x = 0 and y = 4
5y + x = 0 subtract 5y from each side, then x = -5y
If: 4x = 5y Then: x = 5y/4
x - 5y = -4 x + 8y = -4 The first equation gives x = 5y - 4 SUBSTITUTE this value for x in the second equation: (5y - 4) + 8y = -4 Simplify: 13y = 0 so that y = 0 Then x = 5y - 4 gives x = -4
0 = 5y + 20 - xso 5y = x - 20or y = (1/5)x - 40 = 5y + 20 - xso 5y = x - 20or y = (1/5)x - 40 = 5y + 20 - xso 5y = x - 20or y = (1/5)x - 40 = 5y + 20 - xso 5y = x - 20or y = (1/5)x - 4
If you mean x+5y = 4 then (9, -1) is the solution
x = 0 and y = 4
(2, 3) is a solution only if it satisfies the equation. x + 5y = 4 (2) + 5(3) =? 4 2 + 15 =? 4 17 = 4 False Therefore, (2, 3) is not a solution to the given equation.
yes
yes it is..your welcome
5y + x = 0 subtract 5y from each side, then x = -5y
If: 4x = 5y Then: x = 5y/4
x - 5y = -4 x + 8y = -4 The first equation gives x = 5y - 4 SUBSTITUTE this value for x in the second equation: (5y - 4) + 8y = -4 Simplify: 13y = 0 so that y = 0 Then x = 5y - 4 gives x = -4
x = (-5y + 4) / 7
0 = 5y + 20 - xso 5y = x - 20or y = (1/5)x - 40 = 5y + 20 - xso 5y = x - 20or y = (1/5)x - 40 = 5y + 20 - xso 5y = x - 20or y = (1/5)x - 40 = 5y + 20 - xso 5y = x - 20or y = (1/5)x - 4
Without the appropriate mathematical signs the given terms have no solutions.
3x + 5y = -1 2x - 5y = 16 Add the two equations: 5x = 15 or x = 3 Substitute for x in the first equation: 3*3 + 5y = -1 15 + 5y = -1 5y = -10 so y = -10/5 = -2 The solution is (x, y) = (3, -2)