How do you solve $\frac{x + 1}{3} = \frac{x - 1}{4}$ $(x+1)/3 = (x-1)/4$ ?

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How do you solve $\frac{x + 1}{3} = \frac{x - 1}{4}$ $(x+1)/3 = (x-1)/4$ ?

2 Answers

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Answer 1 · 2016-06-05T06:12:06

x=-7

Explanation:

4(x+1)=3(x-1)
4x+4=3x-3
4x-3x=-4-3
x=-7

Answer 2 · 2016-06-05T06:14:59

$x = - 7$ $x=-7$

Explanation:

In the equation $\frac{x + 1}{3} = \frac{x - 1}{4}$ $(x+1)/3=(x-1)/4$ , let us first multiply both sides by LCD of $3$ $3$ and $4$ $4$ i.e. $12$ $12$ to eliminate fractions.

Hence this equation can be written as

$\frac{x + 1}{3} \times 12 = \frac{x - 1}{4} \times 12$ $(x+1)/3xx12=(x-1)/4xx12$

$4 (x + 1) = 3 (x - 1)$ $4(x+1)=3(x-1)$ or

$4 x + 4 = 3 x - 3$ $4x+4=3x-3$

now transposing terms containing variable $x$ $x$ to left hand side and constant terms to right hand side, we get

$4 x - 3 x = - 3 - 4$ $4x-3x=-3-4$ or

$x = - 7$ $x=-7$

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How do you solve $\frac{x + 1}{3} = \frac{x - 1}{4}$ $(x+1)/3 = (x-1)/4$ ?

2 Answers

Explanation:

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