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Unit 4 Seminar Agenda Slope of a Line Writing Equations of Lines Solving Systems by Graphing Solving Systems Algebraically

Slope 4 types of slopes a diagonal line heading uphill (increasing) as you look at it from left to right a diagonal line heading downhill (decreasing) as you look at it from left to right a horizontal line a vertical line

Finding slope given 2 points Find the slope between (1,3) and (4,5)

Example: Find the slope of the line that passes through the points (-2, 6) and (-1, -2).

Example Find the slope of the line that passes through the points (0, 3) and (7, -2).

Example Find the slope of the line that passes through the points (1, 7) and (-3, 7).

Example find the slope of the line that passes through the points (6, 4) and (6, 2).

Finding slope and y-intercept given an equation First solve for y When you have it in the form y = mx + b then m is your slope and b is your y- intercept y = -3x + 7 m = -3 is your slope b = 7 is your y-intercept

y = -3x + 7

Example First move your 2x to the other side simplify We need to isolate the y so divide both sides by 4 simplify

Example First move your -x to the other side simplify We need to isolate the y so divide both sides by- 5 simplify

Parallel and Perpendicular Lines Parallel lines have the same slope, so if one line is y = 2x + 7, then the slope of a line parallel to this one is also 2. Perpendicular lines have “negative reciprocal” slopes, so if one line is y = 2x + 7, then the slope of a line perpendicular would be -1/2

Finding equations of lines using y = mx + b Example: Write the equation of a line when the slope is 2/3 and the y-intercept is (0, 5). From this information, I know that I will use 2/3 for m and 5 for b. Slope-intercept form: y = mx + b y = (2/3)x + (5) y = (2/3)x + 5 I’m leaving the parentheses around the 2/3 so that there is no confusion as to where the x goes (in the numerator).

Finding equations of lines using y – y 1 = m(x – x 1 ) Example: Write the equation of a line whose slope is -2 and passes through the point (3, 5). To prepare for using the point-slope equation, let’s label what we have: m = -2, x1 = 3, and y1 = 5. Plug them in. y – y 1 = m(x – x 1 ) y – (5) = (-2)(x – (3)) y – 5 = -2(x – 3) now distribute through the parentheses y – 5 = -2x + 6 y – = -2x add 5 to both sides y = -2x + 11 this is in SLOPE-INTERCEPT form y + 2x = -2x x add 2x to both sides in order to get xs and y all on the same side 2x + y = 11 as in standard form, the x is first, the y is second, then the equals sign followed by the constant (plain number). This is the form of the choices on your quiz.

Example Example: Write the equation of a line whose slope is 0 and passes through the point (-2, -2). y – y 1 = m(x – x 1 ) y – (-2) = (0)(x – (-2)) y + 2 = 0(x + 2) change minus a negative to plus a positive y + 2 = 0 0 times anything is 0! y + 2 – 2 = 0 – 2 notice that the x just got wiped out; therefore, get y alone y = -2 this is the equation of a horizontal line

Finding equations of the line given two points First find the slope Then use the point-slope equation as in the last problems. If you have the graph of a line, just pick two points and do as stated earlier.

Example Example: write the equation of the line that passes through the points (4, 2) and (7, 3). Step 1: find the slope

Step 2: Use the slope and one of the points (your choice) and plug them into POINT-SLOPE form of a line. (I’m just going to go with the first point.)

Changing to Standard Form

Systems of Equations Systems of equations are simply more than one equation involving more than one variable for which the values associated with the variables work for all equations in the system.

3 Possibilities for Systems of Equations 1. the two lines INTERSECT they have ONE point in common there is ONE unique solution to the system, in the form (x, y) 2. the two lines are PARALLEL to each other they have NO points in common there is NO SOLUTION to the system 3. the two lines are the SAME—called COINCIDENTAL lines they have ALL points in common there are INFINITELY MANY SOLUTIONS to the system

Important Terminology DEPENDENT: the graph of two lines looks like you only graphed one (coincidental lines). INDEPENDENT: the graph of two lines shows up as two lines (parallel and intersecting lines both qualify). CONSISTENT: there is at least one solution to the system (intersecting lines have one and coincidental lines have infinitely many). INCONSISTENT: there is no solution to the system (parallel lines).

Solving Systems of Equations using the Substitution Method Step 1: rearrange one equation in terms of one variable. In other words, get x by itself or get y by itself, whichever is easier at the time. Step 2: take this result and substitute it into the OTHER equation. Step 3: take this solution, substitute it into the equation you used in step #1, and solve for the other variable.

Use the substitution method to find the solution to the system x + y = 6 3x + 4y = 9 Example

Step 1: Rearrange one equation in terms of one variable. In other words, get x by itself or get y by itself, whichever is easier at the time. The first equation looks good to rearrange, as the coefficients of both x and y are 1. x + y = 6 x + y – y = 6 – y x = 6 – y

Step 2: Take this result and substitute it into the OTHER equation. In this case, we’re going to replace x in the other equation with 6 – y. 3x + 4y = 9 3(6 – y) + 4y = 9 Now, it may seem that we just make this equation longer, but in reality we’re in good shape. We changed the equation such that there is now only ONE variable to solve for, something we know how to do! 18 – 3y + 4y = y = y – 18 = 9 – 18 y = -9

Step 3: Take this solution, substitute it into the equation you used in step #1, and solve for the other variable. In this case, replace y in the step #1 equation with –9. x + y = 6 x + (-9) = 6 x – 9 = 6 x – = x = 15

Solving Systems of Equations using the Elimination/Addition Method Step 1: find (because they already exist) or create (by multiplying) additive inverses of one variable Step 2: combine like terms, straight down the columns. If you are in standard form as you should be, the xs are lined up, the ys are lined up, and the numbers are lined up. Step 3: take this solution, substitute it into one of the original equations, and solve for the other variable. For this method, it does not matter which equation you substitute into.

Use the elimination method to find the solution to the system x + y = 4 x – y = 1 **Note: the equations must start in standard form. If they’re not, rearrange them. Example

Step 1: Find (because they already exist) or create (by multiplying) additive inverses of one variable. In this case, they already exist, the ys. (Remember additive inverses are the same numbers with different signs like 2 and -2)

Step 2: Combine like terms, straight down the columns. If you are in standard form as you should be, the xs are lined up, the ys are lined up, and the numbers are lined up. x + y = 4 x – y = 1 2x = 5 The ys “cancelled out”, leaving only x to solve for. Solve for x. 2x = 5 2x/2 = 5/2 x = 5/2

Step 3: Take this solution, substitute it into one of the original equations, and solve for the other variable. For this method, it does not matter which equation you substitute into. x + y = 4 (5/2) + y = 4 5/2 + y – 5/2 = 4 – 5/2 y = 8/2 – 5/2 y = 3/2 Solution: (5/2, 3/2)

Special Cases 1. the two lines INTERSECT they have ONE point in common there is ONE unique solution to the system, in the form (x, y) 2. the two lines are PARALLEL to each other they have NO points in common there is NO SOLUTION to the system 3. the two lines are the SAME—called COINCIDENTAL lines they have ALL points in common there are INFINITELY MANY SOLUTIONS to the system

Example: use the elimination method to find the solution to the system 2x + 5y = 12 2x + 5y = 8 Even though these terms look alike, they are not additive inverses. Additive inverses have opposite signs. Let’s say that I choose to eliminate the xs. I need one positive 2x and one negative 2x. All I have to do to make a –2x is to multiply one of the equations by –1. I’ll do this to the second equation. -1[2x + 5y = 8] becomes –2x – 5y = -8 Here’s the system after that step: 2x + 5y = 12 –2x – 5y = -8 Combine like terms: 2x + 5y = 12 –2x – 5y = -8 0 = 4

Example: use the elimination method to find the solution to the system -3x + 9y = 4 3x – 9y = -4 I’ll make this one easy. We already have additive inverses of the xs…and of the ys, as it works out (hint, hint)…so let’s combine like terms. -3x + 9y = 4 3x – 9y = -4 0 = 0