Solving Linear Equations and Inequalities

Chapter 1 Solving Linear Equations and Inequalities Exercise Set 1.1 RC2.(f) RC4.(c)

CC2.4−

2 3 x= 5 2 a)The LCM of the denominators is 6, so we multiplyby6.b)6

4− 2 3 x ( =6· 5 2 24−4x=15 CC4.0.06−7x=1.2x a)The greatest number of decimal places is 2, so we multiplyby100.

• 100(0.06−7x) = 100(1.2x)

6−700x= 120x 2.47−x=23

47−24?23

23 f f f TRUE 24 is a solution of the equation.

• 3x+14=−27

3(−10)+14 ?−27

−30+14

f f f −16 f f f FALSE −10 is not a solution of the equation.

6.−x 8 =−3 −32 8 ?−3 −4 f f fFALSE 32 is not a solution of the equation.

• 4−5x=59

4−5(−11) ? 59

4+55 f f f 59 f f f TRUE −11 is a solution of the equation.

10.9y+5=86

9·9+5?86

81 + 5

f f f 86 f f f TRUE

• is a solution of the equation.
• x+5=5+ x

−13+5 ? 5+(−13)

−8 f f f−8TRUE −13 is a solution of the equation.

• x+7=14

x+7−7=14−7 x+0=7 x=7

• −27 =y−17

−27+17=y−17 + 17 −10 =y+0 −10 =y 18.−8+r=17 8−8+r=8+17 0+r=25 r=25

• −37 +x=−89

37−37 +x=37−89 0+x=−52 x=−52

• z−14.9=−5.73

z−14.9+14.9=−5.73+14.9 z+0=9.17 z=9.17

• x+

1 12 =− 5 6 x+ 1 12 − 1 12 =− 5 6 − 1 12 x+0=− 10 12 − 1 12 x=− 11 12 Copyrightcf2019 Pearson Education, Inc.(Intermediate Algebra, 13e Marvin Bittinger, Judith Beecher, Barbara Johnson) (Solution Manual, For Complete File, Download link at the end of this File) 1 / 4

12 Chapter 1: Solving Linear Equations and Inequalities

26.5x=30 5x 5 = 30 5 1·x= 30 5 x=6 28.−4x= 124 −4x −4 = 124 −4 1·x= 124 −4 x=−31

30. −

x 3

=−25

− 1 3 x=−25 −3

− 1 3 ( x=−3(−25) x=75 32.−120 =−8y

−120

−8 = −8y −8

−120

−8 =1·y 15 =y 34.0.39t=−2.73 0.39t 0.39 =

−2.73

0.39 1·t=

−2.73

0.39 t=−7

36. −

7 6 y=− 7 8 − 6 7 ) − 7 6 c (y)=− 6 7 ) − 7 8 c 1·y= 42 56 y= 3 4 38.4x−7=81 4x=88 x=22 40.6z−7=11 6z=18 z=3 42.5x+7=−108 5x=−115 x=−23

44.−

9 2 y+4=− 91 2 −9y+8=−91 −9y=−99 y=11 46.2x+x=−1 3x=−1 x=− 1 3 48.−2x+6x−2=11 4x=13 x= 13 4 50.9 5 y+ 4 10 y= 66 10 18y+4y= 66 Multiplying by 10 22y=66 y=3 52.0.8t−0.3t=6.5 0.5t=6.5 t=13 54.15x+40=8x−9 15x=8x−49 7x=−49 x=−7 56.3x−15=15+3x −15 = 15 False equation No solution 58.9t−4=14+15t 9t−18 = 15t −18 = 6t −3=t 60.6−7x=x−14 20−7x=x 20=8x 20 8 =x 5 2 =x 62.5x−8=−8+3x−x 5x−8=−8+2x 3x=0 x=0 64.6y+20=10+3 y+y 6y+20=10+4 y 2y=−10 y=−5 Copyrightcf2019 Pearson Education, Inc. 2 / 4

Exercise Set 1.113 66.−3t+4=5−3t

• = 5 False equation

No solution 68.5−2y=−2y+5

• = 5 True equation

All real numbers are solutions.

70.3 4 − 5 8 m= 1 2 m− 3 8 6−5m=4m−3 Multiplying by 8 9=9m 1=m 72.0.2t+1.7=5.8+0.3t 2t+17=58+3 tMultiplying by 10 −41 =t 74.2−0.01x=0.4x+2.5 200−x=40x+ 250 Multiplying by 100 −50=41x − 50 41 =x 76.2 3 x+1− 1 6 = 1 2 x− 5 6 x 4x+6−1=3x−5x 4x+5=−2x 5=−6x − 5 6 =x 78.3(y+6)=9y 3y+18=9y 18 = 6y 3=y 80.27 = 9(5y−2) 27=45y−18 45=45y 1=y 82.210(x−3) = 840 210x−630 = 840 210x= 1470 x=7 84.8x−(3x−5) = 40 8x−3x+5=40 5x=35 x=7 86.3(4−2x)=4−(6x−8) 12−6x=4−6x+8 12−6x=12−6x 12 = 12 True equation All real numbers are solutions.

88.−40x+ 45 = 3[7−2(7x−4)] −40x+ 45 = 3[7−14x+8] −40x+45=3[−14x+ 15] −40x+45=−42x+45 2x=0 x=0 90.1 6 (12t+ 48)−20 =− 1 8 (24t−144) 2t+8−20 =−3t+18 5t=30 t=6 92.6[4(8−y)−5(9+3y)]−21 =−7[3(7 + 4y)−4] 6[32−4y−45−15y]−21 =−7[21 + 12y−4] 6[−13−19y]−21 =−7[17 + 12y] −78−114y−21 =−119−84y 20 = 30y 2 3 =y 94.3 4 ) 3x− 1 2 c + 2 3 = 1 3 54x−9 + 16 = 8 Multiplying by 24 54x=1 x= 1 54 96.9(4x+7)−3(5x−8)=6

2 3 −x ( −5

3 5 +2x ( 36x+63−15x+24=4−6x−3−10x 21x+87=−16x+1 37x=−86 x=− 86 37 98.a −9a 23 =a −32 = 1 a 32 100.−2x 8 y 3 102.−5+6x 104.−10x+35y−20 106.4(−x−6y), or−4(x+6y) 108.5(−2x+7y−4), or−5(2x−7y+4)

110.{−8,−7,−6,−5,−4,−3,−2,−1};

{x|xis a negative integer greater than−9} 112.−0.00458y+1.7787 = 13.002y−1.005 −13.00658y=−2.7837 y≈0.214 Copyrightcf2019 Pearson Education, Inc. 3 / 4

14 Chapter 1: Solving Linear Equations and Inequalities

114.2x−5 6 + 4−7x 8 = 10+6x 3 4(2x−5) + 3(4−7x) = 8(10 + 6x) Multiplying by 24 8x−20+12−21x=80+48x −88 = 61x − 88 61 =x 116.23−2{4+3(x−1)}+5{x−2(x+3)}= 7{x−2[5−(2x+ 3)]} 23−2{4+3x−3}+5{x−2x−6}= 7{x−2[5−2x−3]} 23−2{3x+1}+5{−x−6}= 7{x−2[−2x+2]} 23−6x−2−5x−30 = 7{x+4x−4} −11x−9= 7{5x−4} −11x−9= 35x−28 19=46x 19 46 =x Exercise Set 1.2 RC2.Bodymass index is calculated using an individual’s weight and height.RC4.The formulaA= 1 2 h(a+b) represents the relationship between the area of a trapezoid, its height, and the lengths of its bases.CC2.qs+4r=t qs=t−4r q= t−4r s The answer is (b).CC4.4q=7r q= 7r 4 ,or 7 4 r The answer is (a).CC6.7r−t=4s 7r=4s+t r= 4s+t 7 The answer is (e).

2.d=rt d r =t 4.V= 4 3 πr 3 3V 4π =r 3

• P=2w+2l

P−2w=2l P−2w 2 =l,or P 2 −w=l 8.A= 1 2 bh 2A=bh 2A b =h

10. A=

a+b 2 2A=a+b 2A−a=b 12.F=ma F m =a 14.I=Prt I rt =P 16.E=mc 2 Ec 2 =m 18.Q= p−q 2 2Q=p−q q=p−2Q 20.Ax+By=c Ax=c−By x= c−By A 22.F= mv

• r

Fr m =v 2 Multiplying by r m

24. N=

1 3 M(t+w) 3N=M(t+w) 3N M =t+w 3N M −t=w,or 3N−Mt M =w

• t=

1 6 (x−y+z) 6t=x−y+z 6t−x+y=z Copyrightcf2019 Pearson Education, Inc.

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