Sorry cant help you, I assume this is missing '=' signs
Sorry,
Jayson :)
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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
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 = 12When x = 0, y = -4, so that y-intercept is -4.2x - 3y = 122(0) - 3y = 12-3y = 12 (divide both sides by -3)y = -4When y = 0, x = 6, so that x-intercept is 6.2x - 3y = 122x - 3(0) = 122x= 12 (divide both sides by 2)x = 6
x + 3y = 2 2x - 3y = 22 Add the two equations: 3x = 24 Divide both sides by 3 to give x= 8. Substitue this value of x in the first equation to give 8 + 3y = 2 Subtract 8 from both sides: 3y = -6 Divide both sides by 3: y = -2 Solution: (X, y) = (8, -2)
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).