Can you help me with my algebra homework please?

Question

The first problem is

y+3
____ =7
4

y+3 over 4 = 7


the second problem is

(7divided by 3) + x(x-1)=x-2)

couldnt find a sign for division but you get it right?
please help me i can find the answers for these

Answer 1

I believe that you have cancel out the four by means of inverse operation. In other words, you are getting rid of 4


y+3/ 4 (4) = 7 (4)

Because multiplication is the inverse of division, we can use it to remove or cancel out numbers.

Inverse Property = A property of real numbers that states that the result of two real numbers that when combined will result in the identity element; when a number is added to its additive inverse, the sum is always zero; (e.g., 8 + -8 = 0); when a number is multiplied by its multiplicative inverse, the product is always 1. (See additive inverse and multiplicative inverse)

Inverse Operation = An operation that is the opposite of, or undoes, another operation; addition and subtraction are inverse operations; multiplication and division are inverse operations.

What is the next operation?
y+3= 7(4)

y+3-3= 7(4) - 3

y= 7(4) -3

You do the math; plug the numbers in the format and see if they fit.

Answer 2

ok - dealing with eqautions (not speling)
what we do to one side = the other

so lets get rid of the denomenator by multiplying each by 4
therefore y+3= 28
now subtract 3 from both sides
y = 25

wow we're done already - - - -- N E X T !!!

ok 7/3 (can be expressed as a fraction) + x^2 - x = x - 2
(how we got there) x(x-1) = x^2 (x squared) -x
so
x^2 - x + 7/3 = x - 2
subtract x from both sides
x^2 -2x +7/3 = -2
x^2 -2x + 2 1/3 = -2
add 2 to both sides
x^2 -2x + 4 1/3 = 0
seems the closest thing to this
is (x+1 1/3) (x - 3 1/3)

is your problem copied correctly??


all the best

Answer 3

for the first one you do 4 over 1 times y+3 over four and then you do 7 times 4 also so you can get rid of the 4 on the bottom and make it not a fraction then you do y+3=7 *4
y+3=28 then -3 from both sides and you get y= 25

Answer 4

(y+3)/4=7
y+3=28
y=25


7/3 + x(x-1) = x-2
7/3 + x^2 - x - x + 2 = 0
x^2 - 2x + 13/3 =0

If you solve for x, you won't have any real numbers.