Question

solve the following equation ​

Answer 1

What could be easier than this.

Put ((3x-1)/(2x+3))² = a

The equation becomes,

a² - 5a + 4 = 0

a² - 4a - a +4 = 0

a (a-4) -1 (a-4) = 0

(a-1)(a-4) = 0

a = 1 or 4

Substituting original value for `a`
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((3x-1)/(2x+3))² = 1

((3x-1)/(2x+3)) = ± 1
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Case 1, ((3x-1)/(2x+3)) = 1

Step 1: Multiply both sides by 2x+3.

3x−1=2x+3

3x−1−2x=2x+3−2x(Subtract 2x from both sides)

x−1=3

x−1+1=3+1(Add 1 to both sides)

x=4
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Case 2, ((3x-1)/(2x+3)) = -1

=> x = -2/5
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((3x-1)/(2x+3))² = 4

((3x-1)/(2x+3)) = ±2
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Case 1, ((3x-1)/(2x+3)) = 2

Step 1: Multiply both sides by 2x+3.

3x−1=4x+6

3x−1−4x=4x+6−4x(Subtract 4x from both sides)

−x−1=6

−x−1+1=6+1(Add 1 to both sides)

−x=7 (Divide both sides by -1)

x=−7
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Case 2, ((3x-1)/(2x+3)) = -2

Step 1: Multiply both sides by 2x+3.

3x−1=−4x−6

3x−1+4x=−4x−6+4x (Add 4x to both sides)

7x−1=−6

7x−1+1=−6+1(Add 1 to both sides)

7x=−5

x = -5/7