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

$\left(- 1 , - 4\right)$
$\left(4 , 1\right)$

Explanation:

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

Substitute first equation into the second

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

Multiply both sides of the equation by x

$x \cdot \left(\frac{4}{x}\right) = x \left(x - 3\right)$

$\cancel{x} \cdot \left(\frac{4}{\cancel{x}}\right) = x \left(x - 3\right)$

$4 = x \left(x - 3\right)$

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

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

We can solve this by factoring method

${x}^{2} - 3 x - 4 = 0$
$\left(x + 1\right) \left(x - 4\right) = 0$

Equate each factor to 0 to find the roots

First factor

$\left(x + 1\right) = 0$

$x = - 1$ 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.