Question

Solve each system by substitution

Answer 1

`\left \{ {{-3x-3y=3\:(I)} \atop {y=-5x-17\:(II)}} \right.`
Replace the equation y in the first equation.

`-3x-3y=3 (I)`

`-3x-3*(-5x-17)=3`

`-3x+15x+51=3`
Now, put the numbers with letter (incognito) to the right and numbers without letter to the left, changing the signal when changing sides.

`-3x+15x=3-51`

`12x = -48`

`x = (-48)/(12)`

`\boxed{\boxed{x = - 4}}\end{array}}\qquad\quad\checkmark`


Now substitute the found value "x" in the equation y (second equation):

`y=-5x-17\:(II)`

`y = - 5*(-4) - 17`

`y = 20 - 17`

`\boxed{\boxed{y=3}}\end{array}}\qquad\quad\checkmark`

Answer:

`\underline{\textbf{The\:values\​\:found\:are:\:X = -4\:and\: Y = 3} }`

Answer 2

-3x-3y=3
y=-5x-17----------use this to substitute for y

-3x-3(-5x-17)=3----------like this
-3x+15x+51=3-------------combine like terms
12x+51=3------------subtract 51 both sides
12x=-48---------divide by 12 each side to get x by itself
x=-4

y=-5(-4)-17--------------------substitute the -4 for x to find y
y=3

now we plug in both to see if it works:
-3(-4)-3(3)=3--------------We plug in to see if both answers work
12-9=3----------------It works
(-4,3)

Hope this helps :)