How do you solve $\frac{x}{2 x + 7} = \frac{x - 5}{x - 1}$ $(x)/(2x+7)=(x-5)/(x-1)$ ?

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How do you solve $\frac{x}{2 x + 7} = \frac{x - 5}{x - 1}$ $(x)/(2x+7)=(x-5)/(x-1)$ ?

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Answer 1 · 2017-06-17T11:00:18

$x = - 5$ $x=-5$ or $x = 7$ $x=7$

Explanation:

Note : when you multiply something with a fraction, you multiply to the numerator.

First, let's simplify the LHS by multiplying $\frac{x - 5}{x - 1}$ $(x-5)/(x-1)$ with $2 x + 7$ $2x+7$

$\frac{x - 5 \cdot 2 x + 7}{x - 1}$ $(x-5*2x+7)/(x-1)$

$\frac{2 x^{2} - 10 x + 7 x - 35}{x - 1}$ $(2x^2-10x+7x-35)/(x-1)$

So we have

$x = \frac{2 x^{2} - 3 x - 35}{x - 1}$ $x=(2x^2-3x-35)/(x-1)$

Then let's simplify the RHS by multiplying $x$ $x$ with $x - 1$ $x-1$ .

$\frac{x}{1} = \frac{2 x^{2} - 3 x - 35}{x - 1}$ $x/1=(2x^2-3x-35)/(x-1)$ since $x = \frac{x}{1}$ $x=x/1$ ,

Then

$x \cdot x - 1 = 2 x^{2} - 3 x - 35$ $x*x-1=2x^2-3x-35$

Then $x^{2} - x = 2 x^{2} - 3 x - 35$ $x^2-x=2x^2-3x-35$

$0 = x^{2} - 2 x - 35$ $0=x^2-2x-35$

What we just did here is called cross multiplying.

Now, we have a quadratic equation.
We can factorise this to

$(x + 5) (x - 7) = 0$ $(x+5)(x-7) =0$

Set each factor equal to $0$ $0$ , since if either $(x + 5)$ $(x+5)$ or $(x - 7)$ $(x-7)$ was 0, the equation would hold, because $0$ $0$ multiplied by any number will always be 0.

$x + 5 = 0$ $x+5=0$
$x = - 5$ $x=-5$

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

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How do you solve $\frac{x}{2 x + 7} = \frac{x - 5}{x - 1}$ $(x)/(2x+7)=(x-5)/(x-1)$ ?

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How do you solve (x)/(2x+7)=(x-5)/(x-1)x2x+7=x−5x−1(x)/(2x+7)=(x-5)/(x-1)?

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How do you solve $\frac{x}{2 x + 7} = \frac{x - 5}{x - 1}$ $(x)/(2x+7)=(x-5)/(x-1)$ ?