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# 5.1: Forms of Linear Equations

Difficulty Level: At Grade Created by: CK-12

## Learning Objectives

• Write equations in slope-intercept form.
• Write equations in point-slope form.
• Write equations in standard form.
• Solve real-world problems using linear models in all three forms.

## Introduction

We saw in the last chapter that many real-world situations can be described with linear graphs and equations. In this chapter, we’ll see how to find those equations in a variety of situations.

## Write an Equation Given Slope and \begin{align*}y-\end{align*}Intercept

You’ve already learned how to write an equation in slope–intercept form: simply start with the general equation for the slope-intercept form of a line, \begin{align*}y=mx+b\end{align*}, and then plug the given values of \begin{align*}m\end{align*} and \begin{align*}b\end{align*} into the equation. For example, a line with a slope of 4 and a \begin{align*}y-\end{align*}intercept of -3 would have the equation \begin{align*}y=4x-3\end{align*}.

If you are given just the graph of a line, you can read off the slope and \begin{align*}y-\end{align*}intercept from the graph and write the equation from there. For example, on the graph below you can see that the line rises by 1 unit as it moves 2 units to the right, so its slope is \begin{align*}\frac{1}{2}\end{align*}. Also, you can see that the \begin{align*}y-\end{align*}intercept is -2, so the equation of the line is \begin{align*}y=\frac{1}{2} x-2\end{align*}.

## Write an Equation Given the Slope and a Point

Often, we don’t know the value of the \begin{align*}y-\end{align*}intercept, but we know the value of \begin{align*}y\end{align*} for a non-zero value of \begin{align*}x\end{align*}. In this case, it’s often easier to write an equation of the line in point-slope form. An equation in point-slope form is written as \begin{align*}y-y_0=m(x-x_0)\end{align*}, where \begin{align*}m\end{align*} is the slope and \begin{align*}(x_0, y_0)\end{align*} is a point on the line.

Example 1

A line has a slope of \begin{align*}\frac{3}{5}\end{align*}, and the point (2, 6) is on the line. Write the equation of the line in point-slope form.

Solution

Start with the formula \begin{align*}y-y_0=m(x-x_0)\end{align*}.

Plug in \begin{align*}\frac{3}{5}\end{align*} for \begin{align*}m\end{align*}, 2 for \begin{align*}x_0\end{align*} and 6 for \begin{align*}y_0\end{align*}.

The equation in point-slope form is \begin{align*}y-6=\frac{3}{5}(x-2)\end{align*}.

Notice that the equation in point-slope form is not solved for \begin{align*}y\end{align*}. If we did solve it for \begin{align*}y\end{align*}, we’d have it in \begin{align*}y-\end{align*}intercept form. To do that, we would just need to distribute the \begin{align*}\frac{3}{5}\end{align*} and add 6 to both sides. That means that the equation of this line in slope-intercept form is \begin{align*}y=\frac{3}{5}x-\frac{6}{5}+6\end{align*}, or simply \begin{align*}y=\frac{3}{5}x+\frac{24}{5}\end{align*}.

## Write an Equation Given Two Points

Point-slope form also comes in useful when we need to find an equation given just two points on a line.

For example, suppose we are told that the line passes through the points (-2, 3) and (5, 2). To find the equation of the line, we can start by finding the slope.

Starting with the slope formula, \begin{align*}m=\frac{y_2-y_1}{x_2-x_1}\end{align*}, we plug in the \begin{align*}x-\end{align*} and \begin{align*}y-\end{align*}values of the two points to get \begin{align*}m=\frac{2-3}{5-(-2)}=\frac{-1}{7}\end{align*}. We can plug that value of \begin{align*}m\end{align*} into the point-slope formula to get \begin{align*}y-y_0=-\frac{1}{7}(x-x_0)\end{align*}.

Now we just need to pick one of the two points to plug into the formula. Let’s use (5, 2); that gives us \begin{align*}y-2=-\frac{1}{7}(x-5)\end{align*}.

What if we’d picked the other point instead? Then we’d have ended up with the equation \begin{align*}y-3=-\frac{1}{7}(x+2)\end{align*}, which doesn’t look the same. That’s because there’s more than one way to write an equation for a given line in point-slope form. But let’s see what happens if we solve each of those equations for \begin{align*}y\end{align*}.

Starting with \begin{align*}y-2=-\frac{1}{7}(x-5)\end{align*}, we distribute the \begin{align*}-\frac{1}{7}\end{align*} and add 2 to both sides. That gives us \begin{align*}y=-\frac{1}{7} x+\frac{5}{7}+2\end{align*}, or \begin{align*}y=-\frac{1}{7}x+\frac{19}{7}\end{align*}.

On the other hand, if we start with \begin{align*}y-3=-\frac{1}{7}(x+2)\end{align*}, we need to distribute the \begin{align*}-\frac{1}{7}\end{align*} and add 3 to both sides. That gives us \begin{align*}y=-\frac{1}{7}x-\frac{2}{7}+3\end{align*}, which also simplifies to \begin{align*}y=-\frac{1}{7}x+\frac{19}{7}\end{align*}.

So whichever point we choose to get an equation in point-slope form, the equation is still mathematically the same, and we can see this when we convert it to \begin{align*}y-\end{align*}intercept form.

Example 2

A line contains the points (3, 2) and (-2, 4). Write an equation for the line in point-slope form; then write an equation in \begin{align*}y-\end{align*}intercept form.

Solution

Find the slope of the line: \begin{align*}m=\frac{y_2-y_1}{x_2-x_1}=\frac{4-2}{-2-3}=-\frac{2}{5}\end{align*}

Plug in the value of the slope: \begin{align*}y-y_0=-\frac{2}{5}(x-x_0)\end{align*}.

Plug point (3, 2) into the equation: \begin{align*}y-2=-\frac{2}{5}(x-3)\end{align*}.

The equation in point-slope form is \begin{align*}y-2=-\frac{2}{5}(x-3)\end{align*}.

To convert to \begin{align*}y-\end{align*}intercept form, simply solve for \begin{align*}y\end{align*}:

\begin{align*}y-2=-\frac{2}{5}(x-3) \rightarrow y-2=-\frac{2}{5}x-\frac{6}{5} \rightarrow y=-\frac{2}{5}x-\frac{6}{5}+2=-\frac{2}{5}x+\frac{4}{5}.\end{align*}

The equation in \begin{align*}y-\end{align*}intercept form is \begin{align*}y=-\frac{2}{5}x+\frac{4}{5}\end{align*}.

## Graph an Equation in Point-Slope Form

Another useful thing about point-slope form is that you can use it to graph an equation without having to convert it to slope-intercept form. From the equation \begin{align*}y-y_0=m(x-x_0)\end{align*}, you can just read off the slope \begin{align*}m\end{align*} and the point \begin{align*}(x_0, y_0)\end{align*}. To draw the graph, all you have to do is plot the point, and then use the slope to figure out how many units up and over you should move to find another point on the line.

Example 5

Make a graph of the line given by the equation \begin{align*}y+2=\frac{2}{3}(x-2)\end{align*}.

Solution

To read off the right values, we need to rewrite the equation slightly: \begin{align*}y-(-2)=\frac{2}{3}(x-2)\end{align*}. Now we see that point (2, -2) is on the line and that the slope is \begin{align*}\frac{2}{3}\end{align*}.

First plot point (2, -2) on the graph:

A slope of \begin{align*}\frac{2}{3}\end{align*} tells you that from that point you should move 2 units up and 3 units to the right and draw another point:

Now draw a line through the two points and extend it in both directions:

## Linear Equations in Standard Form

You’ve already encountered another useful form for writing linear equations: standard form. An equation in standard form is written \begin{align*}ax+by=c\end{align*}, where \begin{align*}a, b\end{align*}, and \begin{align*}c\end{align*} are all integers and \begin{align*}a\end{align*} is positive. (Note that the \begin{align*}b\end{align*} in the standard form is different than the \begin{align*}b\end{align*} in the slope-intercept form.)

One useful thing about standard form is that it allows us to write equations for vertical lines, which we can’t do in slope-intercept form.

For example, let’s look at the line that passes through points (2, 6) and (2, 9). How would we find an equation for that line in slope-intercept form?

First we’d need to find the slope: \begin{align*}m=\frac{9-6}{0-0}=\frac{3}{0}\end{align*}. But that slope is undefined because we can’t divide by zero. And if we can’t find the slope, we can’t use point-slope form either.

If we just graph the line, we can see that \begin{align*}x\end{align*} equals 2 no matter what \begin{align*}y\end{align*} is. There’s no way to express that in slope-intercept or point-slope form, but in standard form we can just say that \begin{align*}x+0y=2\end{align*}, or simply \begin{align*}x=2\end{align*}.

## Converting to Standard Form

To convert an equation from another form to standard form, all you need to do is rewrite the equation so that all the variables are on one side of the equation and the coefficient of \begin{align*}x\end{align*} is not negative.

Example 1

Rewrite the following equations in standard form:

a) \begin{align*}y=5x-7\end{align*}

b) \begin{align*}y-2=-3(x+3)\end{align*}

c) \begin{align*}y=\frac{2}{3}x+\frac{1}{2}\end{align*}

Solution

We need to rewrite each equation so that all the variables are on one side and the coefficient of \begin{align*}x\end{align*} is not negative.

a) \begin{align*}y=5x-7\end{align*}

Subtract \begin{align*}y\end{align*} from both sides to get \begin{align*}0=5x-y-7\end{align*}.

Add 7 to both sides to get \begin{align*}7=5x-y\end{align*}.

Flip the equation around to put it in standard form: \begin{align*}5x-y=7\end{align*}.

b) \begin{align*}y-2=-3(x+3)\end{align*}

Distribute the –3 on the right-hand-side to get \begin{align*}y-2=-3x-9\end{align*}.

Add \begin{align*}3x\end{align*} to both sides to get \begin{align*}y+3x-2=-9\end{align*}.

Add 2 to both sides to get \begin{align*}y+3x=-7\end{align*}. Flip that around to get \begin{align*}3x+y=-7\end{align*}.

c) \begin{align*}y=\frac{2}{3}x+\frac{1}{2}\end{align*}

Find the common denominator for all terms in the equation – in this case that would be 6.

Multiply all terms in the equation by 6: \begin{align*}6 \left(y=\frac{2}{3}x+\frac{1}{2}\right) \Rightarrow 6y=4x+3\end{align*}

Subtract \begin{align*}6y\end{align*} from both sides: \begin{align*}0=4x-6y+3\end{align*}

Subtract 3 from both sides: \begin{align*}-3=4x-6y\end{align*}

The equation in standard form is \begin{align*}4x-6y=-3\end{align*}.

## Graphing Equations in Standard Form

When an equation is in slope-intercept form or point-slope form, you can tell right away what the slope is. How do you find the slope when an equation is in standard form?

Well, you could rewrite the equation in slope-intercept form and read off the slope. But there’s an even easier way. Let’s look at what happens when we rewrite an equation in standard form.

Starting with the equation \begin{align*}ax+by=c\end{align*}, we would subtract \begin{align*}ax\end{align*} from both sides to get \begin{align*}by=-ax+c\end{align*}. Then we would divide all terms by \begin{align*}b\end{align*} and end up with \begin{align*}y=-\frac{a}{b}x+\frac{c}{b}\end{align*}.

That means that the slope is \begin{align*}-\frac{a}{b}\end{align*} and the \begin{align*}y-\end{align*}intercept is \begin{align*}\frac{c}{b}\end{align*}. So next time we look at an equation in standard form, we don’t have to rewrite it to find the slope; we know the slope is just \begin{align*}-\frac{a}{b}\end{align*}, where \begin{align*}a\end{align*} and \begin{align*}b\end{align*} are the coefficients of \begin{align*}x\end{align*} and \begin{align*}y\end{align*} in the equation.

Example 2

Find the slope and the \begin{align*}y-\end{align*}intercept of the following equations written in standard form.

a) \begin{align*}3x+5y=6\end{align*}

b) \begin{align*}2x-3y=-8\end{align*}

c) \begin{align*}x-5y=10\end{align*}

Solution

a) \begin{align*}a=3, b=5\end{align*}, and \begin{align*}c=6\end{align*}, so the slope is \begin{align*}-\frac{a}{b}=-\frac{3}{5}\end{align*}, and the \begin{align*}y-\end{align*}intercept is \begin{align*}\frac{c}{b}=\frac{6}{5}\end{align*}.

b) \begin{align*}a=2, b=-3\end{align*}, and \begin{align*}c=-8\end{align*}, so the slope is \begin{align*}-\frac{a}{b}=\frac{2}{3}\end{align*}, and the \begin{align*}y-\end{align*}intercept is \begin{align*}\frac{c}{b}=\frac{8}{3}\end{align*}.

c) \begin{align*}a=1, b=-5\end{align*}, and \begin{align*}c=10\end{align*}, so the slope is \begin{align*}-\frac{a}{b}=\frac{1}{5}\end{align*}, and the \begin{align*}y-\end{align*}intercept is \begin{align*}\frac{c}{b}=\frac{10}{-5}=-2\end{align*}.

Once we’ve found the slope and \begin{align*}y-\end{align*}intercept of an equation in standard form, we can graph it easily. But if we start with a graph, how do we find an equation of that line in standard form?

First, remember that we can also use the cover-up method to graph an equation in standard form, by finding the intercepts of the line. For example, let’s graph the line given by the equation \begin{align*}3x-2y=6\end{align*}.

To find the \begin{align*}x-\end{align*}intercept, cover up the \begin{align*}y\end{align*} term (remember, the \begin{align*}x-\end{align*}intercept is where \begin{align*}y = 0\end{align*}):

\begin{align*}3x=6 \Rightarrow x=2\end{align*}

The \begin{align*}x-\end{align*} intercept is (2, 0).

To find the \begin{align*}y-\end{align*}intercept, cover up the \begin{align*}x\end{align*} term (remember, the \begin{align*}y-\end{align*}intercept is where \begin{align*}x = 0)\end{align*}:

\begin{align*}-2y=6 \Rightarrow y=-3\end{align*}

The \begin{align*}y-\end{align*}intercept is (0, -3).

We plot the intercepts and draw a line through them that extends in both directions:

Now we want to apply this process in reverse—to start with the graph of the line and write the equation of the line in standard form.

Example 3

Find the equation of each line and write it in standard form.

a)

b)

c)

Solution

a) We see that the \begin{align*}x-\end{align*}intercept is \begin{align*}(3, 0) \Rightarrow x=3\end{align*} and the \begin{align*}y-\end{align*}intercept is \begin{align*}(0, -4) \Rightarrow y=-4\end{align*}

We saw that in standard form \begin{align*}ax+by=c\end{align*}: if we “cover up” the \begin{align*}y\end{align*} term, we get \begin{align*}ax = c\end{align*}, and if we “cover up” the \begin{align*}x\end{align*} term, we get \begin{align*}by = c\end{align*}.

So we need to find values for \begin{align*}a\end{align*} and \begin{align*}b\end{align*} so that we can plug in 3 for \begin{align*}x\end{align*} and -4 for \begin{align*}y\end{align*} and get the same value for \begin{align*}c\end{align*} in both cases. This is like finding the least common multiple of the \begin{align*}x-\end{align*} and \begin{align*}y-\end{align*}intercepts.

In this case, we see that multiplying \begin{align*}x=3\end{align*} by 4 and multiplying \begin{align*}y=-4\end{align*} by –3 gives the same result:

\begin{align*}(x=3) \times 4 \Rightarrow 4x=12 \quad \text{and} \quad (y=-4) \times (-3) \Rightarrow -3y=12\end{align*}

Therefore, \begin{align*}a = 4, b = -3\end{align*} and \begin{align*}c = 12\end{align*} and the equation in standard form is \begin{align*}4x-3y=12\end{align*}.

b) We see that the \begin{align*}x-\end{align*}intercept is \begin{align*}(3, 0) \Rightarrow x=3\end{align*} and the \begin{align*}y-\end{align*}intercept is \begin{align*}(0, 3) \Rightarrow y=3\end{align*}

The values of the intercept equations are already the same, so \begin{align*}a = 1, b = 1\end{align*} and \begin{align*}c = 3\end{align*}. The equation in standard form is \begin{align*}x+y=3\end{align*}.

c) We see that the \begin{align*}x-\end{align*}intercept is \begin{align*}\left(\frac{3}{2}, 0 \right) \Rightarrow x=\frac{3}{2}\end{align*} and the \begin{align*}y-\end{align*}intercept is \begin{align*}(0, 4) \Rightarrow y=4\end{align*}

Let’s multiply the \begin{align*}x-\end{align*}intercept equation by \begin{align*}2 \Rightarrow 2x=3\end{align*}

Then we see we can multiply the \begin{align*}x-\end{align*}intercept again by 4 and the \begin{align*}y-\end{align*}intercept by 3, so we end up with \begin{align*}8x=12\end{align*} and \begin{align*}3y=12\end{align*}.

The equation in standard form is \begin{align*}8x+3y=12\end{align*}.

## Solving Real-World Problems Using Linear Models in Point-Slope Form

Let’s solve some word problems where we need to write the equation of a straight line in point-slope form.

Example 4

Marciel rented a moving truck for the day. Marciel only remembers that the rental truck company charges $40 per day and some number of cents per mile. Marciel drives 46 miles and the final amount of the bill (before tax) is$63. What is the amount per mile the truck rental company charges? Write an equation in point-slope form that describes this situation. How much would it cost to rent this truck if Marciel drove 220 miles?

Solution

Let’s define our variables:

\begin{align*}x = \text{distance in miles}\!\\ y = \text{cost of the rental truck}\end{align*}

Peter pays a flat fee of 40 for the day; this is the \begin{align*}y-\end{align*}intercept. He pays63 for 46 miles; this is the coordinate point (46,63).

Start with the point-slope form of the line: \begin{align*}y-y_0=m(x-x_0)\end{align*}

Plug in the coordinate point: \begin{align*}63-y_0=m(46-x_0)\end{align*}

Plug in the point (0, 40): \begin{align*}63-40=m(46-0)\end{align*}

Solve for the slope: \begin{align*}23=46m \rightarrow m=\frac{23}{46}=0.5\end{align*}

The slope is 0.5 dollars per mile, so the truck company charges 50 cents per mile (0.5 = 50 cents). Plugging in the slope and the \begin{align*}y-\end{align*}intercept, the equation of the line is \begin{align*}y=0.5x+40\end{align*}. To find out the cost of driving the truck 220 miles, we plug in \begin{align*}x=220\end{align*} to get \begin{align*}y-40=0.5(220) \Rightarrow y= \ 150\end{align*}. Driving 220 miles would cost150.

Example 5

Anne got a job selling window shades. She receives a monthly base salary and a $6 commission for each window shade she sells. At the end of the month she adds up sales and she figures out that she sold 200 window shades and made$2500. Write an equation in point-slope form that describes this situation. How much is Anne’s monthly base salary?

Solution

Let’s define our variables:

\begin{align*}x = \text{number of window shades sold}\!\\ y = \text{Anne’s earnings}\end{align*}

We see that we are given the slope and a point on the line:

Nadia gets $6 for each shade, so the slope is 6. She made$2500 when she sold 200 shades, so the point is (200, 2500).

Start with the point-slope form of the line: \begin{align*}y-y_0=m(x-x_0)\end{align*}

Plug in the slope: \begin{align*}y-y_0=6(x-x_0)\end{align*}

Plug in the point (200, 2500): \begin{align*}y-2500=6(x-200)\end{align*}

To find Anne’s base salary, we plug in \begin{align*}x = 0\end{align*} and get \begin{align*}y-2500=-1200 \Rightarrow y=\ 1300\end{align*}.

Anne’s monthly base salary is $1300. ## Solving Real-World Problems Using Linear Models in Standard Form Here are two examples of real-world problems where the standard form of an equation is useful. Example 6 Nadia 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

Let’s define our variables:

\begin{align*}x = \text{pounds of oranges}\!\\ y = \text{pounds of cherries}\end{align*}

The equation that describes this situation is \begin{align*}2x+3y=12\end{align*}.

If she buys 4 pounds of oranges, we can plug \begin{align*}x = 4\end{align*} into the equation and solve for \begin{align*}y\end{align*}:

\begin{align*}2(4)+3y=12 \Rightarrow 3y=12-8 \Rightarrow 3y=4 \Rightarrow y=\frac{4}{3}\end{align*}

Nadia can buy \begin{align*}1 \frac{1}{3}\end{align*} pounds of cherries.

Example 7

Peter skateboards part of the way to school and walks 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 he skateboards for \begin{align*}\frac{1}{2}\end{align*} an hour, how long does he need to walk to get to school?

Solution

Let’s define our variables:

\begin{align*}x = \text{time Peter skateboards}\!\\ y = \text{time Peter walks}\end{align*}

The equation that describes this situation is: \begin{align*}7x+3y=6\end{align*}

If Peter skateboards \begin{align*}\frac{1}{2}\end{align*} an hour, we can plug \begin{align*}x = 0.5\end{align*} into 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*}

Peter must walk \begin{align*}\frac{5}{6}\end{align*} of an hour.

## Further Practice

Now that you’ve worked with equations in all three basic forms, check out the Java applet at http://www.ronblond.com/M10/lineAP/index.html. You can use it to manipulate graphs of equations in all three forms, and see how the graphs change when you vary the terms of the equations.

Another applet at http://www.cut-the-knot.org/Curriculum/Calculus/StraightLine.shtml lets you create multiple lines and see how they intersect. Each line is defined by two points; you can change the slope of a line by moving either of the points, or just drag the whole line around without changing its slope. To create another line, just click Duplicate and then drag one of the lines that are already there.

## Review Questions

Find the equation of each line in slope–intercept form.

1. The line has a slope of 7 and a \begin{align*}y-\end{align*}intercept of -2.
2. The line has a slope of -5 and a \begin{align*}y-\end{align*}intercept of 6.
3. The line has a slope of \begin{align*}-\frac{1}{4}\end{align*} and contains the point (4, -1).
4. The line has a slope of \begin{align*}\frac{2}{3}\end{align*} and contains the point \begin{align*}\left(\frac{1}{2}, 1 \right)\end{align*}.
5. The line has a slope of -1 and contains the point \begin{align*}\left(\frac{4}{5}, 0 \right)\end{align*}.
6. The line contains points (2, 6) and (5, 0).
7. The line contains points (5, -2) and (8, 4).
8. The line contains points (3, 5) and (-3, 0).
9. The line contains points (10, 15) and (12, 20).

Write the equation of each line in slope-intercept form.

Find the equation of each linear function in slope–intercept form.

1. \begin{align*}m=5, f(0)=-3\end{align*}
2. \begin{align*}m=-7, f(2)=-1\end{align*}
3. \begin{align*}m=\frac{1}{3}, f(-1)=\frac{2}{3}\end{align*}
4. \begin{align*}m=4.2, f(-3)=7.1\end{align*}
5. \begin{align*}f \left(\frac{1}{4}\right)=\frac{3}{4}, f(0)=\frac{5}{4}\end{align*}
6. \begin{align*}f(1.5)=-3, f(-1)=2\end{align*}

Write the equation of each line in point-slope form.

1. The line has slope \begin{align*}-\frac{1}{10}\end{align*} and goes through the point (10, 2).
2. The line has slope -75 and goes through the point (0, 125).
3. The line has slope 10 and goes through the point (8, -2).
4. The line goes through the points (-2, 3) and (-1, -2).
5. The line contains the points (10, 12) and (5, 25).
6. The line goes through the points (2, 3) and (0, 3).
7. The line has a slope of \begin{align*}\frac{3}{5}\end{align*} and a \begin{align*}y-\end{align*}intercept of -3.
8. The line has a slope of -6 and a \begin{align*}y-\end{align*}intercept of 0.5.

Write the equation of each linear function in point-slope form.

1. \begin{align*}m=-\frac{1}{5}\end{align*} and \begin{align*}f(0)=7\end{align*}
2. \begin{align*}m=-12\end{align*} and \begin{align*}f(-2)=5\end{align*}
3. \begin{align*}f(-7)=5\end{align*} and \begin{align*}f(3)=-4\end{align*}
4. \begin{align*}f(6)=0\end{align*} and \begin{align*}f(0)=6\end{align*}
5. \begin{align*}m=3\end{align*} and \begin{align*}f(2)=-9\end{align*}
6. \begin{align*}m=-\frac{9}{5}\end{align*} and \begin{align*}f(0)=32\end{align*}

Rewrite the following equations in standard form.

1. \begin{align*}y=3x-8\end{align*}
2. \begin{align*}y-7=-5(x-12)\end{align*}
3. \begin{align*}2y=6x+9\end{align*}
4. \begin{align*}y=\frac{9}{4}x+\frac{1}{4}\end{align*}
5. \begin{align*}y+\frac{3}{5}=\frac{2}{3}(x-2)\end{align*}
6. \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*}

Find the equation of each line and write it in standard form.

1. 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 in his $6 per hour job in order to achieve his goal? 2. 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?

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CK.MAT.ENG.SE.2.Algebra-I.5.1