x = 3y - 3 so 6 (3y - 3) + 2y = -12 ie 18y - 18 + 2y = -12 ie 20y = 6 so y = 0.3 and x = -2.1
Solving for y: 5y = 9 + 2y 5y - 2y = 9 3y = 9 y = 9/3 = 3
5x - 3y -7x + 2y +3x = (5-7+3)x + (-3+2)y = x - y
Given: 2x - 3y = 2 3x + 2y = 3 Take the first equation, and solve for x: x = (2 + 3y) / 2 Now plug it into the second equation: ∴ 3(2 + 3y) / 2 + 2y = 3 ∴ 3 + 9y/2 + 2y = 3 ∴ 9y/2 + 2y = 0 ∴ 22y = 0 ∴ y = 0 Then you can take that value of y, and plug it into either of our first equations to find x; 2x - 3y = 2 ∴ 2x - 3(0) = 2 ∴ 2x = 2 ∴ x = 1 So x is equal to one, and y is equal to zero.
2x+3y=40-2x+2y=20Since 2x does not contain the variable to solve for, move it to the right-hand side of the equation by subtracting 2x from both sides.3y=-2xDivide each term in the equation by 3.(3y)/(3)=-(2x)/(3)Simplify the left-hand side of the equation by canceling the common terms.y=-(2x)/(3)if you were solving for x It would be x=-(3y)/2
2y + 3 = 3y + 9 2y - 3y = 9 - 3 or -y = 6 so y = -6
y = -1
(-1, 0)
are you talking about a system of equations where x+3y=-7 -x+2y=-8 if so, then the answer is {2,-3}
x = 3y - 3 so 6 (3y - 3) + 2y = -12 ie 18y - 18 + 2y = -12 ie 20y = 6 so y = 0.3 and x = -2.1
X + 3y - 2y + 3y = X + (3-2+3)y = X + 4y
Solving for y: 5y = 9 + 2y 5y - 2y = 9 3y = 9 y = 9/3 = 3
perpendicular
5y = 2y+9 5y-2y = 9 3y = 9 y = 3
x = 3 and y = 2
5y + (4 - 2y) = 9 5y + 4 - 2y = 9 3y + 4 = 9 3y = 5 y = 5/3
Multiply the bottom equation by -1