# 2.4 – Linear Equations and Inequalities

## Objectives

• Write the equation of a line in three forms.
• Identify key components of a line from a given equation.
• Express the solution to a linear inequality graphically.

## Key Terms

• Coefficients – Numbers that multiply variables.
• The number in front of a variable
• Can be negative or positive
• Ex. 4x: coef is 4
• Ex. -x: coef is -1
• Ex. -3/5: coef is -3/5
• Coordinates – locations on a map
• In math, (x,y) are the coordinates that represent a point located on the Cartesian plane
• (x,y) are in alphabetical order, and you start plotting points with the x-axis first, then the y-axis
• Linear Inequality – An inequality in which the variable is of degree one.
• The graph of this line is shaded above or below the line
• The graph may or may not include the line itself (depends if the inequality has an equal bar under it or not)
• Point-Slope Form – The equation of a straight line in the form:
• $y-{y}_{1}=m\left(x-{x}_{1}\right)$
• The x and y in the equation remain part of the equation.  Do not substitute a value for these.
• The $({x}_{1},{y}_{1})$ represent ANY point on the line.  You choose!
• Slope-Intercept Form – A form of a linear equation that includes the slope of the line and the value of they-intercept.
• Form:  y = mx + b
• m:  slope (the coefficient of x)
• b:  y-intercept (written as (0,b))
• Standard Form of a Linear Equation – Form:  Ax + By + C = 0.
• Undefined – A value that cannot be computed.
• The slope of the vertical line below is undefined (ex. x=3) because it is ALL rise and NO run.

• Zero Slope – A horizontal line has no slope, also known as a zero slope (ex. y=2) because it is NO rise and ALL run.

## Notes

Graphing Inequalities
• To graph a linear inequality
• Solve for y and shade the solution area
• A linear equality has a border line (since it may or may not be included in the solution).
• Ask yourself, “are the y-values above (greater than) or below (less than) the line?”  (hint: look at the inequality sign)
• If the sign is $y \geq$ or $y \textgreater$, then you shade above the line
• All points above the line are included in the solution
• If the sign is $y \leq$ or $y \textless$, then you shade below the line
• All points below the line are included in the solution
• If the sign has a bar (equal to), then you use a solid line
• This means the line is included in the solution
• If the sign does not have a bar (not equal to), then you use a dashed line
• This means the line is NOT included in the solution

• Ex 1. $y\leq\frac{2}{3}x+1$
• Step 1: Graph the line, but ask yourself, “Does the inequality have a bar (equal to)?”
• If yes, graph the line as a solid line
• If no, graph the line as a dashed line
• Step 2: Shade the half-plane (above or below the line), but ask yourself, “Are my y-values greater or less than the line?”

Answer: There is a bar, so the border line is solid.  The y-coordinates are BELOW ($y\leq$) the line, so shade below the border line.

## Review

Linear Equations in the Real World
• Ex 1. Dianne pays $28 to enter a state fair, plus$2 for each ride. Write the equation that represents her total cost?
• Answer:  y = 2x + 28
• Reason: $28 is a constant price that everyone MUST pay to enter the fair.$2 is the cost of EACH ride.  Each means multiply.  We don’t know the number of rides she will go on, so we use x to represent the number of rides.
• Ex 2. A consultant needs to make at least $800 this week. She earns$80 for each new written piece and $40 for each review. Write an inequality that represents the possible combinations of reviews and new written pieces that she must complete? • Assign variables: x for written pieces and y for reviews • Decide on the inequality sign: “at least” means the work she does is valued “greater than or equal to”$800, so $\geq800$.
• Combinations means the addition of both pieces of work (written pieces and reviews), so use a plus (+) sign for the operator
• Answer: $80x+40y\geq800$
• Ex 3. What will the graph of $y+2=\frac{1}{5}(x+1)$ look like?
• Since the point-slope equation is $y-y_1=m(x-x_1)$, and our equation has plus signs, the points are both negative (-1, -2).
• The slope is $\frac{1}{5}$, so the rise is 1 and the run is 5.
• It’s easier to create a line from slope-intercept form, so distribute and use inverse operations to convert the form.
• $y+2=\frac{1}{5}x+\frac{1}{5}$
• $y=\frac{1}{5}x+\frac{1}{5}-2$, convert -2 to a fraction with 5 in the denominator: $-2=\frac{-10}{5}$
• $y=\frac{1}{5}x+\frac{1}{5}-\frac{10}{5}$
• $y=\frac{1}{5}x+\frac{-9}{5}$, -9 divided by 5 = -1.8 which is easy to graph.
• The graph would look like this:

• Ex 4. What will the graph of $y \textgreater \frac{1}{3}x-2$ look like?
• The slope is $\frac{1}{3}$, so the rise is 1 and the run is 3.
• The y-intercept is -2. Start there, plot a point at (0,-2).
• Follow the slope up 1 and right 3, plot another point.
• Since there is NO bar under the inequality sign, connect the points with a DASHED line.
• Since $y \textgreater$, shade ABOVE the border line.
• The graph would look like this: