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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
1-3: Solving Equations Solve 13y + 48 = 8y – 47
13y + 48 = 8y – 47 – 48 – 48 (subtract 48 from both
sides) 13y = 8y – 95
1-3: Solving Equations Solve 13y + 48 = 8y – 47
13y + 48 = 8y – 47 – 48 – 48 (subtract 48 from both
sides) 13y = 8y – 95
–8y –8y (subtract 8y from both sides)
5y = – 95
1-3: Solving Equations Solve 13y + 48 = 8y – 47
13y + 48 = 8y – 47 – 48 – 48 (subtract 48 from both
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)
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)
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) -11x + 91 = -6x + 27 (combine like
terms)
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) -11x + 91 = -6x + 27 (combine like
terms) - 91 - 91 (subtract 91
from both sides) -11x = -6x – 64
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) -11x + 91 = -6x + 27 (combine like
terms) - 91 - 91 (subtract 91
from both sides) -11x = -6x – 64
+6x +6x (add 6x to both sides) -5x = -64
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) -11x + 91 = -6x + 27 (combine like
terms) - 91 - 91 (subtract 91
from both sides) -11x = -6x – 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
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)x2 x2 (multiply each side by 2, the reciprocal of ½)
2A = h(b1 + b2)
h
b
b
2
1
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)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
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 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
1-3: Solving Equations Solve for x
(multiply both sides by a, to clear the first denominator)
(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
1-3: Solving Equations Solve for x
(multiply both sides by a, to clear the first denominator)
(multiply both sides by b, to clear the second denominator)
(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
1-3: Solving Equations Solve for x
(multiply both sides by a, to clear the first denominator)
(multiply both sides by b, to clear the second denominator)
(subtract bx on both sides, to get the x terms together)
(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
1-3: Solving Equations Solve for x
(multiply both sides by a, to clear the first denominator)
(multiply both sides by b, to clear the second denominator)
(subtract bx on both sides, to get the x terms together)
(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
1-3: Solving Equations Solve for x
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
1-3: Solving Equations Solve for x
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
1-3: Solving Equations Solve for x
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
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
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
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
1-3: Solving Equations Perimeter = 2 • width + 2 • length
100 = 2 • x + 2 • 5x 100 = 2x + 10x
1-3: Solving Equations Perimeter = 2 • width + 2 • length
100 = 2 • x + 2 • 5x 100 = 2x + 10x 100 = 12x
1-3: Solving Equations Perimeter = 2 • width + 2 • length
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
1-3: Solving Equations Perimeter = 2 • width + 2 • length
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
1-3: Solving Equations Writing Equations to Solve Problems
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
1-3: Solving Equations Writing Equations to Solve Problems
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
1-3: Solving Equations Writing Equations to Solve Problems
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 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
1-3: Solving Equations Perimeter = s1 + s2 + s3
18 = 3x + 4x + 5x 18 = 12x
1-3: Solving Equations Perimeter = s1 + s2 + s3
18 = 3x + 4x + 5x 18 = 12x
12 12 1.5 = x
Determine both of your variables (unknowns) from the beginning of the problem
1-3: Solving Equations Perimeter = s1 + s2 + s3
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