Standard Form

Site: Saylor Academy
Course: GKT101: General Knowledge for Teachers – Math
Book: Standard Form
Printed by: Guest user
Date: Tuesday, 20 May 2025, 8:15 AM

Description

When a linear equation is written in standard form, both variables x and y are on the same side of the equation. Watch this lecture series and practice converting equations to standard form.

Graphing a linear equation: 5x+2y=20

Clarifying standard form rules

Converting from slope-intercept to standard form

Graph from linear standard form - Questions

1. Graph \(12 x-9 y=36\).


2. Graph \(3 x+4 y=12\).


3. Graph \(x+3 y=6\).


4. Graph \(-14 x+21 y=84\).

Answers

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:

Convert linear equations to standard form - Questions

1. What is \(y-8=3(x+1)\) written in standard form?

Choose 1 answer:

A. \(y-8=3 x+3\)

B. \(y=3 x+11\)

C. \(-3 x+y-11=0\)

D. \(-3 x+y=11\)


2. What is \(y=\frac{4}{5} x+2\) written in standard form?

Choose 1 answer:

A. \(-\frac{4}{5} x+y-2=0\)

B. \(y=\frac{4}{5}\left(x+\frac{5}{2}\right)\)

C. \(5 y=4 x+10\)

D. \(-4 x+5 y=10\)


3. What is \(y+5=7(x-8)\) written in standard form?

Choose 1 answer:

A. \(x-7 y=-61\)

B. \(7 x+y=-61\)

C. \(-7 x+y=-61\)

D. \(x+7 y=-61\)


4. What is \(y=-\frac{3}{10} x-8\) written in standard form?

Choose 1 answer:

A. \(10 x+3 y=-80\)

B. \(10 x-3 y=-80\)

C. \(-3 x+10 y=-80\)

D. \(3 x+10 y=-80\)

Answers

1. D. \(-3 x+y=11\)

Standard linear equations are in the general form \(A x+B y=C\) where \(A\), \(B\), and \(C\) are constants.

Usually, \(A\), \(B\), and \(C\) are integers.

\(y-8=3(x+1)\)
\(y-8=3 x+3\) Distribute.
\(y=3 x+11\) Collect constants.
\(-3 x+y=11\) Bring to standard form.

\(y-8=3(x+1)\) written in standard form is \(-3 x+y=11\).


2. D. \(-4 x+5 y=10\)

Standard linear equations are in the general form \(A x+B y=C\) where \(A\), \(B\), and \(C\) are constants.

Usually, \(A\), \(B\), and \(C\) are integers.

\(y=\frac{4}{5} x+2\)
\(5 y=4 x+10\) Multiply by denominator.
\(-4 x+5 y=10\) Bring to standard form.


\(y=\frac{4}{5} x+2\) written in standard form is \(-4 x+5 y=10\).


3. C. \(-7 x+y=-61\)

Standard linear equations are in the general form \(A x+B y=C\) where \(A\), \(B\), and \(C\) are constants.

Usually, \(A\), \(B\), and \(C\) are integers.

\(y+5=7(x-8)\)
\( y+5=7 x-56\) Distribute.
\(y=7 x-61\) Collect constants.
\(-7 x+y=-61\) Bring to standard form.


\(y+5=7(x-8)\) written in standard form is \(-7 x+y=-61\).


4. D. \(3 x+10 y=-80\)

Standard linear equations are in the general form \(A x+B y=C\) where \(A\), \(B\), and \(C\) are constants.

Usually, \(A\), \(B\), and \(C\) are integers.

\(y=-\frac{3}{10} x-8\)
\(10 y=-3 x-80\) Multiply by denominator.
\(3 x+10 y=-80\) Bring to standard form.


\(y=-\frac{3}{10} x-8\) written in standard form is \(3 x+10 y=-80\).