I understand this to be 4x + 3y - 5x + 2y. -x + 5y
3x-5y=7 5x-2y=-1
2x + 5y = 16 + (-5x) - 2y = 2 That gives 2x + 5y = 2 . . . . . . . . . . (I) and -5x - 2y = - 14 or 5x + 2y = 14 . . . . (II) (I)*5: 10x + 25y = 10 (II)*2: 10x + 4y = 28 Subtracting the second from the first, 21y = -18 so y = -18/21 = -6/7 and then by (I) x = 1 - 5y/2 = 1 - 5/2*(-6/7) = 1 + 15/7 = 22/7 So the ordered pair is (22/7, -6/7)
8840-026
5(x-y) is also equivalent to 5x-5y
5y explanaition 5x-5x=0, 0-2y=-2y, -2y+7y=5y
I understand this to be 4x + 3y - 5x + 2y. -x + 5y
25xy is an equivalent expression.
2X + 5Y = 6 5Y = - 2X + 6 Y = - 2/5X + 6/5 ----------------------------------and 5X - 2Y = 8 2Y = - 5X + 8 Y = - 5/2 + 4 ------------------- Look like intersecting lines, though not perpendicular
0
It is: 6x-5y simplified
5x - 2y = 20Solve for x:5x = 2y + 20x = 2/5y + 4Solve for y:-2y = -5x + 20y = 5/2x - 10Neither equation can be simplified further without one of the variables already being known.
3x-5y=7 5x-2y=-1
2x + 5y = 16 + (-5x) - 2y = 2 That gives 2x + 5y = 2 . . . . . . . . . . (I) and -5x - 2y = - 14 or 5x + 2y = 14 . . . . (II) (I)*5: 10x + 25y = 10 (II)*2: 10x + 4y = 28 Subtracting the second from the first, 21y = -18 so y = -18/21 = -6/7 and then by (I) x = 1 - 5y/2 = 1 - 5/2*(-6/7) = 1 + 15/7 = 22/7 So the ordered pair is (22/7, -6/7)
Yes
If the slopes of the lines are negative reciprocals, then the lines are perpendicular.2x - 5y = -3 subtract 2x to both sides-5y = -2x - 3 divide both sides by -5y = (2/5)x +3/55x + 2y = 6 subtract 5x to both sides2y = -5x + 6 divide both sides by 2y = (-5/2)x + 3Yes, the lines are perpendicular.
This is a system of equations:-x + 5y = 10-5x + 2y = 27Multiply the first equation by 5.-5x + 25y = 50Subtract the 2 equations:23y = 23Solve:y = 1-x + 5(1) = 10-x = 5x = -5Thus, the ordered pair is (-5, 1).