# Week 6 Review: Answers and Solutions

PCSR WEEK 6 Review: Solving Equations

Practice Exercise: Find the value of x.

1.) x + 5 = 8 => x = 8 – 5 => x = 3

2.) x – 3 = 6 => x = 6 + 3 => x = 9

3.) x + 8 = 0 => x = 0 – 8 => x = -8

4.) 4x = 12 => x = 12/4 => x = 3

5.) x/2 = -6 => x = -6(2) => x = -12

PCSR WEEK 6 Review: Solving Equations. In each equation, find the value of x.

1.) 2x – 1 = 5

2x = 5 + 1

2x = 6

x = 6/2

x = 3

2.) x – 12 = – 2x

x + 2x = 12

3x = 12

x = 12/3

x = 4

3.) x + 6 = 3x – 5

x – 3x = -5 – 6

-2x = -11

x = -11/-2

x = 11/2 or 5 1/2

4.) 5x + 12 = 3x – 6

5x – 3x = -6 – 12

2x = -18

x = -18/2

x = -9

5.) 2(5 – x) = 13

By distributive property, (2)(5) – (2)(x) = 13

10 – 2x = 13

-2x = 13 – 10

-2x = 3

x = 3/(-2)

x = -1 1/2

6.) 3(x + 8) = 15 + 6x

(3)(x) + (3)(8) = 15 + 6x

3x + 24 = 15 + 6x

3x – 6x = 15 – 24

-3x = -9

x = -9/-3

x = 9/3 or 3

7.) -2(3x – 4) = 2(1 – x)

(-2)(3x) – (-2)(4) =(2)(1) -(2)(x)

-6x – (-8) = 2 – 2x

-6x + 8 = 2 – 2x

-6x + 2x = 2 – 8

-4x = – 6

x = -6/(-4)

x = 6/4 or 3/2 or 1 1/2

8.) 4(x + 2) – 5 = x + 6

4(x) + 4(2) – 5 = x + 6

4x + 8 – 5 = x + 6

4x + 3 = x + 6

4x – x = 6 – 3

3x = 3

x = 3/3 or 1

9.) 3x/4 = 18

3x = 18(4)

3x = 72

x = 72/3

x = 24

10.) x/4 + 6 = 16

x/4 = 16 – 6

x/4 = 10

x = 10(4)

x = 40

11.) x/2 – 7 = 5 – 2x

To eliminate the fraction, we multiply both sides of the equation by 2.

2(x/2 – 7) = 2(5 – 2x)

2(x/2) – 2(7) = 2(5) – 2(2x)

x – 14 = 10 – 4x

x + 4x = 10 + 14

5x = 24

x = 24/5 or 4 4/5

12.) (x + 5)/2 = x – 3

To eliminate the fraction, we multiply both sides of the equation by 2.

2[(x + 5)/2] = 2(x – 3)

x + 5 = 2(x) – 2(3)

x + 5 = 2x – 6

x – 2x = -6 – 5

-x = -11

x = -11/-1

x = 11

13.) (2x – 3)/2 = (x + 2)/3

To eliminate the fraction, we multiply both sides of the equation by the LCM of 2 and 3 which is 6.

6[(2x – 3)/2] = 6[(x + 2)/3]

(6/2) (2x – 3) = (6/3) (x + 2)

(3)(2x – 3) = (2)(x + 2)

(3)(2x) – (3)(3) = (2)(x) + (2)(2)

6x – 9 = 2x + 4

6x – 2x = 4 + 9

4x = 13

x = 13/4

x = 3 1/4

14.) 8 – (x + 3)/4 = (x + 8)

4(8) – 4[(x + 3)/4] = 4(x + 8)

(32) – (x + 3) = (4)(x) + (4)(8)

32 – x – 3 = 4x + 32

29 – x = 4x + 32

-x – 4x = 32 – 29

-5x = 3

x = 3/(-5)

x = – 3/5

15.) 3(x -9)/4 = 2(x + 6)/5

[(3)(x) – (3)(9)]/4 = [(2)(x) + (2)(6)]/5

(3x – 27)/4 = (2x + 12)/5

To eliminate the fraction, we multiply both sides of the equation by the LCM of 2 and 3 which is 6.

20 [(3x – 27)/4 = (2x + 12)/5]

(20/4)(3x – 27) = (20/5)(2x + 12)

5(3x – 27) = 4(2x + 12)

(5)(3x) – (5)(27) = (4)(2x) + (4)(12)

15x – 135 = 8x + 48

15x – 8x = 48 + 135

7x = 183

x = 183/7 or 26 1/7

Iba ung paraang natutunan ko from the tutorials. Ang alam ko kung ano ang ibabawas o idadagdag sa kanan, ganun din sa kaliwa. This is not how the example problems were solved in the video tutorials by sipnayan. Iba tuloy ung nakuha kong mga sagot.