Standard Form

1. This is a linear equation given in standard form: $A x+B y=C$. A common way of graphing an equation of this form is to find the $x$- and $y$-intercepts of the graph.

To find the $y$-intercept, let's substitute $x=0$ into the equation and solve for $y$:

$\begin{aligned} 12 x-9 y &=36 \\ 12 \cdot 0-9 y &=36 \\ -9 y &=36 \\ y &=-4 \end{aligned}$

So the $y$-intercept is $(0,-4)$.

To find the $x$-intercept, let's $y=0$ into the equation and solve for $x$:

$\begin{aligned} 12 x-9 y &=36 \\ 12 x-9 \cdot 0 &=36 \\ 12 x &=36 \\ x &=3 \end{aligned}$

So the $x$-intercept is $(3,0)$.

We can graph the linear equation using these two points, as shown below:

2. This is a linear equation given in standard form: $A x+B y=C$. A common way of graphing an equation of this form is to find the $x$- and $y$-intercepts of the graph.

To find the $y$-intercept, let's substitute $x=0$ into the equation and solve for $y$:

$\begin{array}{r} 3 x+4 y=12 \\ 3 \cdot 0+4 y=12 \\ 4 y=12 \\ y=3 \end{array}$

So the $y$-intercept is $(0,3)$.

To find the $x$-intercept, let's $y=0$ into the equation and solve for $x$:

$\begin{array}{r} 3 x+4 y=12 \\ 3 x+4 \cdot 0=12 \\ 3 x=12 \\ x=4 \end{array}$

So the $x$-intercept is $(4,0)$.

We can graph the linear equation using these two points, as shown below:

3. This is a linear equation given in standard form: $A x+B y=C$. A common way of graphing an equation of this form is to find the $x$- and $y$-intercepts of the graph.

To find the $y$-intercept, let's substitute $x=0$ into the equation and solve for $y$:

$\begin{array}{r} x+3 y=6 \\ 0+3 y=6 \\ 3 y=6 \\ y=2 \end{array}$

So the $y$-intercept is $(0,2)$.

To find the $x$-intercept, let's $y=0$ into the equation and solve for $x$:

$\begin{array}{r} x+3 y=6 \\ x+3 \cdot 0=6 \\ x=6 \end{array}$

So the $x$-intercept is $(6,0)$.

We can graph the linear equation using these two points, as shown below:

4. This is a linear equation given in standard form: $A x+B y=C$. A common way of graphing an equation of this form is to find the $x$- and $y$-intercepts of the graph.

To find the $y$-intercept, let's substitute $x=0$ into the equation and solve for $y$:

$\begin{aligned} -14 x+21 y &=84 \\ -14 \cdot 0+21 y &=84 \\ 21 y &=84 \\ y &=4 \end{aligned}$

So the $y$-intercept is $(0,4)$.

To find the $x$-intercept, let's $y=0$ into the equation and solve for $x$:

$\begin{array}{r} -14 x+21 y=84 \\ -14 x+21 \cdot 0=84 \\ -14 x=84 \\ x=-6 \end{array}$

So the $x$-intercept is $(-6,0)$.

We can graph the linear equation using these two points, as shown below: