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

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Answer 1 · 2016-03-27T05:42:02

$(- 1, - 4)$ $(-1, -4)$
$(4, 1)$ $(4, 1)$

We have two equations
$y = \frac{4}{x}$ $y=4/x" "$ first equation
$y = x - 3$ $y=x-3" "$ second equation

Substitute first equation into the second

$y = x - 3$ $y=x-3" "$ second equation
$\frac{4}{x} = x - 3$ $4/x=x-3" "$ second equation

Multiply both sides of the equation by x

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

$cancelx*(4/cancelx)=x(x-3)$

$4=x(x-3)$

$4=x^2-3x$

We have a quadratic equation

$x^2-3x-4=0$

We can solve this by factoring method

$x^2-3x-4=0$
$(x+1 )(x-4)=0$

Equate each factor to 0 to find the roots

First factor

$(x+1 )=0$

$x=-1$ first root
and $y=-4$

Second factor

$x-4=0$

$x=4$ second root
and $y=1$

God bless....I hope the explanation is useful.

How do you solve y=4/xy=4xy=4/x and y=x-3y=x−3y=x-3 using substitution?