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linear system
- Thread starter tinkwink
- Start date
tinkwink said:solve the linear system for:
4x-4y=y
3x+4y=-18
Please explain.
If you've typed the first equation correctly (and I wonder if you have.....), then the first step should be to get the x and y terms in the first equation on the same side of the equals sign. Add -y to both sides of the first equation:
4x - 4y + (-y) = y + (-y)
4x - 5y = 0
Now, your system looks like this:
4x - 5y = 0
3x + 4y = -18
If you multiply both sides of the first equation by 4, and both sides of the second equation by 5, the system will look like this:
16x - 20y = 0
15x + 20y = -90
Now, add the two equations....do you see that you'll end up with 0y?
31x = -90
You can solve this for x. Once you have the value of x, you can substitute it into either one of the original equations, and then solve for y.
tinkwink said:solve the linear system for:
4x-4y=y
3x+4y=-18
Please explain.
For two variable systems, it might be faster to use cramer's rule, even if by hand.