Algebra 2 problems?

Question

Solving rational equations by cross multiplying
1. -2/(x+4) = -3/(x+1)
2. 3/(2x-1) = 4/(3x+1)
3. 2/(x-2)=3/(x+5)+10/(x+5)(x-2)
4. 6/(x-1)+3/(x-1)(x+3)=4/(x+3)
Solving linear equations
5. 6-x=2x+9
6. 4(3-2x)= -7(x+3)
Solving absolute value equation
7. |1-4t|=5
8. |t-2z|=3

Answer 1

(1).

The rational equation is - 2/(x + 4) = - 3/(x + 1).

Apply Property of Proportions : If a/b = c/d, then ad = bc.

- 2(x + 1) = - 3(x + 4)

Apply distributive Property : a(b + c) = ab + ac.

- 2(x) - 2(1) = - 3(x) - 3(4)

- 2x - 2 = - 3x - 12

Add 2 to each side.

- 2x - 2 + 2 = - 3x - 12 + 2

- 2x  = - 3x - 10

Add 3x to each side.

- 2x  + 3x = - 3x - 10 + 3x

x = - 10.

The solution x = - 10.

Answer 2

(2).

The rational equation is 3/(2x - 1) = 4/(3x + 1).

Apply Property of Proportions : If a/b = c/d, then ad = bc.

3(3x + 1) = 4(2x - 1)

Apply distributive Property : a(b + c) = ab + ac.

3(3x) + 3(1) = 4(2x) + 4(-1)

9x + 3 = 8x - 4

Subtract 3 from each side.

9x + 3 - 3 = 8x - 4 - 3

9x = 8x - 7

Subtract 8x from each side.

9x - 8x = 8x - 7 - 8x

x = - 7.

The solution x = - 7.

Answer 3

(3).

The rational equation is 2/(x - 2) = 3/(x + 5) + 10/[(x + 5)(x - 2)].

First add the rational expressions in the right hand side expressions.

The Least common multiply of (x + 5) and (x + 5)(x - 2) is (x + 5)(x - 2).

2/(x - 2) = [3(x - 2) + 10]/[(x + 5)(x - 2)]

2 / (x - 2) = (3x + 4) / [(x + 5)(x - 2)]

Multiply each side by [(x + 5)(x - 2)].

{2 / (x - 2)}*[(x + 5)(x - 2)] = {(3x + 4) / [(x + 5)(x - 2)]}*[(x + 5)(x - 2)]

2(x + 5) = (3x + 4)

Apply distributive Property : a(b + c) = ab + ac.

2(x) + 2(5) = 3x + 4

2x + 10 = 3x + 4

Subtract 4 from each side.

2x + 10 - 4 = 3x + 4 - 4

2x + 6 = 3x

Subtract 2x from each side.

2x + 6 - 2x = 3x - 2x

6 = x.

The solution x = 6.

Answer 4

(4).

The rational equation is 6/(x - 1) + 3/(x - 1)(x + 3) = 4/(x + 3).

First add the rational expressions in the left hand side expressions.

The Least common multiply of (x - 1) and (x - 1)(x + 3) is (x - 1)(x + 3).

[6(x + 3) + 3]/(x - 1)(x + 3) = 4/(x + 3)

(6x + 21)/(x - 1)(x + 3) = 4/(x + 3)

Multiply each side by [(x - 1)(x + 3)].

[(6x + 21) / (x - 1)(x + 3)] * [(x - 1)(x + 3)] = [4 / (x + 3)] * [(x - 1)(x + 3)]

(6x + 21) = 4(x - 1)

Apply distributive Property : a(b - c) = ab - ac.

6x + 21 = 4(x) - 4(1)

6x + 21 = 4x - 4

Subtract 21 from each side.

6x + 21 - 21 = 4x - 4 - 21

6x = 4x - 25

Subtract 4x from each side.

6x - 4x = 4x - 25 - 4x

2x = - 25

Divide each side by 2.

2x/2 = - 25/2

x = - 25/2

The solution x = - 25/2.

Answer 5

(5).

The linear equation is 6 - x = 2x + 9.

Subtract 6 from each side.

6 - x - 6 = 2x + 9 - 6

- x = 2x + 3

Subtract 2x from each side.

- x - 2x = 2x + 3 - 2x

- 3x = 3

Divide each side by negative 3.

(-3x)/(-3) = 3/(-3)

x = - 1

The solution x = - 1.

Answer 6

(6).

The linear equation is 4(3 - 2x) = - 7(x + 3).

Apply distributive Property : a(b + c) = ab + ac.

4(3) + 4(-2x) = -7(x) + (-7)(3)

12 - 8x = -7x - 21

Add 8x to each side.

12 - 8x + 8x = -7x - 21 + 8x

12 = x - 21

Add 21 to each side.

12 + 21 = x - 21 +21

33 = x

x = 33

The solution x = 33.

Answer 7

(7).

The absolute value equation is |1 - 4t| = 5.

For any real numbers a and b, where b ≥ 0, if  | a |  = b, then a = b or -a = b. This second case is often written as a = -b

Case 1 : 1 - 4t = 5.

Subtract 1 from each side.

1 - 4t - 1 = 5 - 1

- 4t = 4

Divide each side by negative 4.

(-4t)/(-4) = 4/(-4)

t = - 1.

 

Case 2 : 1 - 4t = - 5.

Subtract 1 from each side.

1 - 4t - 1 = - 5 - 1

- 4t = - 6

Divide each side by negative 4.

(-4t)/(-4) = (-6)/(-4)

t = 3/2.

The solution set {- 1, 3/2}.

Answer 8

(8).

Let assume that the absolute value equation is |t - 2| = 3.

For any real numbers a and b, where b ≥ 0, if  | a | = b, then a = b or - a = b. This second case is often written as a = - b.

Case 1 : t - 2 = 3.

Add 2 to each side.

t - 2 + 2 = 3 + 2

t = 5.

 

Case 2 : t - 2 = - 3.

Add 2 to each side.

t - 2 + 2 = - 3 + 2

t = - 1.

The solution set {- 1, 5}.