Linear System PLEASE HELP!!!!

[snow123]

Hi,
I am attaching a picture of the question i am stuck on. The question has fractions and I don't know how to get rid of them. Please help me.
x=5 and y=3 but i am not getting the right answer. Please help me.
Thank you.

 

[Denis]

Multiply 1st one by 6: 2(x - 2) - 3(y + 5) = -18
Multiply 2nd one by 3: 9x - 2y = 39

You ok now?

 

[lookagain]

snow123,

when you handwrite or type your equations, you must not write
subtraction signs so they are (or appear to be) negative signs,
especially with fractions/algebraic fractions.

The subtraction (or addition signs) are to be out in front of the number,
and with fractions, are to be on the same level as the fraction
bars, not up with a numerator. For example, your equations
should appear as:


\(\displaystyle \frac{x - 2}{3} - \frac{y + 5}{2} = -3\)

\(\displaystyle 3x - \frac{2y}{3} = 13\)

 

[snow123]

so i get:
2x-4-3y+15=-18 or 2x-3y=-29 i multiply it by 2 so i get the same number in front of y for both equations
9x-2y=39 i multiply it by 3 so i get the same number in front of y for both equations
so i get:
4x-6y=-58
27x-6y=117 i subtract 27x -4x= 23x and -6-(-6)=0 and 117-(-58)=175
so i get 23x=175
x= 7.61
when the right answer is x=5
so i did something wrong.
Please help i am really confused.

 

[lookagain]

so i get:
2x-4-3y+15=-18 or 2x-3y=-29 i multiply it by 2 so i get the same number in front of y for both equations

No, look at my post. The subtraction sign will distribute to
all terms of that numerator, so the equation will be:

\(\displaystyle 2x - 4 - 3(y + 5) = -18\), which becomes

\(\displaystyle 2x - 4 - 3y - 15 = -18\), which becomes

\(\displaystyle 2x - 3y - 19 = -18\), which becomes

\(\displaystyle 2x - 3y = 1\)

 

[Denis]

so i get:
2x-4-3y+15=-18

NO...2x - 4 - 3y - 15 = -18 : -3(y - 5) = -3y - 15 ; kapish?

SLOW down Snow White!!

 

[snow123]

Thank you everyone!
i got the right answer!