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# 5.3: Linear Equations in Standard Form

Difficulty Level: At Grade Created by: CK-12

As the past few lessons of this chapter have shown, there are several ways to write a linear equation. This lesson introduces another method: standard form. You have already seen examples of standard form equations in a previous lesson. For example, here are some equations written in standard form.

0.75(h)+1.25(b)7x3y2x+3y=30=21=6\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 Ax+By=C\begin{align*}Ax+By=C\end{align*}, where A,B\begin{align*}A,B\end{align*}, and C\begin{align*}C\end{align*} are integers and A\begin{align*}A\end{align*} and B\begin{align*}B\end{align*} are not both zero.

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 into an alternate form.

Slopeintercept formSlopeintercept formPointslope formStandard formPointslope formStandard 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 1: Rewrite 34(h)+54(b)=30\begin{align*}\frac{3}{4}(h)+\frac{5}{4}(b)=30\end{align*} in standard form.

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.

34h+54b3h+5b=304(34h+54b)=4(30)=120\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, A=3,B=5\begin{align*}A=3, B=5\end{align*}, and C=120\begin{align*}C=120\end{align*}.

Example 2: Rewrite y5=3(x2)\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

y5=3x6\begin{align*}y-5=3x-6\end{align*}

Rewrite this equation so the variables x\begin{align*}x\end{align*} and y\begin{align*}y\end{align*} are on the same side of the equation.

y5+6yy+11=3x6+6=3xy=3xy, where A=3, B=-1, and C=1.\begin{align*}y-5+6& =3x-6+6\\ y-y+1& =3x-y\\ 1 &=3x-y, && \ \text{where A=3, B=-1, and C=1}.\end{align*}

Example 3: Rewrite 5x7=y\begin{align*}5x-7=y\end{align*} in standard form.

Solution: Rewrite this equation so the variables x\begin{align*}x\end{align*} and y\begin{align*}y\end{align*} are on the same side of the equation.

5x7+75xy5xy=y+7=yy+7=7, where A=5, B=-1, and C=7.\begin{align*}5x-7+7& =y+7\\ 5x-y& =y-y+7\\ 5x-y& =7, && \ \text{where A=5, B=-1, and C=7}.\end{align*}

## Finding Slope and y−\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: Ax+By=C\begin{align*}Ax+By=C\end{align*}.

If you rewrite this equation in slope-intercept form, it becomes:

AxAx+ByByBy=CAx=Ax+CB=ABx+CB\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, y=mx+b\begin{align*}y=mx+b\end{align*}, you can see that the slope of a standard form equation is AB\begin{align*}\frac{-A}{B}\end{align*} and the y\begin{align*}y-\end{align*}intercept is CB\begin{align*}\frac{C}{B}\end{align*}.

The standard form of a linear equation Ax+By=C\begin{align*}Ax+By=C\end{align*} has the following:

slope=AB\begin{align*}slope=\frac{-A}{B}\end{align*} and yintercept=CB\begin{align*}y-intercept=\frac{C}{B}\end{align*}.

Example 4: Find the slope and y\begin{align*}y-\end{align*}intercept of 2x3y=8\begin{align*}2x-3y=-8\end{align*}.

Solution: Using the definition of standard form, A=2,B=3,\begin{align*}A=2, B=-3,\end{align*} and C=8\begin{align*}C=-8\end{align*}.

slopeyintercept=AB=2323=CB=8383\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 23\begin{align*}\frac{2}{3}\end{align*} and the y\begin{align*}y-\end{align*}intercept is 83\begin{align*}\frac{8}{3}\end{align*}.

Example 5: Nimitha buys fruit at her local farmer’s market. This Saturday, oranges cost $2 per pound and cherries cost$3 per pound. She has 12 to spend on fruit. Write an equation in standard form that describes this situation. If she buys 4 pounds of oranges, how many pounds of cherries can she buy? Solution: Define the variables: x=\begin{align*}x=\end{align*} pounds of oranges and y=\begin{align*}y=\end{align*} pounds of cherries. The equation that describes this situation is: 2x+3y=12\begin{align*}2x+3y=12\end{align*} If she buys 4 pounds of oranges, we substitute x=4\begin{align*}x=4\end{align*} in the equation and solve for y\begin{align*}y\end{align*}. 2(4)+3y=123y=1283y=4y=43\begin{align*}2(4)+3y=12 \Rightarrow 3y=12-8 \Rightarrow 3y=4 \Rightarrow y=\frac{4}{3}\end{align*}. Nimitha can buy 113\begin{align*}1\frac{1}{3}\end{align*} pounds of cherries. Example 6: Jethro skateboards part of the way to school and walks for the rest of the way. He can skateboard at 7 miles per hour and he can walk at 3 miles per hour. The distance to school is 6 miles. Write an equation in standard form that describes this situation. If Jethro skateboards for 12\begin{align*}\frac{1}{2}\end{align*} an hour, how long does he need to walk to get to school? Solution: Define the variables: x=\begin{align*}x=\end{align*} hours Jethro skateboards and y=\begin{align*}y=\end{align*} hours Jethro walks. The equation that describes this situation is 7x+3y=6\begin{align*}7x+3y=6\end{align*}. If Jethro skateboards 12\begin{align*}\frac{1}{2}\end{align*} hour, we substitute \begin{align*}x=0.5\end{align*} in the equation and solve for \begin{align*}y\end{align*}. \begin{align*}7(0.5)+3y=6 \Rightarrow 3y=6 - 3.5 \Rightarrow 3y=2.5 \Rightarrow y=\frac{5}{6}\end{align*}. Jethro must walk \begin{align*}\frac{5}{6}\end{align*} of an hour. ## Practice Set 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. 1. What is the standard form of a linear equation? What do \begin{align*}A,B\end{align*}, and \begin{align*}C\end{align*} represent? 2. What is the meaning of “clear the fractions”? How would you go about doing so? 3. 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. 1. \begin{align*}y=3x-8\end{align*} 2. \begin{align*}y=-x-6\end{align*} 3. \begin{align*}y=\frac{5}{3} x-4\end{align*} 4. \begin{align*}0.30x+0.70y=15\end{align*} 5. \begin{align*}5= \frac{1}{6} x-y\end{align*} 6. \begin{align*}y-7=-5(x-12)\end{align*} 7. \begin{align*}2y=6x+9\end{align*} 8. \begin{align*}y=\frac{9}{4}x+\frac{1}{4}\end{align*} 9. \begin{align*}y+\frac{3}{5}=\frac{2}{3}(x-2) \end{align*} 10. \begin{align*}3y+5=4(x-9)\end{align*} Find the slope and \begin{align*}y-\end{align*}intercept of the following lines. 1. \begin{align*}5x-2y=15\end{align*} 2. \begin{align*}3x+6y=25\end{align*} 3. \begin{align*}x-8y=12\end{align*} 4. \begin{align*}3x-7y=20\end{align*} 5. \begin{align*}9x-9y=4\end{align*} 6. \begin{align*}6x+y=3\end{align*} 7. \begin{align*}x-y=9\end{align*} 8. \begin{align*}8x+3y=15\end{align*} 9. \begin{align*}4x+9y=1\end{align*} In 23 – 27, write each equation in standard form by first writing it in point-slope form. 1. \begin{align*}Slope = -1\end{align*} through point (–3, 5) 2. \begin{align*}Slope = -\frac{1}{4}\end{align*} through point (4, 0) 3. Line through (5, –2) and (–5, 4) 4. Line through (–3, –2) and (5, 1) 5. Line through (1, –1) and (5, 2) 6. The farmer’s market sells tomatoes and corn. Tomatoes are selling for1.29 per pound and corn is selling for $3.25 per pound. If you buy 6 pounds of tomatoes, how many pounds of corn can you buy if your total spending cash is$11.61?
7. The local church is hosting a Friday night fish fry for Lent. They sell a fried fish dinner for $7.50 and a baked fish dinner for$8.25. The church sold 130 fried fish dinners and took in $2,336.25. How many baked fish dinners were sold? 8. Andrew has two part-time jobs. One pays$6 per hour and the other pays $10 per hour. He wants to make$366 per week. Write an equation in standard form that describes this situation. If he is only allowed to work 15 hours per week at the $10 per hour job, how many hours does he need to work per week at his$6 per hour job in order to achieve his goal?
9. Anne invests money in two accounts. One account returns 5% annual interest and the other returns 7% annual interest. In order not to incur a tax penalty, she can make no more than $400 in interest per year. Write an equation in standard form that describes this problem. If she invests$5000 in the 5% interest account, how much money does she need to invest in the other account?

Mixed Review

1. Write the following equation in slope-intercept form: \begin{align*}y-2=6(x-3)\end{align*}.
2. Solve for \begin{align*}p:\frac{p-2}{7}=\frac{p+1}{6}\end{align*}.
3. Describe the graph \begin{align*}x=1.5\end{align*}.
4. Tell whether (4, –3) is a solution to \begin{align*}5x+3y=9\end{align*}.
5. Give the coordinates of a point located in quadrant III.
6. Find the slope between (6, 6) and (16, 6).
7. Graph the equation \begin{align*}y=\frac{5}{9} x-7\end{align*}.

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