# Question 6b8f6

Mar 21, 2017

Slope is $\frac{2}{3}$
y_("intercept")" "=" "4/3

x_("intercept")" "=-2

#### Explanation:

$\textcolor{b r o w n}{\text{Extensive explanation given - using first principles method}}$

$\textcolor{p u r p \le}{\text{Note that the shortcut method is based on the outcome of first principle method}}$

Target: Manipulate the given equation into the form of $y = m x + c$
$\text{ }$where m is the gradient and c is the y-intercept.

The $\frac{3}{2} y$ is positive so we need to keep the $y$ on that side of the equation.

To get rid of the $x$ on the left we change it to 0
What we do to one side of the equation we also do to the other.

Add $\textcolor{red}{x}$ to both sides.

color(green)(3/2y-xcolor(red)(+x)" "=" "2color(red)(+x)

But $- x + x = 0$

$\textcolor{g r e e n}{\frac{3}{2} y + 0 = 2 + x}$

Note that $x + 2$ has the same value as $2 + x$

color(green)(3/2y=x+2

To 'get rid' of the $\frac{3}{2}$ change it to 1 as $1 \times y = y$

Multiply both sides by $\textcolor{red}{\frac{2}{3}}$

color(green)(3/2color(red)(xx2/3)xxy=color(red)(2/3)(x+2)

color(green)(y=2/3x+4/3 rarr" "2/3" is the gradient (slope)"
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The y-axis crosses the x-axis at $x = 0$ so substitute 0 for $x$

y_("intercept")=2/3(0)+4/3" "=" "4/3#

The x-axis cross the y-axis at y=0 so substitute 0 for y

$0 = \frac{2}{3} x + \frac{4}{3}$

Subtract from both sides $\frac{4}{3}$

$- \frac{4}{3} = \frac{2}{3} x$

Multiply both sides by $\frac{3}{2}$

$x = \frac{3}{2} \times \left(- \frac{4}{3}\right)$

$x = - \left(\frac{3}{3} \times \frac{4}{2}\right)$

${x}_{\text{intercept}} = - 2$