Review of Solving Combined Operations 8-5
Follow these general procedures to solving equations:
Step 1: Simplify each side of the equation
Step 2: If there are still indicated additions or subtrations, use the inverse operations to undo them
Step 3: If there are indicated multiplication or divisions involving the variable, use the the inverse operations to undo them
YOU MUST ALWAYS PERFORM THE SAME OPERATION ON BOTH SIDES OF THE EQUATION.
DO ONTO ONE SIDE WHAT YOU WOULD DO TO THE OTHER... hmmm... where have you heard that before??
Balance, balance , balance... it's all a balancing act!!
Solve the equation 3/2(n) + 7 = 22
subtract 7 FROM BOTH SIDES
(3/2)n = 15
Multiply BOTH SIDES by 2/3, the reciprocal of 3/2
(2/3)(3/2)n = 15(2/3)
simplify before you multiply out
n = 10
40 - (5/3)n = 15
There are two ways to solve this one
add (5/3)n to both sides
40 = (5/3)n + 15
then subtract 15 from BOTH SIDES
25 = 5/3(n)
Multiply both sides by 3/5, the reciprocal of 5/3
(3/5)25 = 5/3(n)(3/5)
Simplify before multiplying
15 = n
OR
40 - (5/3)n = 15
subtract 30 from both sides
- (5/3)n = 15 -40
need to ask yourself who wins? the negative... and by how much 25
- (5/3)n = -25
multiply by -3/5, which is the reciprocal of -5/3
(-3/5)-(5/3)n = -25(-3/5)
n = 15
we arrived at the same solution
Thursday, April 23, 2009
Wednesday, April 22, 2009
Algebra Period 3 (Tues/Wed)
Solving Quadratics by Completing the Square 13-3
Okay, up until now the methods of solving quadratics should have been fairly familiar to you—but this method is a NEW strategy!!
When does the "square root = +/- square root" method work well?
When the side with the variable is a PERFECT SQUARE!
So what if that side is not a perfect BINOMIAL SQUARED?
You can follow steps to make it into one!
This method is great because then you can just square root each side to find the roots!
THIS METHOD ALWAYS WORKS!
EXAMPLE: x2 - 10x = 0
Not a TRINOMIAL SQUARE so it would not factor to a BINOMIAL SQUARED. (However, we know that we could just factor out an x to solve—I want you to see how completing the square works—even with easy ones)
Here's how you can make this into a trinomial square
Step 1: b/2
Take half of the b coefficient in this case (- 10/2 = -5)
Step 2: Square b/2
(-5 x -5 = 25)
Step 3: Add (b/2)2 to both sides of the equation
(x2 - 10x + 25 = +25)
Step 4: Factor to a binomial square
(x - 5)2 = 25
Step 5: Square root each side and solve
SQRT (x - 5)2 = SQRT 25
x - 5 = + and - 5
ADD 5 TO BOTH SIDES
x = 5 + and - 5
x = 5 + 5 and x = 5 - 5
x = 10 and x = 0
Now the check for this one—just because we are learning this strategy--is to factor and make sure you get the same results.
that is
x2 - 10x = 0 we know becomes
x(x-10) = 0
x = 0 and x = 10
x2-4x -7 = 0
add 7 to both sides
x2-4x =7
take –b/2 or -4/2 = -2 now square that (-2),sup>2 = 4
add 4 to both sides
x2-4x + 4 = 7 +4
x2- 4x + 4 = 11
(x -2)2 = 11
take the square root of both sides
SQRT(x -2)2 = + or – SQRT 11
x - 2 = + or - SQRT 11
IF THERE IS AN "a" COEFFICIENT, YOU MUST DIVIDE EACH TERM BY IT BEFORE YOU CAN COMPLETE THE SQUARE:
Example: 2x2 - 3x - 1 = 0
Move the 1 to the other side of the equation:
2x2 - 3x = 1
Divide each term by the "a" coefficient:
x2 - 3/2 x = 1/2
Now follow the step to find the completing the square term and add it to both sides:
-b/2 = (-3/2)(1/2) remember (instead of dividing by 2, when you have a fraction, multiply by 1/2)= – 3/4
NOTE: REMEMBER this term you will use it again!!
square that [(-3/2)(1/2)]2 = (-3/4)2
= 9/16
x2 - 3/2 x + 9/16 = 1/2 + 9/16
(x - 3/4)2 = 8/16 + 9/16
(x - 3/4)2 = 17/16
SQRT[(x - 3/4)2 ] = + or - SQRT [17/16]
x - 3/4 = + or -[SQRT 17] /4
x = 3/4 + or -[SQRT 17] /4
x = 3 + or - [SQRT 17] /4
When our textbook ask you to complete the square
as in x2 - 6x
all it is asking is that you follow the steps
-b/2 is -6/2 = -3
then square that (-3)2 = 9
so the answer is
x2 - 6x + 9
Okay, up until now the methods of solving quadratics should have been fairly familiar to you—but this method is a NEW strategy!!
When does the "square root = +/- square root" method work well?
When the side with the variable is a PERFECT SQUARE!
So what if that side is not a perfect BINOMIAL SQUARED?
You can follow steps to make it into one!
This method is great because then you can just square root each side to find the roots!
THIS METHOD ALWAYS WORKS!
EXAMPLE: x2 - 10x = 0
Not a TRINOMIAL SQUARE so it would not factor to a BINOMIAL SQUARED. (However, we know that we could just factor out an x to solve—I want you to see how completing the square works—even with easy ones)
Here's how you can make this into a trinomial square
Step 1: b/2
Take half of the b coefficient in this case (- 10/2 = -5)
Step 2: Square b/2
(-5 x -5 = 25)
Step 3: Add (b/2)2 to both sides of the equation
(x2 - 10x + 25 = +25)
Step 4: Factor to a binomial square
(x - 5)2 = 25
Step 5: Square root each side and solve
SQRT (x - 5)2 = SQRT 25
x - 5 = + and - 5
ADD 5 TO BOTH SIDES
x = 5 + and - 5
x = 5 + 5 and x = 5 - 5
x = 10 and x = 0
Now the check for this one—just because we are learning this strategy--is to factor and make sure you get the same results.
that is
x2 - 10x = 0 we know becomes
x(x-10) = 0
x = 0 and x = 10
x2-4x -7 = 0
add 7 to both sides
x2-4x =7
take –b/2 or -4/2 = -2 now square that (-2),sup>2 = 4
add 4 to both sides
x2-4x + 4 = 7 +4
x2- 4x + 4 = 11
(x -2)2 = 11
take the square root of both sides
SQRT(x -2)2 = + or – SQRT 11
x - 2 = + or - SQRT 11
IF THERE IS AN "a" COEFFICIENT, YOU MUST DIVIDE EACH TERM BY IT BEFORE YOU CAN COMPLETE THE SQUARE:
Example: 2x2 - 3x - 1 = 0
Move the 1 to the other side of the equation:
2x2 - 3x = 1
Divide each term by the "a" coefficient:
x2 - 3/2 x = 1/2
Now follow the step to find the completing the square term and add it to both sides:
-b/2 = (-3/2)(1/2) remember (instead of dividing by 2, when you have a fraction, multiply by 1/2)= – 3/4
NOTE: REMEMBER this term you will use it again!!
square that [(-3/2)(1/2)]2 = (-3/4)2
= 9/16
x2 - 3/2 x + 9/16 = 1/2 + 9/16
(x - 3/4)2 = 8/16 + 9/16
(x - 3/4)2 = 17/16
SQRT[(x - 3/4)2 ] = + or - SQRT [17/16]
x - 3/4 = + or -[SQRT 17] /4
x = 3/4 + or -[SQRT 17] /4
x = 3 + or - [SQRT 17] /4
When our textbook ask you to complete the square
as in x2 - 6x
all it is asking is that you follow the steps
-b/2 is -6/2 = -3
then square that (-3)2 = 9
so the answer is
x2 - 6x + 9
Friday, April 17, 2009
Algebra Period 3 (Monday)
Introduction to Quadratic Equations 13-1
We learned from Chapter 12 that a quadratic function is a function that can be defined by an equation of the form
ax2 + bx + c = y where a is not equal to 0. This is a parabola when the domain is the set of REAL numbers.
When y = 0 in the quadratic function ax2 + bx + c = y we have an equation of the form ax2 + bx + c = 0. An equation that can be written in this form is called a quadratic equation.
STANDARD FORM is ax2 + bx + c = 0
4x2 + 7x = 5 write in standard form and determine a, b, and c
4x2 + 7x – 5 = 0
a= 4
b = 7
c = -5
CHAPTER 13 gives you several different ways to SOLVE QUADRATICS
Solving a quadratic means to find the x intercepts of a parabola.
There are different ways of asking the exact same question:
Find the.....
x intercepts = the roots = the solutions = the zeros of a quadratic
We'll answer this question one of the following ways:
1) Read them from the graph (read the x intercepts) That’s were y = 0 or where the parabola crosses the x-axis!!
but...graphing takes time and sometimes the intercepts are not integers
2) Set y or f(x) = 0 and then factor (we did this in Chapter 6)
but...some quadratics are not factorable
3) Square root each side (+ or - square root on the answer side)
but...sometimes the variable side is not a perfect square (it's irrational)
4) If not a perfect square on the variable side, complete the square, then solve using #3 method
Now this method ALWAYS works, but...it takes a lot of time and can get complicated
5) Quadratic Formula (works for EVERY quadratic)
Really easy if you just memorize the formula and how to use it! :)
Section 13-1
Reading the x intercepts from a graph or factoring and solving using the zero products property.
METHOD 1:
Where the graph crosses the x axis is/are the x intercepts. (Remember, y = 0 here!)
The x intercepts are the two solutions or roots of the quadratic.
METHOD 2:
When we factored in Chapter 6 and set each piece equal to zero, we were finding the x value when y was zero.
That means we were finding these two roots!
y2 – 5y = 6 = 6y – 18
first put this in standard form
y2 – 11y + 24 = 0
(y -8) (y-3) = 0
y = 8 or y = 3
Substitute to verify that 8 and 3 are solutions!!
More Solving Quadratic Equations 13-2
You did this in Chapter 11 for Pythagorean Theorem!
If there is no x term, it's easiest to just square root both sides to solve!
DIFFERENT FROM PYTHAGOREAN: NOT LOOKING FOR JUST THE PRINCIPAL SQUARE ROOT ANYMORE. NEED THE + OR - SYMBOL!!
This is also different from what we did when we had radical equations. Before we squared both sides to solve. It looks like these, but only after we squared both sides. Before we had to carefully check each answer—we had changed the equations by squaring. However, always check your solutions for any mistakes!!
3x2 = 18
divide both sides by 3 and get: x2 = 6
square root each side and get x = + SQRT 6 or - SQRT 6
(x - 5)2 = 9
SQRT each side and get: x - 5 = + or - 3
+ 5 to both sides: x = 5 + 3 or x = 5 - 3
So, the 2 roots are x = 8 or x = 2
(x + 2)2 = 7
SQRT each side and get: x + 2 = + or - SQRT of 7
-2 to both sides: x = -2+ SQRT 7 or x = -2 - SQRT 7
FORMULAS THAT ARE QUADRATICS:
Many formulas have a variable that is squared: compounded interest, height of a projectile (ball)
The formula for a projectile is h = -5t2 + v0t
We can find when a projectile is a ground level ( h= 0) by solving for
0 =-5t2 + v0t \If the projectile begins its flight at height c, its approximate height at time t is h = -5t2+v0t + c We can find when it hits the ground by solving 0 = -5t2 + v0t + c
For example: a slow-pitch softball player hits a pitch when the ball is 2 m above the ground. The ball pops up with an initial velocity of 9m/s If the ball is allowed to drop to the ground, how long will it be in the air?
When the ball hits the ground h = 0 so
0 = -5t2 + 9t + 2 or
-5t2 + 9t + 2 = 0
( 5t + 1)(t -2) = 0
5t = 1 and t = 2
t = -1/5 can’t be a solution since the answer should be positive t must be 2 seconds
Check out Purple Math for great help on quadratics
We learned from Chapter 12 that a quadratic function is a function that can be defined by an equation of the form
ax2 + bx + c = y where a is not equal to 0. This is a parabola when the domain is the set of REAL numbers.
When y = 0 in the quadratic function ax2 + bx + c = y we have an equation of the form ax2 + bx + c = 0. An equation that can be written in this form is called a quadratic equation.
STANDARD FORM is ax2 + bx + c = 0
4x2 + 7x = 5 write in standard form and determine a, b, and c
4x2 + 7x – 5 = 0
a= 4
b = 7
c = -5
CHAPTER 13 gives you several different ways to SOLVE QUADRATICS
Solving a quadratic means to find the x intercepts of a parabola.
There are different ways of asking the exact same question:
Find the.....
x intercepts = the roots = the solutions = the zeros of a quadratic
We'll answer this question one of the following ways:
1) Read them from the graph (read the x intercepts) That’s were y = 0 or where the parabola crosses the x-axis!!
but...graphing takes time and sometimes the intercepts are not integers
2) Set y or f(x) = 0 and then factor (we did this in Chapter 6)
but...some quadratics are not factorable
3) Square root each side (+ or - square root on the answer side)
but...sometimes the variable side is not a perfect square (it's irrational)
4) If not a perfect square on the variable side, complete the square, then solve using #3 method
Now this method ALWAYS works, but...it takes a lot of time and can get complicated
5) Quadratic Formula (works for EVERY quadratic)
Really easy if you just memorize the formula and how to use it! :)
Section 13-1
Reading the x intercepts from a graph or factoring and solving using the zero products property.
METHOD 1:
Where the graph crosses the x axis is/are the x intercepts. (Remember, y = 0 here!)
The x intercepts are the two solutions or roots of the quadratic.
METHOD 2:
When we factored in Chapter 6 and set each piece equal to zero, we were finding the x value when y was zero.
That means we were finding these two roots!
y2 – 5y = 6 = 6y – 18
first put this in standard form
y2 – 11y + 24 = 0
(y -8) (y-3) = 0
y = 8 or y = 3
Substitute to verify that 8 and 3 are solutions!!
More Solving Quadratic Equations 13-2
You did this in Chapter 11 for Pythagorean Theorem!
If there is no x term, it's easiest to just square root both sides to solve!
DIFFERENT FROM PYTHAGOREAN: NOT LOOKING FOR JUST THE PRINCIPAL SQUARE ROOT ANYMORE. NEED THE + OR - SYMBOL!!
This is also different from what we did when we had radical equations. Before we squared both sides to solve. It looks like these, but only after we squared both sides. Before we had to carefully check each answer—we had changed the equations by squaring. However, always check your solutions for any mistakes!!
3x2 = 18
divide both sides by 3 and get: x2 = 6
square root each side and get x = + SQRT 6 or - SQRT 6
(x - 5)2 = 9
SQRT each side and get: x - 5 = + or - 3
+ 5 to both sides: x = 5 + 3 or x = 5 - 3
So, the 2 roots are x = 8 or x = 2
(x + 2)2 = 7
SQRT each side and get: x + 2 = + or - SQRT of 7
-2 to both sides: x = -2+ SQRT 7 or x = -2 - SQRT 7
FORMULAS THAT ARE QUADRATICS:
Many formulas have a variable that is squared: compounded interest, height of a projectile (ball)
The formula for a projectile is h = -5t2 + v0t
We can find when a projectile is a ground level ( h= 0) by solving for
0 =-5t2 + v0t \If the projectile begins its flight at height c, its approximate height at time t is h = -5t2+v0t + c We can find when it hits the ground by solving 0 = -5t2 + v0t + c
For example: a slow-pitch softball player hits a pitch when the ball is 2 m above the ground. The ball pops up with an initial velocity of 9m/s If the ball is allowed to drop to the ground, how long will it be in the air?
When the ball hits the ground h = 0 so
0 = -5t2 + 9t + 2 or
-5t2 + 9t + 2 = 0
( 5t + 1)(t -2) = 0
5t = 1 and t = 2
t = -1/5 can’t be a solution since the answer should be positive t must be 2 seconds
Check out Purple Math for great help on quadratics
Algebra Period 3 ( Review)
Quadratic Equations Review
Maybe another teacher reviewing how to graph quadratics might be just what you need:
So did that help?
Maybe another teacher reviewing how to graph quadratics might be just what you need:
So did that help?
Thursday, April 9, 2009
Math 6 Honors Periods 1, 6 & 7 (Thursday)
Combined Operations 8-5
In order to solve an equation of the form
ax + b = c or ax –b = c or b – ax = c
where a, b, c are given numbers and x is the variable, we must use more than one transformation
Solve the equation 3n - 5 = 10 + 6
Simplify the numerical expression
3n - 5 = 10 + 6
3n – 5 = 16
add 5 to both sides
3n – 5 + 5 = 16 + 5
or
3n – 5 = 16
+ 5 = +5
3n = 21
divide both sides by 3 (or multiply each side by the reciprocal of 3)
3n/3 = 21/3
n = 7
General procedures for solving equations
Simplify each side of the equation
If there are still indicated additions or subtractions, use the inverse operation to undo them
If there are indicated multiplications or division involving the variable, use the inverse operations to undo them
The books says you must always perform the same operation on both sides of the equation. I say, “do to one side what you have done to the other side.”
Solve the equation
(3/2)n + 7 = 22
subtract 7 from both sides
(3/2)n + 7 - 7 = 22 - 7
(3/2)n =15
multiply both sides by 2/3, the reciprocal of 3/2
(2/3)(3/2)n = 15(2/3)
n = 10
Solve the equation
40 – (5/3)n = 15
add (5/3)n to both sides
40 – (5/3)n + (5/3)n = 15 + (5/3)n
40 = 15 + (5/3)n
subtract 15 from both sides
40 – 15 = 15-15 + (5/3)n
25 = (5/3)n multiply both sides by 3/5
(3/5)(25) = (5/3)n (3/5)
15 = n
In order to solve an equation of the form
ax + b = c or ax –b = c or b – ax = c
where a, b, c are given numbers and x is the variable, we must use more than one transformation
Solve the equation 3n - 5 = 10 + 6
Simplify the numerical expression
3n - 5 = 10 + 6
3n – 5 = 16
add 5 to both sides
3n – 5 + 5 = 16 + 5
or
3n – 5 = 16
+ 5 = +5
3n = 21
divide both sides by 3 (or multiply each side by the reciprocal of 3)
3n/3 = 21/3
n = 7
General procedures for solving equations
Simplify each side of the equation
If there are still indicated additions or subtractions, use the inverse operation to undo them
If there are indicated multiplications or division involving the variable, use the inverse operations to undo them
The books says you must always perform the same operation on both sides of the equation. I say, “do to one side what you have done to the other side.”
Solve the equation
(3/2)n + 7 = 22
subtract 7 from both sides
(3/2)n + 7 - 7 = 22 - 7
(3/2)n =15
multiply both sides by 2/3, the reciprocal of 3/2
(2/3)(3/2)n = 15(2/3)
n = 10
Solve the equation
40 – (5/3)n = 15
add (5/3)n to both sides
40 – (5/3)n + (5/3)n = 15 + (5/3)n
40 = 15 + (5/3)n
subtract 15 from both sides
40 – 15 = 15-15 + (5/3)n
25 = (5/3)n multiply both sides by 3/5
(3/5)(25) = (5/3)n (3/5)
15 = n
Wednesday, April 8, 2009
Math 6 H Periods 1, 6 & 7 (Wednesday)
Equations: Decimals and Fractions 8-4
You can use transformations to solve equations which involve decimals or fractions
Solve 0.42 x = 1.05
Divide both sides by 0.42
.42x/.42 = 1.05/.42
now, do side bar and actually divide carefully and you will arrive at
x = 2.5
Solve: n/.15 = 92
multiply both sides by .15 to undo the division
(n/.15)(.15) = 92 (.15)
Again, do a sidebar for your calculations and you will arrive at
n = 13.80
How would we solve the following: (2/3)x = 6?
Let’s look at 2x = 6. What do we do?
We divide both sides by 2—or multiply both sides by the reciprocal of 2—which is ½
Remember the product of a number and its reciprocal is 1
Reminder: the ultimate objective is applying transformations to an equation is to obtain an equivalent equation in the form x = c
(where c is a constant.) Also remember that the understood (invivisble) coefficient of x in the equation x = c is 1. [Can you picture the poster in the front of the room?]
So to solve (2/3)x = 6 you would divide both sides by 2/3 but that is the same as multiplying by the reciprocal of 2/3, which is 3/2.
If an equation has the form
(a/b)(x) = c,
where both a and c are nonzero,
multiply both sides by b/a, the reciprocal of a/b
Solve (1/3)y = 18
(3/1)(1/3)y = 18(3/1)
y =18(3)
y = 54
Solve the equation: (6/7)n = 8
(7/6)(6/7)n = 8(7/6)
n = 8(7/6)
simplify first , then multiply
n = 28/3
n = 9 1/3
Let’s check
(6/7)n = 8
well, we said that n = 9 1/3 so substitute back, but change to 28/3 first
(6/7)(28/3) ?=? 8
[read ?=? as ‘does that equal?’]
Now really do a side bar with the left side of the equation to see what
(6/7)(28/3) really equals. Simplify before you multiply
2(4) = 8 so
8 = 8
Try:
1. (1/7)a = 13
2. b/8 = 16
3. 3.6d = 0.9
4. (3/8)f = 129
You can use transformations to solve equations which involve decimals or fractions
Solve 0.42 x = 1.05
Divide both sides by 0.42
.42x/.42 = 1.05/.42
now, do side bar and actually divide carefully and you will arrive at
x = 2.5
Solve: n/.15 = 92
multiply both sides by .15 to undo the division
(n/.15)(.15) = 92 (.15)
Again, do a sidebar for your calculations and you will arrive at
n = 13.80
How would we solve the following: (2/3)x = 6?
Let’s look at 2x = 6. What do we do?
We divide both sides by 2—or multiply both sides by the reciprocal of 2—which is ½
Remember the product of a number and its reciprocal is 1
Reminder: the ultimate objective is applying transformations to an equation is to obtain an equivalent equation in the form x = c
(where c is a constant.) Also remember that the understood (invivisble) coefficient of x in the equation x = c is 1. [Can you picture the poster in the front of the room?]
So to solve (2/3)x = 6 you would divide both sides by 2/3 but that is the same as multiplying by the reciprocal of 2/3, which is 3/2.
If an equation has the form
(a/b)(x) = c,
where both a and c are nonzero,
multiply both sides by b/a, the reciprocal of a/b
Solve (1/3)y = 18
(3/1)(1/3)y = 18(3/1)
y =18(3)
y = 54
Solve the equation: (6/7)n = 8
(7/6)(6/7)n = 8(7/6)
n = 8(7/6)
simplify first , then multiply
n = 28/3
n = 9 1/3
Let’s check
(6/7)n = 8
well, we said that n = 9 1/3 so substitute back, but change to 28/3 first
(6/7)(28/3) ?=? 8
[read ?=? as ‘does that equal?’]
Now really do a side bar with the left side of the equation to see what
(6/7)(28/3) really equals. Simplify before you multiply
2(4) = 8 so
8 = 8
Try:
1. (1/7)a = 13
2. b/8 = 16
3. 3.6d = 0.9
4. (3/8)f = 129
Tuesday, April 7, 2009
Math 6 H Periods 1, 6 & 7 (Tuesday)
Equations: Multiplication and Division 8-3
If an equation involves multiplication or division, the following transformations are used to solve the equation:
Transformation by multiplication: Multiply both sides of the equation by the same nonzero number.
Transformations by division: Divide both sides of the equation by the same nonzero number.
Remember: Do undo on one side what you would do undo the other!!
Our goal is to get the variable alone and to find an equivalent equation of the form
“n = a number”
Our goal is to arrive at the “world’s easiest equation”
Solve 3n = 24
Use the fact that multiplication and division are inverse operations
3n/3 = 24/3
n = 8
Or you could have use the reciprocal of 3-- which is 1/3 and multiplied both sides by 1/3
(1/3)(3n) = 24(1/3)
n = 8 and still arrived at the SAME solution
Solve 5x/5 = 53/5
x = 10 3/5
Solve n/4 = 7
(4)(n/4) = 7(4)
n = 28
How could you know for sure your answer is correct?
Substitute your solution into the ORIGINAL equation
Try:
1. 3r = 57
2. 714 = 7t
3. Solve A = bh for h
4. Solve P = 4s for s
5. Solve C = 2πr for r
If an equation involves multiplication or division, the following transformations are used to solve the equation:
Transformation by multiplication: Multiply both sides of the equation by the same nonzero number.
Transformations by division: Divide both sides of the equation by the same nonzero number.
Remember: Do undo on one side what you would do undo the other!!
Our goal is to get the variable alone and to find an equivalent equation of the form
“n = a number”
Our goal is to arrive at the “world’s easiest equation”
Solve 3n = 24
Use the fact that multiplication and division are inverse operations
3n/3 = 24/3
n = 8
Or you could have use the reciprocal of 3-- which is 1/3 and multiplied both sides by 1/3
(1/3)(3n) = 24(1/3)
n = 8 and still arrived at the SAME solution
Solve 5x/5 = 53/5
x = 10 3/5
Solve n/4 = 7
(4)(n/4) = 7(4)
n = 28
How could you know for sure your answer is correct?
Substitute your solution into the ORIGINAL equation
Try:
1. 3r = 57
2. 714 = 7t
3. Solve A = bh for h
4. Solve P = 4s for s
5. Solve C = 2πr for r
Algebra Period 3 (Monday)
Direct Variation, Indirect Variation and Joint Variation: 12-5 TO 12-7
DIRECT VARIATION
As x increases, y also increases (graphs as a LINE)
or as x decreases, y also decreases
x and y go in the SAME DIRECTION
For example: y = 2x
the "2" is called the constant of variation and is "k" in the formula:
y = kx
To find k, you need one (x, y) coordinate to solve.
It's like solving for "m" in y = mx + b, but there is no b!
Example: y varies directly with x. When y = 8, x = 2. Find y when x = 3.
FORMULA FOR DIRECT VARIATION: y = kx
PLUG IN THE (x, y):
8 = k(2)
SOLVE FOR k:
k = 4
WRITE THE EQUATION:
y = 4x
PLUG IN VARIABLE GIVEN:
y = 4(3)
SOLVE FOR MISSING VARIABLE:
y = 12
INDIRECT VARIATION
As x increases, y decreases (graphs as a RATIONAL function...I'll show you it in class)
x and y go in OPPOSITE DIRECTIONS
For example: y = 2/x
the "2" is still called the constant of variation and is "k" in the formula:
y = k/x
To find k, you need one (x, y) coordinate to solve.
Example: y varies indirectly with x. When y = 8, x = 2. Find y when x = 3.
FORMULA FOR INDIRECT VARIATION:
y = k/x
PLUG IN THE (x, y):
8 = k/2
SOLVE FOR k:
k = 16
WRITE THE EQUATION:
y = 16/x
PLUG IN VARIABLE GIVEN:
y = 16/3
SOLVE FOR MISSING VARIABLE:
y = 5 1/3
JOINT VARIATION
TWO VARIABLES VARY WITH y AT THE SAME TIME!
As x AND z increase, y also increases (graphs as a LINEAR function)
The PRODUCT xz and y go in the SAME DIRECTION
For example: y = 2xz
the "2" is still called the constant of variation and is "k" in the formula:
y = kxz
To find k, you need one (x, y, z) coordinate to solve.
Example: y varies jointly with x and z. When y = 12, x = 2, and z = 3. Find y when x = 3 and z = 5
FORMULA FOR JOINT VARIATION:
y = kxz
PLUG IN THE (x, y, z):
12 = k(2)(3)
SOLVE FOR k:
k = 2
WRITE THE EQUATION:
y = 2xz
PLUG IN VARIABLE GIVEN:
y = 2(3)(5)
SOLVE FOR MISSING VARIABLE:
y = 30
DIRECT VARIATION
As x increases, y also increases (graphs as a LINE)
or as x decreases, y also decreases
x and y go in the SAME DIRECTION
For example: y = 2x
the "2" is called the constant of variation and is "k" in the formula:
y = kx
To find k, you need one (x, y) coordinate to solve.
It's like solving for "m" in y = mx + b, but there is no b!
Example: y varies directly with x. When y = 8, x = 2. Find y when x = 3.
FORMULA FOR DIRECT VARIATION: y = kx
PLUG IN THE (x, y):
8 = k(2)
SOLVE FOR k:
k = 4
WRITE THE EQUATION:
y = 4x
PLUG IN VARIABLE GIVEN:
y = 4(3)
SOLVE FOR MISSING VARIABLE:
y = 12
INDIRECT VARIATION
As x increases, y decreases (graphs as a RATIONAL function...I'll show you it in class)
x and y go in OPPOSITE DIRECTIONS
For example: y = 2/x
the "2" is still called the constant of variation and is "k" in the formula:
y = k/x
To find k, you need one (x, y) coordinate to solve.
Example: y varies indirectly with x. When y = 8, x = 2. Find y when x = 3.
FORMULA FOR INDIRECT VARIATION:
y = k/x
PLUG IN THE (x, y):
8 = k/2
SOLVE FOR k:
k = 16
WRITE THE EQUATION:
y = 16/x
PLUG IN VARIABLE GIVEN:
y = 16/3
SOLVE FOR MISSING VARIABLE:
y = 5 1/3
JOINT VARIATION
TWO VARIABLES VARY WITH y AT THE SAME TIME!
As x AND z increase, y also increases (graphs as a LINEAR function)
The PRODUCT xz and y go in the SAME DIRECTION
For example: y = 2xz
the "2" is still called the constant of variation and is "k" in the formula:
y = kxz
To find k, you need one (x, y, z) coordinate to solve.
Example: y varies jointly with x and z. When y = 12, x = 2, and z = 3. Find y when x = 3 and z = 5
FORMULA FOR JOINT VARIATION:
y = kxz
PLUG IN THE (x, y, z):
12 = k(2)(3)
SOLVE FOR k:
k = 2
WRITE THE EQUATION:
y = 2xz
PLUG IN VARIABLE GIVEN:
y = 2(3)(5)
SOLVE FOR MISSING VARIABLE:
y = 30
Monday, April 6, 2009
Math 6 H Periods 1, 6 & 7 (Monday)
Equations: Addition and Subtraction 8-2
If the replacement set for an equation is the set of whole numbers, it is not practical to use substitution to solve the equation. Instead we transform or change the given equation into a simpler, equivalent equation. When we transform the given equation, our goal is to arrive at an equivalent equation of the form
variable = number
for example
n = 5
the number 5 is then, the solution of the original equation. The following transformations can be used to solve equation
Transformation by addition: add the same number to both sides
Transformation by subtraction: subtract the same number from both sides
solve x – 2 = 8
our goal is to find an equivalent equation of the form
x = a number
The left side of the given equation is x – 2. Recall that addition and subtraction are inverse operations. If we add 2 to both sides the left sides simplifies to x
x-2 = 8
x – 2 + 2 = 8 + 2 (We usually show the +2 right below each side of the equation)
x = 10
The solution is 10
Solve x + 6 = 17
Subtract 6 from both sides of the equation to get an equivalent equation of the form “ x = a number”
x + 6 = 17
x + 6 – 6 = 17 – 6 (Again, we usually show the -6 right below each side of the equation)
x = 11
the solution is 11
In equations involving a number of steps, it is a good idea to check your answer. This can be done easily by substituting the answer in the original equation.
What about the following
34 – x = 27
add x to both sides
34 – x + x = 27 + x
34 = 27 + x
subtract 27 from both sides
34 – 27 = 27 – 27 + x
7 = x
You may be able to solve some of the equations in the homework without pencil and paper. Nevertheless, it is important to show all the steps in your work and to make sure you can tell which transformation you are using in each step.
Which transformation was used to transform the first equation into the second
x – 3 = 5
x – 3 + 3 = 5 + 3
Remember when we stated the properties as well-- back in our 1st quarter!!
If the replacement set for an equation is the set of whole numbers, it is not practical to use substitution to solve the equation. Instead we transform or change the given equation into a simpler, equivalent equation. When we transform the given equation, our goal is to arrive at an equivalent equation of the form
variable = number
for example
n = 5
the number 5 is then, the solution of the original equation. The following transformations can be used to solve equation
Transformation by addition: add the same number to both sides
Transformation by subtraction: subtract the same number from both sides
solve x – 2 = 8
our goal is to find an equivalent equation of the form
x = a number
The left side of the given equation is x – 2. Recall that addition and subtraction are inverse operations. If we add 2 to both sides the left sides simplifies to x
x-2 = 8
x – 2 + 2 = 8 + 2 (We usually show the +2 right below each side of the equation)
x = 10
The solution is 10
Solve x + 6 = 17
Subtract 6 from both sides of the equation to get an equivalent equation of the form “ x = a number”
x + 6 = 17
x + 6 – 6 = 17 – 6 (Again, we usually show the -6 right below each side of the equation)
x = 11
the solution is 11
In equations involving a number of steps, it is a good idea to check your answer. This can be done easily by substituting the answer in the original equation.
What about the following
34 – x = 27
add x to both sides
34 – x + x = 27 + x
34 = 27 + x
subtract 27 from both sides
34 – 27 = 27 – 27 + x
7 = x
You may be able to solve some of the equations in the homework without pencil and paper. Nevertheless, it is important to show all the steps in your work and to make sure you can tell which transformation you are using in each step.
Which transformation was used to transform the first equation into the second
x – 3 = 5
x – 3 + 3 = 5 + 3
Remember when we stated the properties as well-- back in our 1st quarter!!
Saturday, April 4, 2009
Algebra Period 3 ( Review)
Review of How to Graph a Quadratic
Find the vertex
first find the x value of this ordered pair. (x = -b/2a)
then (using that value) plug in to the equation to find the y coordinate
Find the axis (line) of symmetry – just use the x value of the vertex that is
x = -b/2a
draw the line of symmetry using a dotted or dashed line
Find 2 more points in an x y table (go either left OR right of the vertex and use 2 points close to the line of symmetry for the x, then plug into the equation to find the y values)
Find the 2 shadow (mirror) points by counting over from the axis at the exact same y value
Draw a U (not a V) through the points. Extend it past and put arrows at the end
Remember to label the x and y axes and put arrows on the axes.
Function vs. Relations: Functions are special relations where there is a unique x for each y
Therefore, there will never be 2 x’s that will repeat and if you use a vertical line on the graph, it will only hit one point on the graph.
Domain vs. Range:
DOMAIN:
The domain for any quadratic function is ALL REAL NUMBERS
RANGE:
The range depends on where the vertex is and whether the quadratic is a smile or a frown.
Generally, it will be of the form: y such that y is either greater than or equal to (≥ ) or less than or equal to (≤) the y value of the vertex.
f(x) is just a different way of saying y. What is better about it? Without seeing the work before the solution, you can actually tell the x value as well as the y value in the solution.
For example, if the solution is f(-3) = 12 you know that the point is (-3, 12)
Compare that to the solution y = 12. For that solution, you would not know the x value unless you look back in the problem.
Check out Purplemath.com Great website for help
Find the vertex
first find the x value of this ordered pair. (x = -b/2a)
then (using that value) plug in to the equation to find the y coordinate
Find the axis (line) of symmetry – just use the x value of the vertex that is
x = -b/2a
draw the line of symmetry using a dotted or dashed line
Find 2 more points in an x y table (go either left OR right of the vertex and use 2 points close to the line of symmetry for the x, then plug into the equation to find the y values)
Find the 2 shadow (mirror) points by counting over from the axis at the exact same y value
Draw a U (not a V) through the points. Extend it past and put arrows at the end
Remember to label the x and y axes and put arrows on the axes.
Function vs. Relations: Functions are special relations where there is a unique x for each y
Therefore, there will never be 2 x’s that will repeat and if you use a vertical line on the graph, it will only hit one point on the graph.
Domain vs. Range:
DOMAIN:
The domain for any quadratic function is ALL REAL NUMBERS
RANGE:
The range depends on where the vertex is and whether the quadratic is a smile or a frown.
Generally, it will be of the form: y such that y is either greater than or equal to (≥ ) or less than or equal to (≤) the y value of the vertex.
f(x) is just a different way of saying y. What is better about it? Without seeing the work before the solution, you can actually tell the x value as well as the y value in the solution.
For example, if the solution is f(-3) = 12 you know that the point is (-3, 12)
Compare that to the solution y = 12. For that solution, you would not know the x value unless you look back in the problem.
Check out Purplemath.com Great website for help
Thursday, April 2, 2009
Algebra Period 3 (Wednesday)
Quadratic Functions: 12- 4
A QUADRATIC FUNCTION is not y = mx + b
(which is a LINEAR function),
but instead is
y = ax2 + bx + c
OR
f(x) = ax2 + bx + c
where a, b, and c are all real numbers and
a cannot be equal to zero because
it must have a variable that is squared ( degree of 2)
[If a = 0, then we would end up with y = bx + c which is really y= mx + b]
Quadratics have a squared term, so they have TWO possible solutions also called roots. You already saw this in Chapter 6 when you factored the trinomial and used zero products property!! ( CHAPTER 6-- again)
If the domain is all real numbers, then you will have a PARABOLA which looks like
a smile when the a coefficient is positive or
looks like a frown when the a coefficient is negative.
Graphing quadratics:
You can graph quadratics exactly the way you graphed lines
by plugging in your choice of an x value and using the equation to find your y value.
Because it's a U shape, you should graph 5 points as follows:
STEP 1: determine Point 1: the vertex -
the minimum value of the smile or
the maximum value of the frown
The x value of the VERTEX = -b/2a
We get the values for a and b from the actual equation
f(x) = ax2 + bx + c
just plug in the b and the a value from your equation into -b/2a and you have the x-value of the vertex.
Now to find the y value -- take that x value and PLUG it into the equation
STEP 2: Next, draw the AXIS OF SYMMETRY : x = -b/2a
a line through the vertex parallel to the y axis . Draw this line as a dashed line. REMEMBER: It will be a dashed line parallel to the y-axis
STEP 3: Point 2- Pick an x value to the right or left of the axis and find its y by plugging into the equation.
STEP 4: Point 3- Graph its mirror image on the other side of the axis of symmetry by counting from axis of symmetry
STEP 5: Points 4 and 5- Repeat point 2 and 3 directions with another point even farther from the vertex
JOIN YOUR 5 POINTS IN A "U" SHAPE AND EXTEND LINES WITH ARROWS ON END
Parabolas that are functions have domains that are ALL REAL NUMBERS
Their ranges depend on where the vertex is and also if the a coefficient is positive or negative
EXAMPLE: f(x) = -3x2 (or y = -3x2)
the a coefficient is negative so it is a frown face
the x value of the vertex (maximum) is -b/2a or 0/2(-3) = 0
the y value of the vertex is 0
So the vertex is (0, 0)
To graph this function:
1) graph vertex (0, 0)
2) Draw dotted line x = 0 (actually this is the y axis!)
3) Pick x value to the right of axis of symmetry, say x = 1
Plug it in the equation: y = -3(1) = -3
Plot (1, -3)
4) count steps from axis of symmetry and place another point to the LEFT of axis in same place
5) pick another x value to the right of the axis of symmetry, say x = 2
plug it in the equation y = -3(2) = -6
plot ( 2, -6). Count the steps from the axis of symmetry and place another point to the LEFT of the axis in the same place.
The domain is all real numbers.
The range is y is less than or equal to zero
A QUADRATIC FUNCTION is not y = mx + b
(which is a LINEAR function),
but instead is
y = ax2 + bx + c
OR
f(x) = ax2 + bx + c
where a, b, and c are all real numbers and
a cannot be equal to zero because
it must have a variable that is squared ( degree of 2)
[If a = 0, then we would end up with y = bx + c which is really y= mx + b]
Quadratics have a squared term, so they have TWO possible solutions also called roots. You already saw this in Chapter 6 when you factored the trinomial and used zero products property!! ( CHAPTER 6-- again)
If the domain is all real numbers, then you will have a PARABOLA which looks like
a smile when the a coefficient is positive or
looks like a frown when the a coefficient is negative.
Graphing quadratics:
You can graph quadratics exactly the way you graphed lines
by plugging in your choice of an x value and using the equation to find your y value.
Because it's a U shape, you should graph 5 points as follows:
STEP 1: determine Point 1: the vertex -
the minimum value of the smile or
the maximum value of the frown
The x value of the VERTEX = -b/2a
We get the values for a and b from the actual equation
f(x) = ax2 + bx + c
just plug in the b and the a value from your equation into -b/2a and you have the x-value of the vertex.
Now to find the y value -- take that x value and PLUG it into the equation
STEP 2: Next, draw the AXIS OF SYMMETRY : x = -b/2a
a line through the vertex parallel to the y axis . Draw this line as a dashed line. REMEMBER: It will be a dashed line parallel to the y-axis
STEP 3: Point 2- Pick an x value to the right or left of the axis and find its y by plugging into the equation.
STEP 4: Point 3- Graph its mirror image on the other side of the axis of symmetry by counting from axis of symmetry
STEP 5: Points 4 and 5- Repeat point 2 and 3 directions with another point even farther from the vertex
JOIN YOUR 5 POINTS IN A "U" SHAPE AND EXTEND LINES WITH ARROWS ON END
Parabolas that are functions have domains that are ALL REAL NUMBERS
Their ranges depend on where the vertex is and also if the a coefficient is positive or negative
EXAMPLE: f(x) = -3x2 (or y = -3x2)
the a coefficient is negative so it is a frown face
the x value of the vertex (maximum) is -b/2a or 0/2(-3) = 0
the y value of the vertex is 0
So the vertex is (0, 0)
To graph this function:
1) graph vertex (0, 0)
2) Draw dotted line x = 0 (actually this is the y axis!)
3) Pick x value to the right of axis of symmetry, say x = 1
Plug it in the equation: y = -3(1) = -3
Plot (1, -3)
4) count steps from axis of symmetry and place another point to the LEFT of axis in same place
5) pick another x value to the right of the axis of symmetry, say x = 2
plug it in the equation y = -3(2) = -6
plot ( 2, -6). Count the steps from the axis of symmetry and place another point to the LEFT of the axis in the same place.
The domain is all real numbers.
The range is y is less than or equal to zero
Tuesday, March 31, 2009
Math 6 H Periods 1, 6 & 7 (Tuesday)
Scale Drawing 7-9
Opening your books to page 237, you will notice a drawing of a house. In this drawing of the house, the actual height of 9 meters is represented by a length of 3 centimeters, and the actual length of 21 meters is represented by a length of 7 centimeters.
This means that 1 cm in the drawing represents 3 m in the actual building. Such a drawing in which all lengths are in the same ratio to actual lengths is called a scale drawing.
The relationship of length in the drawing to actual length is called the scale. In the drawing of the house the scale is 1cm: 3m
We can express the scale as a ratio, called the scale ratio, if a common unit of measure is used. Since 3 m = 300 cm, the scale ratio is 1/300
Using the book’s drawing on page 237, find the length and width of the room shown, if the scale of the drawing is 1cm: 1.5 m
Measuring the drawing, we find that it has a length of 4 cm and a width of 3 cm
Method 1: write a proportion for the length
Let l = the actual length
1/1.5 = 4/ l
l= 4 (1.5)
l = 6
The room is 6 m long
Write a proportion for the width
Let w = the actual width
1/1.5 = 3/w
w = 3 (1.5)
w = 4.5 m
The room is 4.5 m wide
Method 2 : Use the scale ratio
1 cm/ 1.5 m = 1/cm/150 cm = 1/150
You need to change the units to the same and then set up a ratio
The actual length is 150 times the length in the drawing so
l =150 (4) = 600 cm = 6 m
w =150 (3) = 450 cm = 4.5 m
The scale on a map is 1 cm to 240 m
The distance from Ryan’s house to his school is 10 cm on the map. What is the actual distance?
let d = the distance from Ryan’s house to school
1/240 = 10/d so d = 240(10)
d = 2400m or
The distance from Ryan's house to school is 2.4 km
A picture of an insect has a scale 7 to 1. The length of the insect in the picture is 5.6 cm. What is the actual length of the insect?
Let l = the actual length of the insect
7/1 = 5.6/ l
7l = 5.6
l = .8 cm
The actual length of the insect is 0.8 cm which is 8 mm
Opening your books to page 237, you will notice a drawing of a house. In this drawing of the house, the actual height of 9 meters is represented by a length of 3 centimeters, and the actual length of 21 meters is represented by a length of 7 centimeters.
This means that 1 cm in the drawing represents 3 m in the actual building. Such a drawing in which all lengths are in the same ratio to actual lengths is called a scale drawing.
The relationship of length in the drawing to actual length is called the scale. In the drawing of the house the scale is 1cm: 3m
We can express the scale as a ratio, called the scale ratio, if a common unit of measure is used. Since 3 m = 300 cm, the scale ratio is 1/300
Using the book’s drawing on page 237, find the length and width of the room shown, if the scale of the drawing is 1cm: 1.5 m
Measuring the drawing, we find that it has a length of 4 cm and a width of 3 cm
Method 1: write a proportion for the length
Let l = the actual length
1/1.5 = 4/ l
l= 4 (1.5)
l = 6
The room is 6 m long
Write a proportion for the width
Let w = the actual width
1/1.5 = 3/w
w = 3 (1.5)
w = 4.5 m
The room is 4.5 m wide
Method 2 : Use the scale ratio
1 cm/ 1.5 m = 1/cm/150 cm = 1/150
You need to change the units to the same and then set up a ratio
The actual length is 150 times the length in the drawing so
l =150 (4) = 600 cm = 6 m
w =150 (3) = 450 cm = 4.5 m
The scale on a map is 1 cm to 240 m
The distance from Ryan’s house to his school is 10 cm on the map. What is the actual distance?
let d = the distance from Ryan’s house to school
1/240 = 10/d so d = 240(10)
d = 2400m or
The distance from Ryan's house to school is 2.4 km
A picture of an insect has a scale 7 to 1. The length of the insect in the picture is 5.6 cm. What is the actual length of the insect?
Let l = the actual length of the insect
7/1 = 5.6/ l
7l = 5.6
l = .8 cm
The actual length of the insect is 0.8 cm which is 8 mm
Monday, March 30, 2009
Pre Algebra Period 2 (Monday)
Scientific Notation 4-9
You've had this since 6th grade!
You restate very big or very small numbers using powers of 10 in exponential form
Move the decimal so the number fits in this range: less than 10 and greater than or equal to 1
Count the number of places you moved the decimal and make that your exponent
Very big numbers - exponent is positive
Very small numbers (decimals) - exponent is negative (just like a fraction!)
Remember that STANDARD notation is what you expect (the normal number)
When you multiply or divide scientific notations, use the power rules!
Just be careful that is your answer does not fit the scientific notation range, that you restate it.
ORDERING SCIENTIFIC NOTATION NUMBERS:
As long as numbers are in scientific notation, they are easy to put in order from least to greatest!
1) If they are all different powers, simply order them by powers
2) If they have the same power, simply order them using your decimal ordering skills.
EXAMPLE 1: Order 3.7 x 108, 4.3 x 10-2, 9.3 x 105, and 8.7 x 10-5
8.7 x 10-5, 4.3 x 10-2, 9.3 x 105, 3.7 x 108
EXAMPLE 2: 3.7 x 108, 4.3 x 108, 9.3 x 108, and 8.7 x 108
3.7 x 108, 4.3 x 108, 8.7 x 108, 9.3 x 108
TRY THESE LINKS:
TRY THIS LINK THAT TAKES YOU FROM LARGE POWERS
TO
LITTLE POWERS (NEGATIVE POWERS OR DECIMALS)
Try this link to practice scientific notation!
You've had this since 6th grade!
You restate very big or very small numbers using powers of 10 in exponential form
Move the decimal so the number fits in this range: less than 10 and greater than or equal to 1
Count the number of places you moved the decimal and make that your exponent
Very big numbers - exponent is positive
Very small numbers (decimals) - exponent is negative (just like a fraction!)
Remember that STANDARD notation is what you expect (the normal number)
When you multiply or divide scientific notations, use the power rules!
Just be careful that is your answer does not fit the scientific notation range, that you restate it.
ORDERING SCIENTIFIC NOTATION NUMBERS:
As long as numbers are in scientific notation, they are easy to put in order from least to greatest!
1) If they are all different powers, simply order them by powers
2) If they have the same power, simply order them using your decimal ordering skills.
EXAMPLE 1: Order 3.7 x 108, 4.3 x 10-2, 9.3 x 105, and 8.7 x 10-5
8.7 x 10-5, 4.3 x 10-2, 9.3 x 105, 3.7 x 108
EXAMPLE 2: 3.7 x 108, 4.3 x 108, 9.3 x 108, and 8.7 x 108
3.7 x 108, 4.3 x 108, 8.7 x 108, 9.3 x 108
TRY THESE LINKS:
TRY THIS LINK THAT TAKES YOU FROM LARGE POWERS
TO
LITTLE POWERS (NEGATIVE POWERS OR DECIMALS)
Try this link to practice scientific notation!
Math 6 H Periods 1, 6 & 7 (Monday)
Problem Solving: Using Proportion 7-8
Proportions can be used to solve word problems. Use the following steps to help you in solving problems using proportions
~ Decide which quantity is to be found and represent it by a variable
~ Determine whether the quantities involved can be compared using ratios (rates)
~ Equate the ratios in a proportion
~ Solve the proportion
Taylor’s mom bought 4 tires for her car at a cost of $264. How much would 5 tires cost at the same rate?
Let c = the cost of 5 tires. Set up a proportion
4/264 = 5/c
Solve the proportion
4c = 5(264)
Now divide both sides by 4
4c/4 = 5(264)/4
c = 330
Therefore, 5 tires would cost $330
Notice, the proportion in this example could also have been written as
2/264 = c/5
In fact--Any of the following proportions can be used to solve the problem
4/5 = 264/c 5/4 = c/264 4/264 = 5/c 264/4 = c/ 5
All of the above proportions result in the same equation
4c = 5(264)
But be careful you need to use a proportion that does relate.
4/5 DOES NOT EQUAL c/264. That is not an accurate proportion and would result in an inaccurate solution.
For every 5 sailboats in a harbor, there are 3 motorboats. If there are 30 sailboats in the harbor, how many motorboats are there?
Let m = the number of motorboats
5/3 = 30/m
5m = 3(30) divide both sides by 5 5m/5 = 3(30)/5
m=18
There are 18 motorboats in the harbor.
Some guidelines you can use to determine when it is appropriate to use a proportion to solve a word problem.
Ask the following questions
If one quantity increase does the other quantity also increase? (If one quantity decreases, does the other quantity decrease?) When the number of tires is increase, the cost is also increased.
Does the amount of change (increase or decrease) o one quantity depend upon the amount of change (increase or decrease) of the other quantity? The amount of increase in the cost depends upon the number of additional tires bought.
Does one quantity equal some constant times the other quantity? The total costs equals the cost of one tire times the number of tires. The cost of one tire is constant.
If the answers to all the questions above is YES, then it is appropriate to use a proportion.
Sometimes setting up a table can be useful
Although the problems in this lesson may be solved without using proportions, I must insist that you write a proportion for each problem and solve using this method. You may check your work using another other method you know.
Proportions can be used to solve word problems. Use the following steps to help you in solving problems using proportions
~ Decide which quantity is to be found and represent it by a variable
~ Determine whether the quantities involved can be compared using ratios (rates)
~ Equate the ratios in a proportion
~ Solve the proportion
Taylor’s mom bought 4 tires for her car at a cost of $264. How much would 5 tires cost at the same rate?
Let c = the cost of 5 tires. Set up a proportion
4/264 = 5/c
Solve the proportion
4c = 5(264)
Now divide both sides by 4
4c/4 = 5(264)/4
c = 330
Therefore, 5 tires would cost $330
Notice, the proportion in this example could also have been written as
2/264 = c/5
In fact--Any of the following proportions can be used to solve the problem
4/5 = 264/c 5/4 = c/264 4/264 = 5/c 264/4 = c/ 5
All of the above proportions result in the same equation
4c = 5(264)
But be careful you need to use a proportion that does relate.
4/5 DOES NOT EQUAL c/264. That is not an accurate proportion and would result in an inaccurate solution.
For every 5 sailboats in a harbor, there are 3 motorboats. If there are 30 sailboats in the harbor, how many motorboats are there?
Let m = the number of motorboats
5/3 = 30/m
5m = 3(30) divide both sides by 5 5m/5 = 3(30)/5
m=18
There are 18 motorboats in the harbor.
Some guidelines you can use to determine when it is appropriate to use a proportion to solve a word problem.
Ask the following questions
If one quantity increase does the other quantity also increase? (If one quantity decreases, does the other quantity decrease?) When the number of tires is increase, the cost is also increased.
Does the amount of change (increase or decrease) o one quantity depend upon the amount of change (increase or decrease) of the other quantity? The amount of increase in the cost depends upon the number of additional tires bought.
Does one quantity equal some constant times the other quantity? The total costs equals the cost of one tire times the number of tires. The cost of one tire is constant.
If the answers to all the questions above is YES, then it is appropriate to use a proportion.
Sometimes setting up a table can be useful
Although the problems in this lesson may be solved without using proportions, I must insist that you write a proportion for each problem and solve using this method. You may check your work using another other method you know.
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