solving system of linear equations by eliminating please help me?!?

Question

solving system of linear equations by eliminating please help me?!?


1) x - y - 6 = 0
2x + 3y + 3 = 0


2) x + 3y + 3 = 0
2x - 3y + 6 = 0




PLEASE HELP ME????? ANY MATH GENIUS HERE!!!!!!!!

Answer 1

solving system of linear equations by elimination.
1) x - y - 6 = 0 (1st equation)
2x + 3y + 3 = 0 (2nd equation)

solution:
multiply the 1st equation by 2 to be able to eliminate x by subtracting the 2 given equations.

2( x-y-6=0) makes it 2x - 2y - 12 = 0 (1st equation)
by elimination:

2x - 2y - 12 = 0
-(2x + 3y + 3 = 0)
______________
-5y - 15 = 0 (derived 3rd equation)

you can now solve for the value of y.

5y = -15
y = -15/5 = -3

and by substituting your derived value of y to either of the 2 given equation you can now solve the value of your x

x - y - 6 = 0
x - (-3) - 6 = 0
x + 3 -6 = 0
x = 6 - 3 = 3

same procedure applies.

2) x + 3y + 3 = 0
2x - 3y + 6 = 0

solution:
2(x + 3y + 3 = 0)
- (2x - 3y + 6 = 0)
_____________
9y = 0

y = 0/9 = 0

substitution:
x + 3(0) +3 = 0
x + 3 = 0
x = -3

i hope this helped you.

Answer 2

1)
x - y - 6 = 0
2x + 3y + 3 = 0
2x - 3y = 12
2x + 3y = -3
4x = 9
x = 9/4
y = -5/2

2)
x + 3y + 3 = 0
2x - 3y + 6 = 0
3x = -9
x = -3
y = 0

Answer 3

Let

Eq’n 1: x – y - 6 = 0
Eq’n 2: 2x + 3y + 3 = 0

Multiply Eq’n 1 by -2 to get

-2x + 2y + 12 = 0

Add the result to Eq’n 2

-2x + 2y + 12 = 0
2x + 3y + 3 = 0
_________________
5y + 15 = 0

Divide by 5

Y + 3 = 0 or y = -3

Substitue value to Eq’n 1

x – (-3) - 6 = 0
x +3 - 6 = 0
x - 3 = 0
x = 3

Problem 2:

Eq’n 1: x + 3y + 3 = 0
Eq’n 2: 2x - 3y + 6 = 0

Multiply Eq’n 1 by -2 to get

-2x – 6y – 3 = 0

Add to Eq’n 2

-2x – 6y – 3 = 0
2x - 3y + 6 = 0
_________________
-9y + 3 = 0

Divide by 9

-y + 1/3 = 0 or y = 1/3

Subsitute value to Eq’n 1

x + 3(1/3) + 3 = 0
x + 1 + 3 = 0
x + 4 = 0 or x = -4

Answer 4

2x+2y+3z=2 -x+3y-6z=4 4x+8y+3z=8 it is basically one conceptual step previous 2 equations and 2 unknowns. I notice that 3z turns up two times, so which you will on the instant do away with z 4x+8y+3z=8 2x+2y+3z=2 subtract to get 2x + 6y = 6 and then via taking a distinctive pair 2x+2y+3z=2 multiply with the aid of via 2 to get 4x + 4y + 6z = 4 and then use the relax equation -x + 3y - 6z = 4 and upload to get 3x + 7y = 8 Now you have 2 equations with 2 unknowns 2x + 6y = 6 3x + 7y = 8 you're taking it from there.

Answer 5

in 1 part its
x=6+y
=> putting value of x in 2x+3y= -3
thus 2(6+y)+3y= -3
12+2y+3y= -3
12+5y= -3
5y= -15
y= -3 (15 divided by 5 is equal to 3)
putting the value of y in x=6+y
we get x=6-3
thus x=3
thus x=3 and y= -3

for second part...
x+3y+3=0
2x- 3y+6=0
x= -(3+3y) (taking - common)
putting value of x in 2nd equation..
-2(3+3y)-3y+6=0
-6-6y-3y+6=0
-9y=0 (-6 and +6 equals to 0)
y=0
putting value in first eq x+3y+6=0
x=-6-3y
x= -6-3(0)
x= -6
thus x= -6 and y=0