How do you solve $y = \frac{4}{x}$ $y=4/x$ and $y = 3 x - 1$ $y=3x-1$ using substitution?

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How do you solve $y = \frac{4}{x}$ $y=4/x$ and $y = 3 x - 1$ $y=3x-1$ using substitution?

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Answer 1 · 2016-07-01T16:18:10

$x = \frac{4}{3}$ $x=4/3$ or $x = - 1$ $x=-1$

Explanation:

To solve $y = \frac{4}{x}$ $y=4/x$ and $y = 3 x - 1$ $y=3x-1$ using substitution, just substitute value of $y$ $y$ from first equation into second, which leads to

$\frac{4}{x} = 3 x - 1$ $4/x=3x-1$ or multiplying each by $x$ $x$ (we assome $x \neq 0$ $x!=0$ , we get

$4 = 3 x^{2} - x$ $4=3x^2-x$ or

$3 x^{2} - x - 4 = 0$ $3x^2-x-4=0$ or

$3 x^{2} + 3 x - 4 x - 4 = 0$ $3x^2+3x-4x-4=0$ or

$3 x (x + 1) - 4 (x + 1) = 0$ $3x(x+1)-4(x+1)=0$ or

$(3 x - 4) (x + 1) = 0$ $(3x-4)(x+1)=0$ , which means that $3 x - 4 = 0$ $3x-4=0$ or $x + 1 = 0$ $x+1=0$

i.e. $x = \frac{4}{3}$ $x=4/3$ or $x = - 1$ $x=-1$

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How do you solve $y = \frac{4}{x}$ $y=4/x$ and $y = 3 x - 1$ $y=3x-1$ using substitution?

1 Answer

Explanation:

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How do you solve $y = \frac{4}{x}$ $y=4/x$ and $y = 3 x - 1$ $y=3x-1$ using substitution?