Chapter 3 Study guide – Algebra II
Graph and Check method:
Steps:
1. Graph the two lines. They must be in y=mx+b (slope intercept form) to graph.
2. Find the ordered pair that the two lines intersect at ( x,y).
3. Use that ordered pair (x,y) to plug into BOTH original equations.
4. If (x,y) is a solution for both equations, then it is a solution.
5. You will get an ordered pair for an answer.
Slope­Intercept Form:
Example:
Graph both equations.
y=2x ­ 2
y = ­3/2x + 5
Check for intersection point. That will be the
solution. (2, 2).
Substitution method:
Steps:
1. Get one of the equations in the form x= something or y=something.
2. Take the x = or y = equation and plug it into the other equation.
3. Solve for the right variable.
4. Plug the answer that you get back into the other equation.
5. Solve for the other variable.
6. You will get an ordered pair for an answer.
Linear combination (elimination) method:
Steps:
1. Line both equations up correctly.
2. Multiply one equation (or maybe both) by a number that will allow you to eliminate one
when you add them together. ( One variable must become negative)
Be sure to multiply the entire equation by that number.
1. Add the two equations together.
2. Solve for one variable.
3. Plug the answer into the other equation.
4. Solve for the other variable.
5. You will get an ordered pair for an answer.
Solving Systems of equations with 3 variables:
Steps:
● Set up the 3 equations as A,B,C.
● You will create two new equations D,E.
● Choose 2 equations. Use elimination to get it down to two variables. Eq D.
● Choose 2 other equations. Use Elimination to get it down to the same two
variables. Eq. E.
● Use elimination with equations D and E to solve for one variable.
● Substitute it back in to get the other variable.
● Plug both of the solutions into an original equation to find the third variable.
● You will get three answers x, y, z.
Example:
A. 2x + y ­ z = ­1
B. 3x ­ y +2z=15
C. x +2y +3z =12
● A + B
2x + y ­ z = ­1 Add equations to eliminate y.
+3x ­ y +2z = 15
5x + 1z = 14 This becomes Equation D
● A + C
2x + y ­ z = ­1 Multiply by ­2 to eliminate y.
­4x ­ 2y +2z = 2
1x +2y +3z=12
­3x + 5z = 14 This becomes Equation E
● Solve D and E by using elimination.
D + E
5x + 1z = 14 Mulitply by ­5 to eliminate x.
­25x +­5z = ­70
­3x + 5z = 14
­28x = ­56 x = 2
● Plug 2 in for x and solve for z. 5(2) + 1z = 14 z = 4.
● Plug 2 and 4 into original equation to solve for y. 2(2) + y ­(4) = ­1 y = ­1
● Solution is x= 2, y = ­1, z = 4.
Story Problem Clues: ( With two variables..ex. I bought ice cream and hotdogs)
○ Determine the question
○ Decide if you should use one or two variables.
○ Define your variables – be sure they line up with whatever the answer should be
in. ( labels..ex. cost à cost)
○ Write one or two equations.
○ Solve for one variable.
○ Put that variable in to solve for the 2nd one.
○ Use elimination or substitution.
More story problem: ( formula ones..ex. How far did he go in this amt of time)
∙ Determine question ­ answer must be in that variable
∙ Decide if there is a formula you could use ex. D = r x t
∙ Draw a picture
∙ Write an equation
∙ Solve for the variable
Systems of inequalities: (system of constraints)
● Graph each line
● Y = is a horizontal line
● X = is a vertical line
● May need to get into slope intercept form y = mx + b
● > is a dashed line, > is a solid line
● Plug ( 0, 0 ) in to equation – shade in the direction of the true answers
● Graph both lines, shade where both solutions are
● The solution to the system is where the two shaded areas overlap
● Example: y = ­2/3x + 2 y = 2x ­3
Graph each inequality as if it was
tated in "y=" form.
If the inequality is < or > a dotted
line must be used to represent the
line. If the inequality is < or >, a solid
line is used.
Choose a test point to determine
which side of the line needs to be
shaded. The test point for this
problem was (0,0). Always pick a
point that is easy to work with.
For the test point (0,0),0 < 2(0)­3 False
0 (­2/3)0+2 False
Since both equations were false, shading
occurred on the other side of the line, not
covering the point.
The solution, S, is where the two shadings
overlap one another
Linear Programming:
Ex.
C = 2x +3y
Constraints:
x< 5 (Vertical line thru 5, shade to left)
y > 3 ( Horizontal line thru 3, shade above)
­3x + 5y < 30 ( plug zero in for x to find y intercept, plot point, plug zero in for y to find x
interecept, plot that point, then connect those two points (­10 on x, 6 on y , connect points, shade
below)
1. Graph each constraint
2. Shade where the true values have to be
3. Choose feasible region where all the shading occurs
4. Find the vertices of the shape formed. (Where the lines intersect).
5. Test the ordered pairs in the original C= equation
6. Whichever ordered pairs yield the highest value is the max, lowest is the min.