Question

Find the solutions for 5x+3y=30 and 3x+6y=12
FAST PLS i will mark !!!!!

Answer 1

`x=(48)/(7)\\ y=-(10)/(7)`

Explanation:

`5x+3y=30\\3x+6y=12`

Let's solve the second equation for x

`3x+6y=12\\3x=12-6y\\x=(12-6y)/(3)`

Now let's replace this in the first equation.

`5x+3y=30\\5((12-6y)/(3))+3y=30`

Distribute the 5

`(60-30y)/(3)+3y=30`

Individually solve the fractions.

`(60)/(3)-(30)/(3)y+3y=30`

Solve;

`20-10y+3y=30\\20-7y=30`

Subtract 20 from both sides.

`20-20-7y=30-20\\-7y=10`

Divide both sides by -7

`(-7y)/(-7)=(10)/(-7)`

`y=-(10)/(7)`

Now replace the value of y in any of the two equations to find x.

`3x+6y=12\\3x+6(-(10)/(7))=12`

`3x-(60)/(7)=12\\`

add
`(60)/(7)` on both sides.

`3x-(60)/(7)+(60)/(7) =12+(60)/(7)`

`3x=((12)(7)+60)/(7)`

`3x=(84+60)/(7)\\ 3x=(144)/(7)`

Multiply by the reciprocal fraction of 3. I suppose you know that 3 is 3/1 hence the inverted fraction would be 1/3

`((1)/(3))(3x)=((144)/(7))((1)/(3))`

`x=(48)/(7)`