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

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1 Response

  1. Hazel Karen says:

    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.

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