Can anyone help me on these algebra 1 problems.?

Question

1) The product of two consecutive odd whole numbers is 143. find the numbers.


2) Twenty more than the square of a number is twelve times the number. Find the number.\


3)the product of three more than a number times two less than the number is fourteen. Find the number.


4) The sum of the squares of two consecutive odd ositive integers is 202. Find the numbers.


5)The product of two consecutive even whole numbers is 288. Find the numbers.

Please helpme on these problems.

Answer 1

The first one:

x + (x+ 1) = 143
2x + 1 = 143
2x = 142
x= 71

so the numbers are 71 and 72.

Answer 2

1) n* (n+2) = 143
n^2 +2n - 143 = 0
(n-11)*(n+13) = 0
(n+13) = 0 n = -13 doesn't make sense
(n-11) = 0 N = 11 N+2 = 13

11 and 13

2) 20 + n^2 = 12*n
n^2 - 12n +20 = 0
(n-2) * (n-10)
n = 2 4 +20 = 12 * 2 yes
n = 10 100 +20 = 12 * 10 yes
both 2 and 10 are good answers

3) ( n+3) * (n-2) = 14
n^2 +n - 20 = 0
(n-4) * (n+5) = 0
n=4
check (4+3) * (4-2) = 14
7 * 2 = 14 yes

the number is 4


4) n^2 + (n+2)^2 = 202
n^2 + n^2 + 4n + 4 = 202
2n^2 + 4n + 4 =202
n^2 +2n + 2 = 101
n^2 +2n -99 = 0
(n+11) * (n-9)=0

n=9 n+2 = 11 check 81 + 121 = 202 yes

5) n* (n+2)) = 288
n^2 + 2n -288 = 0
(n+18) * (n-16) = 0
n = 16 and n+2 = 18 check 16*18 = 288 yes

Answer 3

Most of these are trial and error, you just have to know what the terms mean and a few basic equations. Consecutive means next to each other.
1)xy=143 (11 &)
2)(2)
3)2x+3=14+x (x=11)
4)x^2+y^2=202 (9 &)
5)xy=288 (16 &)
Hope that helps at least.

Answer 4

heres a start:

1)let x be the first ODD integer and x+2 be the second ODD integer. product means multiply.
therefore your new equation is x(x+2)=143.
distribute the x. (x^2+2x) =143
set =0, x^2+2x-143=0
factor. multiples of 143 are (1,143)(11,13)
so the answer is 11,13 because they are consecutive odd integers.


2) let x= a number, therefore your new equation is
x^2+20=12x
set equal to 0 to solve for x...x^2-12x+20=0,
Factor. find multiples of 20...(1, 20)(2,10)(4,5)...
since the equation says +20 that means signs are like (either both + or both -) given that its -12, we know that 2,10 are the only multiples that when added give us -12, therefore you now have
(x-10)(x-2)=0, x-10=0 and x-2=0
therefore x=10 and x=2

3)let x=a number, therefore your new equation is:
(x+3)(x-2)=14
FOIL. x^2+3x-2x-6=14.
combine like terms. x^2+x-20=0
factor. multiples of 20 are (1,20)(2,10)(4,5). since its -20 we know signs are differrent. since its +1 we will use 5 as the positive value.
so after factoring we now have
(x+5)(x-4)=0, set each =0
x+5=0 and x-4=0, therefore x=-5 and x=4

now you try the remaining problems...if you still need help, you can email me at [e-mail] and i will help you.

Answer 5

Suppose x is the required no.
From the given data,
x^2 + 20 = 12x
x^2 - 12x + 20=0
(x-10)(x-2)=0
x=10 or x=2

Answer 6

Henry, to solve these problems, call your unknown "x" and set up an equation to represent each of the problems. I'll do no.1
x (x+2)= 143
x(squared) +2x =143
x (squared)+2x-143=0
(x -11 )(x+13)=0
x=11 (Proof: 11 x 13 =143)

You may try the others on your own. If I do them, it won't help you learn very much.

Answer 7

the answers in

#1 is 71, 72

#4 is 9 & 11

#5 is 14 & 16

i cant understand the 2nd and 3rd questions.. very confusing.

Answer 8

1) 11 and 13;
2) 2 and 10;
3)
4)
5) 16 and 18

Answer 9

cheaters never prosper. even cheaters who use Yahoo! to answer their homework. ive learned my lesson.