Those would be two equations that each describe a different line:
2x + 5y = 2
-3x - y = -2
If you'd like to find the point at which they intersect, that can be done using the substitution method. First, take either of the equations and solve it for one of the two variables. In this case, we'll solve the second equation for y:
-3x - y = -2
∴ y = 2 - 3x
Now take that definition of y, and plug it into the other equation:
2x + 5y = 2
∴ 2x + 5(2 - 3x) = 2
∴ 2x + 10 - 15x = 2
∴ -13x = -8
∴ x = 8/13
Now you can take that value of x, and plug it into either of the original equations to find the value of y:
2x + 5y = 2
∴ 16/13 + 5y = 2
∴ 5y = 26/13 - 16/13
∴ 5y = 10/13
∴ y = 2/13
To be certain that our answer is correct, we should also plug that value into the other original equation:
-3x - y = -2
∴ -24/13 - y = -2
∴ -y = -26/13 + 24/13
∴ y = 2/13
So our answer is indeed correct, and the two lines intersect at the point (8/13, 2/13)
If: 2x+5y = 2 and -3x-y = -3Then it works out that: x = 1 and y = 0
(4, -1)
(2, 3)
x/5y + 2x/3y = 3x/15y + 10x/15y = 13x/15y
x = 1, y = 0
If: 2x+5y = 2 and -3x-y = -3Then it works out that: x = 1 and y = 0
(4, -1)
(2, 3)
x/5y + 2x/3y = 3x/15y + 10x/15y = 13x/15y
x = 1, y = 0
x = 1, y = 0
3x/7 + 5y/14x = 3x*2x/(7*2x) + 5y/14x = (6x2 + 5y)/14x
3x - 8y + 2x - 5 - 5y = 5x -13y - 5
They are: x = -2 and y = -1.
no
x = 4 and y = -1 (4. -1)
3x+5y=48 5y=48-3x-3x+5y=12 -3x+(48-3x)=12-6x=-36x=65y=48-3(6)5y=30y=6(6,6)