Arithmatic english 3 ppsx

Positive directions are to the right and up and negative directions are to the left and down... The slope between two points x1, y1 and x2, y2 can be found by using the following formula

You may be asked to perform various operations on rational expressions See the following examples.

Examples

1 Simplify

2 Simplify

3 Multiply

4 Divide

5 Add

6 Subtract

7 Solve

8 Solve

Answers

1.

2.

3.

4.

5.

3x

1 3x

xy

a 1a 22

1a  121a  22 21a  12

a 1a  2

a

4x1x 42

x 1x

1x

3

x2b

x3b2 1

xb

1

x1

4 1

6

2

3x1

6x1

4

x 6

x –x – 2 3x

1

xy3

y

a2  2a

a2  3a  2 a2– 3a

2a 2

4x

x2 – 16 x 4

2x2

x2

3x

x2b

x3b2

7 Multiply each term by the LCD = 12.

8x + 2x = 3

10x = 3

x =

8 Multiply each term by the LCD = 12x.

3x + 2x = 12

5x = 12

x =

 C o o r d i n a t e G r a p h i n g

The coordinate plane is divided into four quadrants that are created by the intersection of two

perpendicu-lar signed number lines: the x- and y-axes The quadrants are numbered I, II, III, and IV as shown in the

diagram

Each location in the plane is named by a point (x, y) These numbers are called the coordinates of the point Each point can be found by starting at the intersection of the axes, the origin, and moving x units to the right or left and y units up or down Positive directions are to the right and up and negative directions

are to the left and down

When graphing linear equations (slope and y-intercept), use the y = mx + b form, where m represents the slope of the line and b represents the y-intercept.

5 4 3 2 1

-1 -2 -3 -4 -5

-1 -2 -3 -4 -5

y-axis

x-axis

I II

12

5  2.4

10 3

The slope between two points (x1, y1) and (x2, y2) can be found by using the following formula:

Here are a few helpful facts about slope and graphing linear equations:

■ Lines that slant up to the right have a positive slope

■ Lines that slant up to the left have a negative slope

■ Horizontal lines have a slope of zero

■ Vertical lines have an undefined slope or no slope

■ Two lines with the same slope are parallel and will never intersect

■ Two lines that have slopes that are negative reciprocals of each other are perpendicular

To find the midpoint between any two points (x1, y1) and (x2, y2), use the following formula:

To find the distance between any two points (x1, y1) and (x2, y2), use the following formula:

 S y s t e m s o f E q u a t i o n s w i t h Tw o Va r i a b l e s

When solving a system of equations, you are finding the value or values where two or more equations equal each other This can be done in two ways algebraically: by elimination and by substitution

Elimination Method

Solve the system x – y = 6 and 2x + 3y = 7.

Put the equations one above the other, lining up the xs, ys, and the equal sign.

x – y = 6

2x + 3y = 7

2 1x1– x222 1y1 – y222

1x1 x2

2 , y1 y2

change in y

change in x y1 2

x1 2

Multiply the first equation by –2 so that the coefficients of x are opposites This will allow the xs to

can-cel out in the next step Make sure that ALL terms are multiplied by –2 The second equation remains the same

–2 (x – y = 6) ⇒ –2x + 2y = –12

2x + 3y = 7 ⇒ 2x + 3y = 7

Combine the new equations vertically

–2x + 2y = –12

2x + 3y = 7

5y = –5

Divide both sides by 5

To complete the problem, solve for x by substituting –1 for y into one of the original equations.

x – y = 6

x – (–1) = 6

x + 1 = 6

x + 1 – 1 = 6 – 1

x = 5

The solution to the system is x = 5 and y = –1, or (5, –1).

Substitution Method

Solve the system x + 2y = 5 and y = –2x + 7

Substitute the second equation into the first for y.

x + 2(–2x + 7) = 5

Use distributive property to remove the parentheses

x + –4x + 14 = 5

y –1

5y

5 –5

5

Combine like terms Remember x = 1x.

–3x + 14 = 5

Subtract 14 from both sides and then divide by –3

–3x + 14 –14 = 5 – 14

x = 3

To complete the problem, solve for y by substituting 3 for x in one of the original equations.

y = –2x + 7

y = –2 (3) + 7

y = –6 + 7

y = 1

The solution to the system is x = 3 and y = 1, or (3, 1).

 P r o b l e m S o l v i n g w i t h Wo r d P r o b l e m s

You will encounter a variety of different types of word problems on the GMAT quantitative section To help with this type of problem, first begin by figuring out what you need to solve for and defining your variable

as that unknown Then write and solve an equation that matches the question asked

Mixture Problems

How many pounds of coffee that costs $4.00 per pound need to be mixed with 10 pounds of coffee that costs $6.40 per pound to create a mixture of coffee that costs $5.50 per pound?

a 4

b 6

c 8

d 10

e 16

For this type of question, remember that the total amount spent in each case will be the price per pound

times how many pounds are in the mixture Therefore, if you let x = the number of pounds of $4.00 coffee, then $4.00(x) is the amount of money spent on $4.00 coffee, $6.40(10) is the amount spent on $6.40 coffee,

–3x

–3 –9

–3

and $5.50(x + 10) is the total amount spent Write an equation that adds the first two amounts and sets it

equal to the total amount

4.00(x) + 6.40(10) = 5.50(x + 10)

Multiply through the equation: 4x + 64 = 5.5x + 55

Subtract 4x from both sides: 4x – 4x + 64 = 5.5x – 4x + 55

Subtract 55 from both sides: 64 – 55 = 1.5x + 55 – 55

Divide both sides by 1.5:

6 = x

You need 6 pounds of the $4.00 per pound coffee The correct answer is b.

Distance Problems

Most problems that involve motion or traveling will probably use the formula distance = rate×time.

Wendy drove 4 hours in a car to reach a conference she was attending On her return trip, she followed the same route but the trip took her 112hours longer If she drove 220 miles to conference, how much slower was her average speed on the return trip?

a 10

b 15

c 25

d 40

e 55

Use the formula distance = rate × time and convert it to = rate Remember that the distance was

220 miles for each part of the trip Since it took her 4 hours to reach the conference, then 4 + 112= 512hours for the return trip = 40 miles per hour However, the question did not ask for the speed on the way back;

it asked for the difference between the speed on the way there and the speed on the way home The speed on the way there would be = 55 miles per hour and 55 – 40 = 15 miles per hour slower on the return trip

The correct answer is b.

220 4

220 5.5

distance time

9 1.51.5x

1.5

10x = 3< /i>

x =

8 Multiply each term by the LCD = 12x.

3x + 2x = 12

5x = 12

x =

 C... 1

-1 -2 -3 -4 -5

-1 -2 -3 -4 -5

y-axis

x-axis

I II

12... terms Remember x = 1x.

–3x + 14 = 5

Subtract 14 from both sides and then divide by ? ?3

–3x + 14 –14 = – 14

x = 3< /i>

To complete the problem,