# Lin­ear equa­tions

Jamaica Gleaner - - YL - CLE­MENT RAD­CLIFFE Con­trib­u­tor

AS WE pur­sue exam-type ma­te­rial, I ex­pect that you are mak­ing greater ef­fort to un­der­stand what is pro­vided and are prac­tis­ing in earnest.

We will be­gin this week’s les­son by re­view­ing the answers to last week’s home­work. 1. Eval­u­ate: (5n - 2)2

SO­LU­TION

(5n - 2)2 = (5n - 2)(5n - 2) = (25n2 - 10n -10n + 4) = 25n2 - 20n + 4 2. Ex­pand the fol­low­ing: (a) (k + 4) (k - 3) (b) (t - 1) (t + 4)

SO­LU­TION

(a) (k + 4) (k - 3) = k2 + 4k -3k - 12 = k2 + k - 12 (b) (t - 1) (t + 4) = t2 - t + 4t - 4 = t2 + 3t - 4 3. Eval­u­ate: ( 1 - 2p)(5 - 3p)

SO­LU­TION

(1 - 2p)(5 - 3p) = 5 - 10p - 3p + 6p2 = 5 - 13p + 6p2 4. Sim­plify 2y - 1 /6 - y + 3 /5

SO­LU­TION

2y - 1/6 - y + 3/5 The LCM of 6 and 5 is 30. 5(2y - 1) - 6(y + 3) /30 = 10y - 5 - 6y - 18 /30 = 4 y - 23 /30

A BRIEF RE­VIEW OF CHANGE THE SUB­JECT OF THE FOR­MULA

Given the for­mula, A = xy Then, x = A/y is the for­mula ex­pressed in terms of x.

Ex­am­ple: Given the for­mula, V = pi r2h, make r the sub­ject. So­lu­tion : Given that V = pi r2h Then r2 = V/ pi h r = √ V/ pi h We will now con­tinue with lin­ear equa­tion.

LIN­EAR EQUA­TIONS

The in­clu­sion of the EQUAL sign dif­fer­en­ti­ates an EQUA­TION from an al­ge­braic ex­pres­sion. This point is com­monly missed by stu­dents, who some­times at­tempt to solve al­ge­braic ex­pres­sions. Do not make this er­ror. The fol­low­ing points should be noted:

Equa­tions iden­tify ei­ther the re­la­tion­ship be­tween vari­ables or the value of a vari­able.

The value of the vari­able is main­tained by per­form­ing iden­ti­cal op­er­a­tions on both sides of the equa­tion.

The meth­ods of clear­ing brack­ets and sim­pli­fy­ing al­ge­braic ex­pres­sions are usu­ally re­quired to find so­lu­tion of lin­ear equa­tions.

In or­der to solve lin­ear equa­tions, one ap­proach is to sim­plify each side of the equa­tion and then equate both sides. The above is il­lus­trated by the fol­low­ing ex­am­ple:

EX­AM­PLE 1

Given that 2(x - 1) - 3x = 6, then x = a) - 8 b) - 4 c) 4 d) 8

Sim­pli­fy­ing the left-hand side, 2(x - 1) - 3x = 2x - 2 - 3x = - x - 2 Equat­ing - x - 2 = 6 with - x = 6 + 2 x = - 8, the an­swer is a)

EX­AM­PLE 2

Solve 3x/2 + x /4 = 14 Sim­plify the left-hand side: 3 x /2 + x/4 = 6x/4 + x/4 = 7x/4 Equat­ing both sides: 7x /4 = 14 Mul­ti­ply both sides by 4 7x/4 x 4 = 14 x 4 = 56 7x = 56 x = 56/7 = 8 Ans.: 8 Ex­am­ple 3 Solve 4x + 5/4 - 9 + 2x/3 = 0

Con­sid­er­ing the left-hand side, the LCM of 3 and 4 is 12. 3(4x + 5) - 4(9 + 2x) /12 = 12x + 15 - 36 - 8x /12 = 4x - 21/12 Equat­ing both sides: 4x - 21/12 = 0 Mul­ti­ply both sides by 12 4x - 21 = 0 or 4x = 21 x = 21/4

AL­TER­NA­TIVELY, you may mul­ti­ply all terms by the LCM of the de­nom­i­na­tors.

4x + 5/4 - 9 + 2x/3 = 0 Mul­ti­ply both sides by 12: 4x+5 x 12/4 - 9 +2x x 12 /3 = 0x12 3(4x + 5) - 4(9 + 2x) = 0 12x + 15 -36 - 8x = 0 4x - 21 = 0 x = 21/4

We will now con­tinue al­ge­bra with the topic fac­tor­iza­tion.

Note that an al­ge­braic ex­pres­sion is fac­torised when it is ex­pressed as the prod­uct of its sim­plest fac­tors. For ex­am­ple, 6b + 15 is ex­pressed as

3( 2b + 5) The usual meth­ods are: (a) Com­mon fac­tor (b) Group­ing (c) Fac­tor­iz­ing of qua­dratic ex­pres­sions (d) Dif­fer­ence of two squares

The meth­ods are ad­e­quately ex­plained in the text­books and you should use them to aid you as you re­vise for your ex­ams.

It is im­por­tant that you do the fol­low­ing in all cases:

(a) Bring each fac­tor to its sim­plest form. For ex­am­ple, a fac­tor 20x + 12 should be ex­pressed as 4(5x + 3) or 2 - 6y as 2( 1 - 3y).

(b) Check your answers, if you have the time, by ex­pand­ing the fac­tors and com­par­ing the re­sult with the orig­i­nal ex­pres­sion. This week ,we will re­view the first two meth­ods of Fac­tor­iza­tion men­tioned above.

EX­AM­PLES OF COM­MON FAC­TOR METHOD

1. Fac­torise 21ab + 7a 21ab + 7a = 7a ( 3b + 1) NB: 7a is the only fac­tor com­mon to both terms.

Check­ing, mul­ti­ply­ing 7a ( 3b + 1) = 21ab + 7a, the given ex­pres­sion. 2. Fac­torise 16m + 4 The fac­tors of 16m are 2, 4, 8 and m and 4 has fac­tors 2 and 4. The high­est com­mon fac­tor (HCF) is 4. 16m + 4 = 4( 4m + 1 ) An­swer : 4( 4m + 1 ) 3. Fac­torise: 9x2 - 12x The com­mon fac­tor method is used, as 3x is the fac­tor which is com­mon to both terms. Both terms are di­vided by 3x for us to ob­tain the sec­ond fac­tor.

An­swer: 3x(3x - 4) Please note that by ex­pand­ing the an­swer, 3x(3x - 4) = 9x2 - 12x, the given ex­pres­sion 4. Fac­torise: 15x2y -10xy2 Note that the com­mon fac­tor to both terms is 5xy By di­vid­ing each term by 5xy, An­swer: 5xy(3x - 2y2)

Please fac­torise each of the fol­low­ing on your own: 6x2 y +12xy 3 - 12mn 2b3 - 6 b2 - 4b

EX­AM­PLES OF GROUP­ING METHOD

5. Fac­torise ax + ay + bx + by Note that a is the com­mon fac­tor of ax + ay and b the com­mon fac­tor of bx + by

ax + ay + bx + by = a(x + y) + b(x + y)

Do you re­alise that (x + y) is com­mon to both ex­pres­sions? a(x + y) + b(x + y) = (x + y)(a + b) This method could, there­fore, be de­scribed as re­peated com­mon fac­tor method.

This method could, there­fore, be de­scribed as re­peated com­mon fac­tor method. 4. Fac­torise 2ax - 6ay + bx - 3by (2ax - 6ay) + (bx - 3by) Fac­torise each as fol­lows: 2a(x - 3y) + b (x - 3y) = (x - 3y)(2a+ b) As usual, I will close with your home­work. 1. Solve 5x + 7 = 3x/2 2. Solve x - 2 /3 + x + 1/4 = 1 3. Fac­torise: (a) 6x2 - 18x (b) xy2 + x2y 4. Fac­torise: 2mh - 2nh- 3mk+ 3nk 5. Fac­torise: (a) 9k2 - 4 (b) 5pq2 - pq + 5rq - r 6. Fac­torise: 5a2b + 2ab Please iden­tify ad­di­tional ques­tions to be in­cluded in your re­vi­sion pool.