Question 648018
Original equations: {{{x - 3y = 3}}}  and  {{{3x + 5y = -19}}}

Step 1: To solve by substitution, we'd have to get a variable by itself so that we can plug it into the other.

Let's try to get 'x' by itself on the left equation to make it easier, since it doesn't have a number already attached to it.

{{{x - 3y = 3}}}

Step 2: Add '3y' to both sides to get 'x' by itself

{{{x - 3y = 3}}}
{{{x = 3y + 3}}}

Step 3: We now have 'x' by itself and can plug it into the other equation to give us what 'y' is.

{{{x = 3y + 3}}}

Plug in the above into the other equation.

{{{3x + 5y = -19}}}
{{{3(3y+3) + 5y = -19}}}

Step 4: Distribute the '3' to everything in the parenthesis

{{{3(3y+3) + 5y = -19}}}
{{{9y+ 9 + 5y = -19}}}

Step 5: Combine like terms

{{{9y+ 9 + 5y = -19}}}
{{{14y+ 9 = -19}}}

Step 6: Continue combining like terms.. subtract 9 to both sides to get '14'y by itself

{{{14y+ 9 = -19}}}
{{{14y = -28}}}

Step 7: Divide '14' to both sides to get 'y' by itself

{{{14y = -28}}}
{{{y = -2}}}

We now know that the 'y' point to this system of equations is {{{-2}}}

With the 'y' point, you can find the 'x' by plugging in the 'y' to any of the two given equations.

I'm going to use this equation to make it easier {{{x - 3y = 3}}}

Step 1: We know what 'y' is, so plug {{{y = -2}}} into the equation to get the 'x' point

{{{x - 3y = 3}}}
{{{x - 3(-2) = 3}}}

Step 2: Simplify

{{{x - 3(-2) = 3}}}
{{{x + 6 = 3}}}

Step 3: Subtract '6' to both sides to get 'x' by itself

{{{x + 6 = 3}}}
{{{x = -3}}}

We now know that the 'x' point to this system of equations is {{{-3}}}

The solution is {{{x = -3}}} and {{{y = -2}}}