Suppose that you go to a website that allows you to enter two points and that returns the equation of the line that passes through these points. The only problem is that the website returns the equation in the form \begin{align*}Ax+By=C\end{align*}
Guidance
As the previous few lessons have shown, there are several ways to write a linear equation. This Concept introduces another method: standard form. You have already seen examples of standard form equations in a previous Concept. For example, here are some equations written in standard form.
\begin{align*}0.75(h)+1.25(b)& =30\\
7x-3y& =21\\
2x+3y& =-6\end{align*}
The standard form of a linear equation has the form \begin{align*}Ax+By=C\end{align*}
Equations written in standard form do not have fractional coefficients and the variables are written on the same side of the equation.
You should be able to rewrite any of the formulas in an alternate form.
\begin{align*}Slope-intercept \ form & \leftrightarrow Standard \ form\\
Slope-intercept \ form & \leftrightarrow Point - slope \ form\\
Point-slope \ form & \leftrightarrow Standard \ form\end{align*}
Example A
Rewrite \begin{align*}\frac{3}{4}(h)+\frac{5}{4}(b)=30\end{align*}
Solution: According to the definition of standard form, the coefficients must be integers. So we need to clear the fractions of the denominator using multiplication.
\begin{align*}\frac{3}{4} h+\frac{5}{4} b& =30 \rightarrow 4\left (\frac{3}{4} h+\frac{5}{4} b\right )=4(30)\\ 3h+5b& =120\end{align*}
This equation is now in standard form: \begin{align*}A=3, B=5\end{align*}, and \begin{align*}C=120\end{align*}.
Example B
Rewrite \begin{align*}y-5=3(x-2)\end{align*} in standard form.
Solution: Use the Distributive Property to simplify the right side of the equation.
\begin{align*}y-5=3x-6\end{align*}
Rewrite this equation so the variables \begin{align*}x\end{align*} and \begin{align*}y\end{align*} are on the same side of the equation.
\begin{align*}y-5+6& =3x-6+6\\ y-y+1& =3x-y\\ 1 &=3x-y, && \ \text{where } A=3, B=-1, \text{and } C=1.\end{align*}
Finding Slope and \begin{align*}y-\end{align*}Intercept of a Standard Form Equation
Slope-intercept form and point-slope form of a linear equation both contain the slope of the equation explicitly, but the standard form does not. Since the slope is such an important feature of a line, it is useful to figure out how you would find the slope if you were given the equation of the line in standard form.
Begin with standard form: \begin{align*}Ax+By=C\end{align*}.
If you rewrite this equation in slope-intercept form, it becomes:
\begin{align*}Ax-Ax+By& =C-Ax\\ \frac{By}{B}& =\frac{-Ax+C}{B}\\ y& =\frac{-A}{B}x+\frac{C}{B}\end{align*}
When you compare this form to slope-intercept form, \begin{align*}y=mx+b\end{align*}, you can see that the slope of a standard form equation is \begin{align*}\frac{-A}{B}\end{align*} and the \begin{align*}y-\end{align*}intercept is \begin{align*}\frac{C}{B}\end{align*}.
The standard form of a linear equation, \begin{align*}Ax+By=C\end{align*}, has the following:
\begin{align*}slope=\frac{-A}{B}\end{align*} and \begin{align*}y-intercept=\frac{C}{B}\end{align*}.
Example C
Find the slope and \begin{align*}y-\end{align*}intercept of \begin{align*}2x-3y=-8\end{align*}.
Solution: Using the definition of standard form, \begin{align*}A=2, B=-3,\end{align*} and \begin{align*}C=-8\end{align*}.
\begin{align*}slope & =\frac{-A}{B}=\frac{-2}{-3} \rightarrow \frac{2}{3}\\ y-intercept & =\frac{C}{B} =\frac{-8}{-3}\rightarrow \frac{8}{3}\end{align*}
The slope is \begin{align*}\frac{2}{3}\end{align*} and the \begin{align*}y-\end{align*}intercept is \begin{align*}\frac{8}{3}\end{align*}.
Guided Practice
Rewrite the following in standard form.
1. \begin{align*}5x-7=y\end{align*}
2. \begin{align*}0.75h+1.25b =30\end{align*}
Solutions:
1. Rewrite this equation so the variables \begin{align*}x\end{align*} and \begin{align*}y\end{align*} are on the same side of the equation.
\begin{align*}5x-7+7& =y+7\\ 5x-y& =y-y+7\\ 5x-y& =7, && \ \text{where }A=5, B=-1, \text{and } C=7.\end{align*}
2. This equation is almost in the form already, since the two variables are on the same side, and the number is on the opposite side. What this equation needs is to have integer coefficients. In this example, we will clear the decimals, which is similar to clearing fractions. What can we multiply by 0.75 and 1.25 to make them be integers? Since each decimal goes to the hundredths place, we need to multiply each by 100.
\begin{align*}0.75h+1.25b& =30\\ 100 \cdot (0.75h+1.25b)& =100\cdot(30)\\ 75h+125b=3000, && \ \text{where } A=75, B=125, \text{and } C=3000.\end{align*}
Note: Technically, \begin{align*}A\end{align*} and \begin{align*}B\end{align*} can go with either variable, and can just be thought of as the two coefficients in front of the two variables. Usually, we have \begin{align*}A\end{align*} in front of \begin{align*}x\end{align*} and \begin{align*}B\end{align*} in front of \begin{align*}y\end{align*}. But this is just something mathematicians choose to do to be consistent. Since our variables are not \begin{align*}x\end{align*} and \begin{align*}y\end{align*}, \begin{align*}A\end{align*} and \begin{align*}B\end{align*} can go with either variable.
Practice
Sample explanations for some of the practice exercises below are available by viewing the following video. Note that there is not always a match between the number of the practice exercise in the video and the number of the practice exercise listed in the following exercise set. However, the practice exercise is the same in both. CK-12 Basic Algebra: Linear Equations in Standard Form (10:08)
- What is the standard form of a linear equation? What do \begin{align*}A,B\end{align*}, and \begin{align*}C\end{align*} represent?
- What is the meaning of “clear the fractions”? How would you go about doing so?
- Consider the equation \begin{align*}Ax+By=C\end{align*}. What are the slope and \begin{align*}y-\end{align*}intercept of this equation?
Rewrite the following equations in standard form.
- \begin{align*}y=3x-8\end{align*}
- \begin{align*}y=-x-6\end{align*}
- \begin{align*}y=\frac{5}{3} x-4\end{align*}
- \begin{align*}0.30x+0.70y=15\end{align*}
- \begin{align*}5= \frac{1}{6} x-y\end{align*}
- \begin{align*}y-7=-5(x-12)\end{align*}
- \begin{align*}2y=6x+9\end{align*}
- \begin{align*}y=\frac{9}{4}x+\frac{1}{4}\end{align*}
- \begin{align*}y+\frac{3}{5}=\frac{2}{3}(x-2) \end{align*}
- \begin{align*}3y+5=4(x-9)\end{align*}
Find the slope and \begin{align*}y-\end{align*}intercept of the following lines.
- \begin{align*}5x-2y=15\end{align*}
- \begin{align*}3x+6y=25\end{align*}
- \begin{align*}x-8y=12\end{align*}
- \begin{align*}3x-7y=20\end{align*}
- \begin{align*}9x-9y=4\end{align*}
- \begin{align*}6x+y=3\end{align*}
- \begin{align*}x-y=9\end{align*}
- \begin{align*}8x+3y=15\end{align*}
- \begin{align*}4x+9y=1\end{align*}
In 23 – 27, write each equation in standard form by first writing it in point-slope form.
- \begin{align*}Slope = -1\end{align*} through point (–3, 5)
- \begin{align*}Slope = -\frac{1}{4}\end{align*} through point (4, 0)
- Line through (5, –2) and (–5, 4)
- Line through (–3, –2) and (5, 1)
- Line through (1, –1) and (5, 2)