y = (4x+9)/(2x+3)2xy+3y = 4x+93y-9 = 4x-2xy3y-9 = x(2-2y)x = (3y-9)/(2-2y)So, inverse function is 3x minus 9 divided by 2 minus 27.
1.5
X + 3y - 2y + 3y = X + (3-2+3)y = X + 4y
I don't know whether it's y^2-3y+2+3y-4y-5 or 2y-3y+2+3y-4y-5, so I'll do both.y^2-3y+2+3y-4y-5y^2+(-3y+3y-4y)+(2-5); group them togethery^2+(-4y)+(-3)y2-4y-3OR2y-3y+2+3y-4y-5(2y-3y+3y-4y)+(2-5); group them together(-2y)+(-3)-2y-3
2y + 3 = 3y + 9 2y - 3y = 9 - 3 or -y = 6 so y = -6
eq. 1 is x-2y=0 which means x=2y eq. 2 is 4x-3y=15 subst. x=2y in eq. 2 4(2y)-3y=15 8y-3y=15 5y=15 y=3 subst. y=3 in eq. 1 we get x=2(3) x=6 Ans: x=6 and y=3 is the required solution. Thank you.
if you wish to multiply 3y-2y by 3y-y the answer is 2y^2
x+7x+3y-y =8x+2y =2(4x+y)
If you mean: 4x -3y = 1 and x -2y = 4 then the solutions are x = -2 and y = -3
I assume you mean,X - 2Y = 04X - 3Y = 15this is set up for substitution, use ( X - 2Y = 0)X - 2Y = 0X = 2Y----------substitute into ( 4X - 3Y = 15 ) to find Y4(2Y) - 3Y = 158Y - 3Y = 155Y = 15Y = 3---------find X in ( X - 2Y = 0 )X - 2(3) = 0X - 6 = 0X = 6---------------check in both original equation6 - 2(3) = 06 - 6 = 00 = 0-----------checks4(6) - 3(3) = 1524 - 9 = 1515 = 15----------------both check and are consistent(6, 3) is the solution set
y = (4x+9)/(2x+3)2xy+3y = 4x+93y-9 = 4x-2xy3y-9 = x(2-2y)x = (3y-9)/(2-2y)So, inverse function is 3x minus 9 divided by 2 minus 27.
1.5
X + 3y - 2y + 3y = X + (3-2+3)y = X + 4y
I don't know whether it's y^2-3y+2+3y-4y-5 or 2y-3y+2+3y-4y-5, so I'll do both.y^2-3y+2+3y-4y-5y^2+(-3y+3y-4y)+(2-5); group them togethery^2+(-4y)+(-3)y2-4y-3OR2y-3y+2+3y-4y-5(2y-3y+3y-4y)+(2-5); group them together(-2y)+(-3)-2y-3
x - 2y = 04x - 3y = 15From the first equation, x = 2y.SUBSTITUTING this value of x in the second equation,4*(2y) - 3y = 15 ie 8y - 3y = 15 or 5y = 15 so that y = 3and then, x = 2y gives x = 2*3 = 6Answer: x = 6, y = 3
To find the extreme value of the parabola y = x2 - 4x + 3 ...(1) Take the derivative of the equation.y = x2 - 4x + 3y' = 2x - 4(2) Set the derivative = 0 and solve for x.y' = 2x - 40 = 2x - 42x = 4x = 4/2x = 2(3) Plug this x value back into the original equation to find the associated y coordinate.x = 2y = x2 - 4x + 3y = (2)2 - 4(2) + 3y = 4 - 8 + 3y = -1So the vertex is at (2, -1).
3y - 3 = 2y + 5 3y - 2y = 5 + 3 y = 8