Solve the system by substitution.
3x-3y=-12,y=-3x-12 i see its the same problem rewritten but i confused on what i should do any help would be nice
If 3x - 3y = -12, then
x - y = -4. You also can rewrite the second equation to get
3x + y = -12
Therefore, adding the two equations,
4x = -16 , so x = -4.
That tells you that y must be 0.
@drwls: The solution you provided is correct, however, it was asked that this system be solved by means of substitution.
In the second equation, it is given that y=-3x-12. Simply by replacing (i.e. substituting) y in the first equation by the value that was given for y in the second equation, we get:
3x-3(-3x-12)=-12
<=> 3x+9x+36=-12
<=> 12x = -48
<=> x = -4
since y = -3x-12, y=0
To solve the system by substitution, you need to substitute the expression for y in the second equation into the first equation. Let's go step by step:
Given:
Equation 1: 3x - 3y = -12
Equation 2: y = -3x - 12
Step 1: Substitute y in Equation 1 with its value from Equation 2.
Replace y in Equation 1 with (-3x - 12):
3x - 3(-3x - 12) = -12
Step 2: Simplify and solve for x.
Expand the brackets and simplify:
3x + 9x + 36 = -12
12x + 36 = -12
Step 3: Solve for x.
Move the constant term to the other side of the equation:
12x = -12 - 36
12x = -48
Divide both sides by 12 to isolate x:
x = -48/12
x = -4
Step 4: Substitute the value of x back into Equation 2 to solve for y.
Using Equation 2: y = -3x - 12
Substitute x = -4:
y = -3(-4) - 12
y = 12 - 12
y = 0
Therefore, the solution to the system is x = -4 and y = 0.