:   #M–2    Al-Khorezmi and the Dawn of Algebra

Author and Curator:   Dr. David P. Stern
     Mail to Dr.Stern:   stargaze("at" symbol)phy6.org .
Last updated 25 November 2001

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