How do I solve a system of equations?

Question

I need help solving something like:

x=y-3
5x+3y=1

appreciate it.

Answer 1

(x-y= -3)-5
5x+3y =1

-5x+5y =15
5x +3y =1

8y = 16
y=2

5x + 3(2) =1
5x = -5
x=-1

Answer 2

in the 2nd equation replace the x next to the 5 with the 1st equation because you know what x equals so 5(y-3) + 3y=1
then distribute the 5 so 5y-15 + 3y =1 then combine like terms so 8y-15=1 so add 15 to opposite side to cancel so 8y=16 so y=2 and then plug the y answer into the first equation so x=2-3 so x=-1

Answer
X= - 1
Y= 2

Answer 3

1) Plug the solution for x in the first equation into the second Equation and solve for y.

5(y-3) + 3y = 1
5y - 15 + 3y = 1
8y = 16
y = 2

2) Plug the answer for y back into the first equation and solve for x.

x = 2 - 3
x= -1

Answer 4

it is fairly simple. first you need to plug in the first equation to the second equation. so now it should look something like this.

5(y-3)+3y=1

After that, you need to multiply. Therefore...

5y-15+3y=1

Then you need to simplify everything.

8y-15=1

Add 15 on both sides.

8y=16

Divide everything by 8.

therefore, y=2

Since now you know y equals to 2. plug in the 2 to the first equation, which is x=y-3

x=2-3
x=-1

So your answer is y=2
x=-1

Answer 5

Substitute x = y-3 into the second equation,
5(y-3)+3y=1
8y = 16
y = 2
x = -1

Answer 6

x=y-3 ----1
5x+3y=1 ------2

Sub Eqn 1 into Eqn 2

5 (y-3) + 3y = 1
5y - 15 + 3y = 1
8y = 16
y = 2

Sub y=2 into Eqn 1
x = 2-3
x = -1

Answer 7

make them able to be equal to each other

like change the secone one to
5x=-3y+1
x=(-3y+1)/5
then
(-3y+1)/5=y-3
then solve

Answer 8

x - y = - 3

3x - 3y = - 9
5x + 3y = 1----ADD

8x = - 8
x = - 1

y - 3 = x
y - 3 = - 1
y = 2

Solution is x = - 1 , y = 2