Download 2-1-9-day-3

Learning Target:
Continue solving equations for x and begin to consider
special types of solutions such as “all numbers” and “no
solution.” Strengthen simplification and recording skills.
Focus questions:
• How can you simplify?
• Can you get the variable alone?
• Is there more than one way to simplify?
• Is there always a solution?
Equation A mathematical sentence in which two expressions appear on either side of an
“equals” sign (=) between them indicating that they have an equivalent value. A
statement of balance.
Solution The number (or numbers) that when substituted into an equation or inequality
make the equation or inequality true.
Simplify to make an expression as simple as possible using "legal" moves without
changing the value of the expression;
NOTE: the simplification either needs to represent zero (i.e., opposites together
on the same side in the same region) or needs to keep both sides balanced (i.e.,
adding or removing like tiles on both sides).
Solve To find all the solutions to an equation; to find the value(s) of the variable that
will make the statement true
Evaluate To find the numerical value of. To evaluate an expression, substitute the value(s)
given for the variable(s) and perform the operations according to the Order of
Operations. Red Light/Green Light
2-81 On your paper, write the equation represented in each diagram below. For
each equation, simplify as much as possible and then solve for x or y. Be sure
to record your work on your paper. a)
x
x
x
x
x
x
Red Light/Green Light
2-81 On your paper, write the equation represented in each diagram below. For
each equation, simplify as much as possible and then solve for x or y. Be sure
to record your work on your paper. b)
y
y
y
y
Red Light/Green Light
2-82. IS THERE A SOLUTION?
While solving homework last night, Richie came across three homework
questions that he thinks have no solution. Build each equation below and
determine if it has a solution for x. If it has a solution, find it. If it does not
have a solution, explain why not.
a)
x
x2
x
x2
x
Red Light/Green Light
2-82. IS THERE A SOLUTION?
While solving homework last night, Richie came across three homework
questions that he thinks have no solution. Build each equation below and
determine if it has a solution for x. If it has a solution, find it. If it does not
have a solution, explain why not.
b)
x
x
x
Red Light/Green Light
2-82. IS THERE A SOLUTION?
While solving homework last night, Richie came across three homework
questions that he thinks have no solution. Build each equation below and
determine if it has a solution for x. If it has a solution, find it. If it does not
have a solution, explain why not.
c)
x
x
Class Discussion
2-81 and 82. Comparison
• Look at the equations and solutions for 2-81a, 2-81b, 2-82b, and 2-82c.
• How are their solutions different?
• Why are there solutions different?
Pairs Check
2-83. Continue to develop your equation-solving strategies by solving each
equation below (if possible). Remember to build each equation, simplify as
much as possible, and solve for x or y. There are often multiple ways to solve
equations, so remember to justify that each step is “legal.” If you cannot solve
for x, explain why not. Be sure to record your work.
a) −x + 2 = 4
b) 4x − 2 + x = 2x + 8 + 3x
c) 4y − 9 + y = 6
d) 9 − (2 − 3y) = 6 + 2y − (5 + y)
Closure:
2-84: Learning Log: Solutions to an Equation
Discuss the three types of solutions and give examples
Learning Log
2.1.9
Explain when you can solve for x in an equation and when you
cannot. Be sure to give an example of each situation. Situation 1:
Example 1:
Situation 2:
Example 2:
Situation 3:
Example 3:
Using an Equation Mat
An Equation Mat can help you visually represent an equation with
algebra tiles.
For example, the equation: 2x − 1 − (−x + 3) = 6 − 2x can be represented by the
Equation Mat at right. (Note that
there are other possible ways to
represent this equation correctly
on the Equation Mat.)
2-85 thru 2-89
2-82 a) x2 + 2 - (-2x) = x - 2 - (-x2 - 1)
x2 + 2 + 2x = x - 2 + x2 + 1
2x + 2 = x - 1
x = -3
2-82 b) x + 1 - 1 - (- 2) = x + 3 - (-x + 1)
x+1-1+2=x+3 +x-1
x + 2 = 2x + 2
0=x
2-82 c) -x - 4 + 1 - (4 - 1) = -x - 2 - (4)
-x - 4 + 1 - 4 + 1 = -x - 2 - 4
-x - 6 = -x - 6
-6 = -6
True: this is ALWAYS true:
Therefore there are Infinitely Many Solutions!