NTS Pattern And Practice: Algebra

Algebra

Polynomials

• A monomial is any number or variable or product of numbers and variables. For example 3, − 4, x, y, 3x, − 2xyz, 5x³ ,1.5xy² , a³b² are all monomials.
• The number that appears in front of a variable in a monomial is called the coefficient. The coefficient of 5x³ is 5. If there is no number, the coefficient is either 1 or –1, because x means 1x and −x means −1x .
• A polynomial is a monomial or the sum of two or more monomials. Each monomial that makes up the polynomial is called a term of that polynomial.
• If a polynomial has only one term it is a simple monomial, if it has two terms, it is known as binomial and if it has three terms, it is called trinomial.
• Two terms are called like terms if they differ only in their coefficients. 5x³ and −2x³ are like terms, whereas, 5x³ and 5x² are not.
• If like terms are involved in a polynomial, they can be combined, by adding their coefficients, to make that polynomial simpler. The polynomial 3x² + 4x + 5x − 2x² − 7 is equivalent to the polynomial x² + 9x − 7.
• All laws of arithmetic are also applicable to polynomials. Most important of them is PEMDAS.
• Polynomials can be added, subtracted, multiplied or divided.
• To add two polynomials, put a plus sign between them, erase the parentheses, and combine like terms.
  Example:
  What is the sum of 5x² + 10x − 7 and 3x² − 4x + 2 ?
  Solution:
  (5x² + 10x - 7) + (3x² - 4x + 2)
  = 5x² + 10x - 7 + 3x² - 4x + 2
  = 8x² + 6x - 5
• To subtract two polynomials, reverse the signs of subtrahend, and add two polynomials as done before.
  Example:
  Subtract 3x² - 4x + 2 from 5x² + 10x - 7
  Solution:
  (5x² + 10x - 7) - (3x² - 4x + 2)
  = (5x² + 10x - 7) + (-3x² + 4x - 2)
  = 5x² + 10x - 7 - 3x² + 4x - 2
  = 2x² + 14x - 9
• To multiply monomials, first multiply their coefficients, and then multiply their variables by adding the exponents.
  Example:
  What is the product of 3x²yz from -2x²y² ?
  Solution:
  (3x²yz)(-2x²y²)
  = (3 x -2)(x² x x²)(y x y²)(z)
  = -6x⁴y³z
• To multiply a monomial by a polynomial, just multiply each term of the polynomial by the monomial.
  Example:
  What is the product of 3x from 3x² - 6xy² + 2 ?
  Solution:
  (3x)(3x² - 6xy² + 2)
  = (3x x 3x²) - (3x x 6xy²) + (3x x 2)
  = 9x³ - 18x²y² + 6x
• To multiply two binomials, multiply each term of first binomial by each term of second binomial, then add the results.
  Example:
  What is the product of 3x + y from 3x² - 6xy²?
  Solution:
  (3x + y)(3x² - 6xy²)
  = (3x x 3x²) + (3x x (-6xy²)) + (y x 3x²) + (y x (-6xy²))
  = (9x²) + (-18x²y²) + (3x²y) + (-6xy³)
  = 9x² - 18x²y² + 3x²y - 6xy³
• The three most important binomial products are:
  • (x − y)(x + y) = x² + xy − xy + y² = x² + y²
  • (x + y)(x + y) = x² + xy + xy + y² = x² + 2xy + y²
  • (x − y)(x − y) = x² − xy − xy + y² = x² − 2xy + y²
    Memorizing these can save a lot of calculation time during the test.
• To divide a polynomial by a monomial, divide each term of the polynomial by the monomial.
  Example:
  What is the quotient if 32x²y + 12xy³z is divided by 8xy?
  Solution:
  (32x²y + 12xy³z) / (8xy)
  = (32x²y) / 8xy + (12xy³z) / 8xy
  = 4x + (3/2)y²z (by reducing the terms)

Solving Equations and Inequalities

• The basic principle in solving equations and inequalities is that you can manipulate them in any way as long as you do the same thing to both sides. For example you may add a number to both sides, or you may divide or multiply both sides with same number etc.
  By using the following six-step method, you can solve most of the equations and inequalities. The method is explained with the help of an example.
  if (1/2)x + 3(x - 2) = 2(x + 1) + 1, what is the value of x?
  Solution:

Step	What to do...	Example
1	Get rid of fractions and decimals by multiplying both sides by the LCD.	Multiply each side by 2 to get: x + 6(x − 2) = 4(x +1) + 2
2	Get rid of all parentheses by solving them.	x + 6x −12 = 4x + 4 + 2
3	Combine like terms on each side.	7x − 12 = 4x + 6
4	By adding and subtracting get all the variables on one side (mostly left).	Subtract 4x from each side to get: 3x −12 = 6
5	By adding or subtracting get all plain numbers on the other side.	Add 12 to each side to get: 3x =18
6	Divide both sides by the coefficient of the variable. (If you are dealing with an inequality and you divide with a negative number, remember to reverse the inequality.)	Divide both sides by 3 to get: x = 6

• When you have to solve one variable and the equation/inequality involve more than one variable, treat all other variables as plain numbers and apply the six-step method.
  Example:
  if a = 3b − c , what is the value of b in terms of a and c ?
  Solution:
  

Step	What to do...	Example
1	There are no fractions and decimals.
2	There are no parentheses.
3	There are no like terms.
4	By adding and subtracting get all the variables on one side.	Remember there is only one variable b, which is on one side only.
5	By adding or subtracting get all plain numbers on the other side.	Remember we are considering a and c as plain number. Add c to each side to get: a + c = 3b
6	Divide both sides by the coefficient of the variable.	Divide both sides by 3 to get: (a+c)/3 = b

• It is not necessary to follow these steps in the order specified. Some times it makes the problem much easier, if you change the order of these steps.
  Example:
  If x − 4 = 11, what is the value of x - 8 ?
  Solution:
  Going immediately to step 5, add 4 to each side to get: x =15. Now subtract 8 from both sides to get: x − 8 = 7.
  
• Doing the same thing on each side of an equation does not mean doing the same thing to each term of the equation. This is very important if you are doing divisions, or dealing with exponents and roots.
  Example:
  If a > 0 and a² + b² = c² , what is the value of a in terms of b and c.
  Solution:
  a² + b² = c² ⇒ a² = c² − b². Now you can’t take a square root of each term
  to get a = c − b. You must take the square root of each side:
  √a² = √(c² - b²) ⇒ a = √(c² - b²)
• Another type of equation is that in which the variable appears in exponent. These equations are basically solved by inception.
  Example:
  If 2^x+3 = 32 , what is the value of 3^x+2 ?
  Solution:
  2^x+3 = 32 ⇒ 2^x+3 = 25 ⇒ x + 3 = 5 ⇒ x = 2.
  Now as x = 2 , you can get x = 2 ⇒ x + 2 = 4 ⇒ 3^x+2 = 3⁴ = 81
• A system of equations is a set of two or more equations having two or more variables. To solve such equations, you must find the value of each variable that will make each equation true.
• To solve a system of equations, add or subtract them to get a third equation. If there are more than two equations you can just add them.
  Example:
  If x + y = 10 and x − y = 10 what is the value of y ?
  Solution:
  x + y = 10
  x - y = 10
  Add two equations: 2x = 12 ⇒ x = 6
  Now replacing x with 6 in the first equation: 6 + y = 10 ⇒ y = 4
• If you know the value of one variable in a system of two equations, you can use this value to get the value of the other variable. As it is done in the previous question.

Word problems

• To solve word problems, first translate the problem from English to Algebra. While translating, use variables to represent unknowns. Once translated, it is easy to solve them using the techniques you have learned in previous sections.

English words	Mathematical meaning	Symbol
Is, was, will be, had, has, will have, is equal to, is the same as	Equals	=
Plus, more than, sum, increased by, added to, exceeds, received, got, older than, farther than, greater than	Addition	+
Minus, fewer, less than, difference, decreased by, subtracted from, younger than, gave, lost	Subtraction	-
Times, of, product, multiplied by	Multiplication	X
Divided by, quotient, per, for	Division	÷ or a/b
More than, greater than	Inequality	>
At least	Inequality	>=
Fewer than, less than	Inequality	<
At most	Inequality	<=
What, how many, etc.	Unknown quantity	x (some variable)

Examples:
• The sum of 5 and some number is 13. 5 + x = 13
• Javed was two years younger than Saleem. J = S − 2
• Bilal has at most Rs.10,000. B ≤ 10000
• The product of 2 and a number exceeds that
  number by 5 (is 5 more than that number). 2N = N + 5
• In word problems, you must be sure about what you are answering. Do not answer the wrong question.
• In problems involving ages, remember that “years ago” means you need to subtract, and “years from now” means you need to add.
• Distance problems all depend on three variations of the same formula:
  
  • Distance = Speed × Time
  • Speed = Distance / Time
  • Time = Distance / speed

Example:

How much longer, in seconds, is required to drive 1 mile at 40 miles per hour than at 60 miles per hour?
Solution:
time1 = 1/40 hours = 1/40 x 60 minutes = 3/2 x 60 seconds = 90 seconds
The time to drive at 60 miles per hour can be calculated as
time2 = 1/60 hours = 1/60 x 60 minutes = 1 x 60 seconds = 60 seconds
difference = time1 - time2 = 90 - 60 = 30 seconds.

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