Solve the system using substitution?

(1/3) (y + 8) + (1/3) y = 0
y + 8 + y = 0
2y = - 8
y = - 4

x = y + 8
x = 4

x = 4 , y = - 4

no longer complicated: 3x + 4y = -sixteen, 3x + 11 = y... or; 3x - y = -11 Subtract out the 2nd equation from the 1st. 5y = -5 y = -a million And now sparkling up for x: 3x + 4y = -sixteen, 3x + 11 = y 3x + (-4) = -sixteen, 3x + 11 = (-a million) 3x = -12, 3x = -12 x = -4, x = -4 y = -a million and x = -4 No mistaking it from the above. it quite is an ironclad data!

1/3(x) + 1/3(y) = 0
x = y + 8

1/3(x) + 1/3(y) = 0
1/3(y + 8) + y/3 = 0
3[(y + 8)/3 + y/3] = 3[0]
y + 8 + y = 0
y + y = -8
2y = -8
y = -8/2
y = -4

x = y + 8
x = -4 + 8
x = 4

∴ x = 4 , y = -4

What i always like to do is make either x or y the subject 1st;
say we make x the subject for both eqns:

1/3x + 1/3y = 0
1/3x = -1/3y
*multiply both sides with 3 to get rid of the denominator*
x = -y ---> (1)

x = y + 8 ---> (2)

Since both eqns are equal to x it shows that they are equivalent to ech other:

(1) = (2)
-y = y + 8
-y - y = 8
-2y = 8
(-2) ÷ -2y = 8 ÷ (-2)
y = -4
*Subst. in y = -4 in eqn (1) to find x*
x = - y
x = - (-4)
= 4

PROOF:
Eqn (1)
1/3x + 1/3y = 0
1/3(4) + 1/3(-4) = 0
4/3 + (-4/3) = 0
0 = 0 [Proven]

Eqn (2)
x = y + 8
4 = -4 + 8
4 = 4 [Proven]

1/3x+1/3y=0
x=y+8

3x + 3y = 0
3(y+8) + 3y = 0
6y + 24 = 0
y = -4
x = -4 + 8 = 4

Answer: x=4, y = -4

1/3 x + 1/3 y = 0

Multiply both side with 3

x + y = 0

Since x = y + 8 then

y + 8 + y = 0
2y = -8
y = -4

x + (-4) = 0
x = 4

(x,y) = (4,-4)