1.4 Solving EquationsI can solve equations and solve problems by writing

equations

Success Criteria: I can identify an equation as

two expressions that are equal

I can use equations to model and solve problems

Warm Up

1. Do Now 2. Check HW #23. Lesson 1.44. Complete iReady5. HW #3

Evaluate each expression if a =-4, b = -2 and c = -3

1.

2. 4a + b -

Today

Properties of Real Numbers

Reflexive Propertya + b = a + b

Symmetric PropertyIf a = b then b = a

Transitive Property

If a = b and b = c then a = c

Substitution Property If a = b, then a can be replaced by b.

Solve 4 – 6a + 4a = –1 – 5(7 – 2a).

Example 2: Simplifying Each Side Before Solving Equations

Combine like terms.

Distribute –5 to the expression in parentheses.

4 – 6a + 4a = –1 –5(7 – 2a)

4 – 6a + 4a = –1 –5(7) –5(–2a)

4 – 6a + 4a = –1 – 35 + 10a

4 – 2a = –36 + 10a

+36 +36

40 – 2a = 10a+ 2a +2a

40 = 12a

Since –36 is added to 10a, add 36 to both sides.

To collect the variable terms on one side, add 2a to both sides.40 = 12a

Steps to Solving Equations

I. Simplify each side of the equationII. Move variables to one side of the

equation by using the opposite operation of addition or subtraction.

III. Isolate the variable by applying the opposite operation to each side.

I. First, use the opposite operation of addition or subtraction. II. Second, use the opposite operation of multiplication or

division.

IV. Check your answer.

Solve 12x – 3 + x = 5x – 4 + 8x.

Example 3B: Infinitely Many Solutions or No Solutions

Subtract 13x from both sides.

Identify like terms.12x – 3 + x = 5x – 4 + 8x

13x – 3 = 13x – 4

–3 = –4

–13x –13x False statement.

Combine like terms on the left and the right.

12x – 3 + x = 5x – 4 + 8x

The equation 12x – 3 + x = 5x – 4 + 8x is a contradiction. There is no value of x that will make the equation true. There are no solutions.

Solve 10 – 5x + 1 = 7x + 11 – 12x.

Example

Add 5x to both sides.

Identify like terms.10 – 5x + 1 = 7x + 11 – 12x

11 – 5x = 11 – 5x

11 = 11

+ 5x + 5x True statement.

Combine like terms on the left and the right.

10 – 5x + 1 = 7x + 11 – 12x

The equation 10 – 5x + 1 = 7x + 11 – 12x is an identity. All values of x will make the equation true. All real numbers are solutions.

An identity is an equation that is true for all values of the variable. An equation that is an identity has infinitely many solutions.

A contradiction is an equation that is not true for any value of the variable. It has no solutions.

Solve 2c + 7 + c = –14 + 3c + 21.

Check It Out! You Try

2c + 7 + c = –14 + 3c + 21

3c + 7 = 3c + 7

7 = 7

–3c –3c

True statement.

The equation 2c + 7 + c = –14 + 3c + 21 is an identity. All values of c will make the equation true. All real numbers are solutions.

4y + 7 – y = 10 + 3y

3y + 7 = 3y + 10

7 = 10

–3y –3y

False statement.

Solve 4y + 7 – y = 10 + 3y

The equation 4y + 7 – y = 10 + 3y is a contradiction. There is no value of y that will make the equation true. There are no solutions.

HW #3 Pg 30 #12 – 30 x3

1.4 Solving EquationsI can solve a formula for a given variable. I can solve an

equation in two or more variables for one

Success Criteria: I can identify an equation

as two expressions that are equal

I can use equations to model and solve problems

Warm Up – Solve each equation

1. Do Now 2. Check HW #2 HW #33. Finish Lesson 1.4 Literal Equations4. Complete iReady5. HW #4

Today

1.

2. 0.3s + 0.6 = 1.5

3. 10k – 6 = 9k + 2

3

19

8

A formula is an equation that states a rule for a relationship among quantities.

In the formula d = rt, d is isolated. You can "rearrange" a formula to isolate any variable by using inverse operations. This is called solving for a variable.

Solving for a VariableStep 1 Locate the variable you are asked to

solve for in the equation.

Step 2 Identify the operations on this variable and the order in which they are applied.

Step 3 Use inverse operations to undo operations and isolate the variable.

A formula is a type of literal equation. A literal equation is an equation with two or more variables. To solve for one of the variables, use inverse

operations.

Example 2A: Solving Formulas for a Variable

The formula for the area of a triangle is A = bh, where b is the length of the base, and is the height. Solve for h.

Locate h in the equation. A = bh

Since bh is multiplied by , divide both

sides by to undo the multiplication.2A = bh

Since h is multiplied by b, divide both sides by b to undo the multiplication.

Example 2B: Solving Formulas for a Variable

The formula for a person’s typing speed is ,where s is speed in words per minute,

w is number of words typed, e is number of errors, and m is number of minutes typing. Solve for e.

Locate e in the equation.

Since w–10e is divided by m, multiply both sides by m to undo the division.ms = w – 10e

–w –wms – w = –10e Since w is added to –10e,

subtract w from both sides to undo the addition.

Lesson Quiz: Part 1

Solve for the indicated variable.

1.

2.

3. 2x + 7y = 14 for y

4.

P = R – C for C C = R – P

for m m = x(k – 6 )

5. for C C = Rt + S

for h

Assignment #4

Pg 30 #33 – 42 x3, 48, 51, 57, 60, 61