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

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

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Answer 1 · 2016-08-05T10:44:18

$x = \frac{7}{2}$ $x=7/2$

Explanation:

$\frac{x - 3}{x - 4} + 4 = 3 \frac{x}{x}$ $(x-3)/(x-4)+4=3x/x$
or
$\frac{x - 3}{x - 4} + 4 = 3$ $(x-3)/(x-4)+4=3$
or
$\frac{x - 3}{x - 4} = - 4 + 3$ $(x-3)/(x-4)=-4+3$
or
$\frac{x - 3}{x - 4} = - 1$ $(x-3)/(x-4)=-1$
or
$x - 3 = - x + 4$ $x-3=-x+4$
or
$x + x = 4 + 3$ $x+x=4+3$
or
$2 x = 7$ $2x=7$
or
$x = \frac{7}{2}$ $x=7/2$

Answer 2 · 2016-08-05T11:00:30

$x = \frac{7}{2}$ $x=7/2$

Reworked to make it simpler

Explanation:

Note that $\frac{3 x}{x} = 3$ $(3x)/x = 3$ giving

$\frac{x - 3}{x - 4} + 4 = 3$ $(x-3)/(x-4)+4=3$

$\Rightarrow \frac{x - 3}{x - 4} + 1 = 0$ $=>(x-3)/(x-4)+1=0$

But 1 can be written as: $\frac{x - 4}{x - 4}$ $(x-4)/(x-4)$

$\frac{x - 3}{x - 4} + \frac{x - 4}{x - 4} = 0$ $(x-3)/(x-4)+(x-4)/(x-4)=0$

Multiply through out by $(x - 4)$ $(x-4)$

$(x - 3) + (x - 4) = 0$ $(x-3)+(x-4)=0$

$2 x - 7 = 0$ $2x-7=0$

$x = \frac{7}{2}$ $x=7/2$

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

2 Answers

Explanation:

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How do you solve x−3x−4+4=3xx(x-3)/(x-4)+4=(3x)/x?

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