simultaneous equation grade 10

Question

I need help with the working out on this equation because I always end up with -30x=4.

The equations are: -3x-2y=-12 and y=5x-20/3

The answer is meant to be x=4 and y=0

Answer 1

If x = 4 and y = 0 is a solution, then the problem must be -3x-2y=-12 and y=(5x-20)/3. If the 20 is the only thing divided by the 3, it makes a big difference.

Assuming the problem is -3x - 2y = -12 and y = (5x - 20)/3 I would multiply both sides of the second equation by 3 to get:

3y = 5x - 20 then subtract 5x from both sides of the equation to get it in the same form as the first equation:

-5x + 3y = -20

Then you have these 2 equations:

-3x - 2y = -12

-5x + 3y = -20

Notice that one of the y values is positive and one is negative. That means you could eliminate them if the coefficients were slightly different. Multiply the first equation by 3 and the second equation by 2 to get the y values to be plus and minus 6 as follows:

-9x - 6y = -36

-10x + 6y = -40

Add the two equations together:

-9x - 6y = -36

-10x + 6y = -40

--------------------

-19x = -76

divide both sides by -19 to get:

x = 4

The go back to the original equation y=(5x-20)/3 and plug in x = 4:

y=(5(4)-20)/3

y = (20 - 20)/3

y = 0/3

y=0