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: