Hi,
a) y = x + 1
3x + 2y = 17
b) 3x - 2y = 0
7x + y = 23
c) 5x = 1 + 3y
-2y = 8 - 7x
d) 7x - 3y = 0
5x = 6 + 3y
e) 4x - 5y = 7
12x - 3 = 9y
f) x = 2y+1
3x - 4y = 7
thank you im terrible at these
I'll do the first, the rest can be done in the same way.
There are fancy methods to do this and there are simpler ones. It all depends on the type of problem and how you are learning to solve them. This is called the "substitution method."
The top equation is already solved for y, so I'm going to insert this "value" of y into the second equation:
3x + 2(x + 1) = 17
3x + 2x + 2 = 17
5x = 15
x = 3
Now you can put this value of x into either equation. Again for simplicity I'm using the top equation:
y = x + 1 = 3 + 1 = 4.
So the solution is (x, y) = (3, 4)
-Dan