Chapter 1: Tools of Algebra 1-3: Solving Equations Essential Question: What is the procedure to solve an equation for a variable?

Chapter 1: Tools of Algebra1-3: Solving Equations

Essential Question: What is the procedure to solve an equation for a variable?

1-3: Solving Equations The solution of an equation is a number that can

be used in place of the variable that makes the equation true.

You can manipulate equations to help find a solution, so long as you do the same thing to both sides of the equation. Addition Property If a = b, then

a + c = b + c Subtraction Property If a = b, then

a – c = b – c Multiplication Property If a = b, then

ac = bc Division Property If a = b, then

a/c = b/c

1-3: Solving Equations Solve 13y + 48 = 8y – 47


13y + 48 = 8y – 47 – 48 – 48 (subtract 48 from both

sides) 13y = 8y – 95



sides) 13y = 8y – 95

–8y –8y (subtract 8y from both sides)

5y = – 95



sides) 13y = 8y – 95

–8y –8y (subtract 8y from both sides)

5y = – 955 5 (divide both sides by 5)

y = -19

1-3: Solving Equations Solve 3x – 7(2x – 13) = 3(-2x + 9)


3x – 7(2x – 13) = 3(-2x + 9) 3x – 14x + 91 = -6x + 27 (distribute)


3x – 7(2x – 13) = 3(-2x + 9) 3x – 14x + 91 = -6x + 27 (distribute) -11x + 91 = -6x + 27 (combine like

terms)



terms) - 91 - 91 (subtract 91

from both sides) -11x = -6x – 64





+6x +6x (add 6x to both sides) -5x = -64





+6x +6x (add 6x to both sides) -5x = -64

-5 -5 (divide both sides by -5)

x = 12.8

1-3: Solving Equations Solving a Formula for One of Its Variables

The formula for the area of a trapezoid isA = ½ h(b1 + b2). Solve the formula for h.

The goal is to use PEMDAS (in reverse) toget the variable in question alone.

A = ½ h(b1 + b2)

h

b

b

2

1




A = ½ h(b1 + b2)x2 x2 (multiply each side by 2, the reciprocal of ½)

2A = h(b1 + b2)

h

b

b

2

1




A = ½ h(b1 + b2)x2 x2 (multiply each side by 2, the reciprocal of ½)

2A = h(b1 + b2)(b1 + b2) (b1 + b2) (divide each side by b1 + b2)

h

b

b

2

1

1 2

2Ah

b b


The formula for the area of a trapezoid isA = ½ h(b1 + b2).

Solve the formula for b1

h

b

b

2

1

1-3: Solving Equations Solve for x1

x x

a b

1-3: Solving Equations Solve for x

(multiply both sides by a, to clear the first denominator)

1x x

a b

1x xa a

b

x

ax

b

a

aa



(multiply both sides by b, to clear the second denominator)

1x x

a b

1x x

a a aa b

axx a

b

axb bx b a

bx ab ab

x




(subtract bx on both sides, to get the x terms together)

1x x

a b

1x x

a a aa b

axx a

b

axb x b a b

bbx ab ax

ab x x

b bx

b

x

a





(distributive property, backwards)

1x x

a b

1x x

a a aa b

axx a

b

axb x b a b

bbx ab ax

bx bx

ab x xa b

( )ab a xb





(distributive property, backwards) (divide both sides by “a – b”)

1x x

a b

1x x

a a aa b

axx a

b

axb x b a b

bbx ab ax

bx bx

ab ax bx

( )ab a b x

( )

abx

a b


Are there any restrictions on the variables? Is there any number we couldn’t use in place of a, b, or x?

1x x

a b

( )

abx

a b


Are there any restrictions on the variables? Is there any number we couldn’t use in place of a, b, or x? Looking at the beginning problem

a ≠ 0 and b ≠ 0 (can’t have a denominator of 0)

1x x

a b

( )

abx

a b


Are there any restrictions on the variables? Is there any number we couldn’t use in place of a, b, or x? Looking at the beginning problem

a ≠ 0 and b ≠ 0 (can’t have a denominator of 0) Looking at the solution

a – b ≠ 0 (again, denominator can’t be 0) a ≠ b (add b to both sides)

1x x

a b

( )

abx

a b

1-3: Solving Equations Assignment

Page 21 1 – 27, odd problems

Show your work

Tomorrow: Word problems

Chapter 1: Tools of Algebra1-3: Solving Equations (Day 2)

Essential Question: What is the procedure to solve an equation for a variable?

1-3: Solving Equations Writing Equations to Solve Problems

Ex 5: A dog kennel owner has 100 ft of fencing to enclose a rectangular dog run. She wants it to be 5 times as long as it is wide. Find the dimensions of the dog run. (Optional) Draw a diagram Determine the formula to use



Perimeter = 2 • width + 2 • length Determine the unknowns



Perimeter = 2 • width + 2 • length Determine the unknowns

Let perimeter = 100 Let width = x Let length= 5x

Use variable in the equation, and solve

1-3: Solving Equations Perimeter = 2 • width + 2 • length

100 = 2 • x + 2 • 5x


100 = 2 • x + 2 • 5x 100 = 2x + 10x


100 = 2 • x + 2 • 5x 100 = 2x + 10x 100 = 12x


100 = 2 • x + 2 • 5x 100 = 2x + 10x 100 = 12x

12 12 8 1/3 = x

Determine both of your unknowns from the beginning of the problem


100 = 2 • x + 2 • 5x 100 = 2x + 10x 100 = 12x

12 12 8 1/3 = x

Determine both of your unknowns from the beginning of the problem Width = x = 8 1/3 ft Length = 5x = 5 • 8 1/3 = 41 2/3 ft


Ex 6: The lengths of the sides of a triangle are in the ratio 3:4:5. The perimeter of the triangle is 18 in. Find the length of the sides. (Optional) Draw a diagram Determine the formula to use



Perimeter = s1 + s2 + s3

Determine the variables



Perimeter = s1 + s2 + s3

Determine the variables Let perimeter = 18 Let s1 (shortest side) = 3x Let s2 (second side) = 4x Let s3 (third side) = 5x

Use variable in the equation, and solve

1-3: Solving Equations Perimeter = s1 + s2 + s3

18 = 3x + 4x + 5x


18 = 3x + 4x + 5x 18 = 12x


18 = 3x + 4x + 5x 18 = 12x

12 12 1.5 = x

Determine both of your variables (unknowns) from the beginning of the problem


18 = 3x + 4x + 5x 18 = 12x

12 12 1.5 = x

Determine both of your variables (unknowns) from the beginning of the problem s1 = 3x = 3 • 1.5 = 4.5 in s2 = 4x = 4 • 1.5 = 6 in s3 = 5x = 5 • 1.5 = 7.5 in

1-3: Solving Equations Assignment

Page 22 29 – 35, all problems

Skip 35b Show your work

What equation you used to solve the problem Some of the steps you took to find your solution

Chapter 1: Tools of Algebra 1-3: Solving Equations Essential Question: What is the procedure to solve an equation for a variable?

Documents

x slide

x terms

b slide

b c slide

denominator slide

backwards slide

equations day

b c multiplication propertyif