How do I solve by elimination for the following: 3(x-3)-2y = 0 & 2(x-y) = -x-3?

Question

pure example

Answer 1

Hi Man

• Part 1
3(x-3)-2y = 0
[ (3*x) + ( 3*-3) ] - 2y = 0
3x -9 - 2y = 0
3x - 2y = 9

• Part 2
2(x-y) = -x-3
[(2*x) + ( 2*-y) = -x - 3
2x - 2y = -x - 3
2x +x - 2y = -3
3x - 2y = -3

• Part 3
( 3x - 2y = 9 ) - ( 3x - 2y = -3 )
( 3x - 3x ) + ( -2y + 2y ) = ( 9 + 3)
0 + 0 ≠ 12

AS the lines are parallel
so ; The result is " NO SOLUTION "

Good Luck dear...

Answer 2

3(x-3)-2y=0 -->multiply things inside bracket by 3
3x-9-2y=0 -->reorder the terms
3x-2y-9=0

2(x-y)=-x-3 -->multiply things inside bracket by 2
2x-2y=-x-3 -->put all terms on LHS
3x-2y+3=0

that doesn't really work hon... wrong question?
3x-2y can't equal to 9 & -3 at the same time i think?
i mean if you want elimination...
e.g. Eq I minus Eq II will leave you 1 variable instead of 2, then you can solve the question...
a Q like this can't have solution?!

Answer 3

1) 3(x-3)-2y = 0
3x - 9 - 2y = 0
3x - 2y = 9

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

Since the coefficients of x an y are the same in both equations, the slopes of the two lines are equal - the lines are parallel - no solution.

Answer 4

Here we go:

3 (x -3) - 2y = 0
3x - 9 - 2y = 0

And:
2 (x - y) = - x - 3
2x - 2y = - x - 3
2x + x - 2y + 3 = 0
3x - 2y + 3 = 0

Now you have to add them together or solve by substitution.

Answer 5

9x-9-2y=0 & 2x -2y =x-3
x - 2y+3 = 0

so first you have to eliminate one variable.
(9x-9-2y=0)*2 = 18x -18-4y =0
(x-2y+3)*-2 = -2x + 4y -6
___________
16x -24 =0
so 16x=24
2x= 3
x= 1.5
then insert your value for x into the first equations to check.

Answer 6

equation 1=3x-9-2y=0
equatoion 2=2x-2y=-x-3
=>3x-2y+3=0
lines are parallel
no solution

Answer 7

This is unsolvable!!!

3x=2y+9

3x=2y-3

2Y+9 = 2Y-3 which is wrong

X=-1