Proportion math problems- find x?

Question

here are a few problems i am struggling with. can you help?

1.) 3x+1 over 4x+3 = x+1 over x+3

2.) x+1 over x+3 = 2x-1 over x+7

3.) x+1 over x+3 = 3x-5 over 2x+6

4.) 2x+1 over 5x+4 = 11x+7 over 49x+5

you dont have to do them all, but type which number you did. thanksss

Answer 1

Cross multiply to solve each one.

For example:

1.  3x+1 over 4x+3 = x+1 over x+3

     (3x+1) * (x+3) = (x+1) * (4x+3)

     3x^2 + 10x + 3 = 4x^2 + 7x + 3

     x^2 -3x = 0

     x * (x-3) = 0

     x = 0 and 3

2. x+1 over x+3 = 2x-1 over x+7

    (x+1) * (x+7) = (2x-1) * (x+3)

    x^2 + 8x + 7 = 2x^2 + 5x - 3

    x^2 -3x - 10 = 0

    (x-5) * (x+2) = 0

    x = -2 and 5

3 and 4 can be solved in same fashion.

Answer 2

i'm sorry i could help you but its just way to hard to show the work on here, so if you would like i could either just give you the answers, or i could send you something written up, on an easier program

okay well for the left side i minused the top and botton by 3x leaving you with 1 over x+3

and for the right side is minused the top and bottom by 1x, leaving you with 1 over 3

than i flipped the fractions upside down so i had x+3 over 1 = 3 over 1

than i got rid of the denomenators(botton # on fraction)

so i had x+3=3

so then i subract 3 from both sides, x+3-3 = 3-3

so that gave me the answer that x=0

and to check that all you do is write the question out again, and replace the x with the # 0 and solve it like you normally would, and if both sides of the = sign at the end are = than your answer is correct and i will get your answers to the other questions in a couple minutes k

Answer 3

1. 3x+1 over 4x+3 = x+1 over x+3

First step is to get a common denominator. Here's what i did

(3x+1)(x+3) over (4x+3)(x+3)  = (x+1)(4x+3) over (x+3)(4x+3)

after you get this you have to multiply it out. you should get

3x^2 + x + 9x + 3 + 4x^2 + 3x + 4x + 3 over 4x^2 +12x +3x + 9

add like terms and you should get

7x^2 + 17x + 6 over 4x^2 + 15x + 9     The answer for # 1

I did number 2 also and the answer i got is

3x^2 +13x +4 over  x^2 + 10x + 21     Answer for #2

Sorry too lazy to do number 3 & 4

Answer 4

Oh, gosh...I tried to do the first one for you and I got x = -2 and x = -3.

I think it's right, if you're doing some kind of quadratic system of equations thing.

Answer 5

same method for all of this type:

eg Q1

multiply top of side 1 by bottom of side 2 and top side 2 by bottom of 1

(3x+1)(x+3) = (x+1)(4x+3)

expand both sides

3x^2 +10x +3 = 4x^2 +7x +3

collect on one side and get totals

0= x^2 -3x

factorise

0=x(x-3)

so x=0 or x-3=0 x=3

check by putting value of x in original equation