Add the two equations together. This will give you a single equation in one variable. Solve this - it should give you two solutions. Then replace the corresponding variable for each of the solutions in any of the original equations.
3x - 12 = 10x - 15 - 6x -10 3x - 12 = 4x -25 3x + 13 = 4x 13 = 4x - 3x 13 = x None?
so, if 2 minus Ln times 3 minus x equals 0, then 2 minus Ln times 3 equals x, therefore 2 minus Ln equals x divided by three, so Ln + X/3 = 2 therefore, (Ln + [X/3]) = 1
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Dude, stop trying to cheat on your own maths enrichment task loser.
The equation x2 - 5x - 24 = 0 has the two solutions 8 and -3 for x.
The solutions are: x = -2 and y = 4
The solutions are: x = 4, y = 2 and x = -4, y = -2
{-1,-2}
16
You can write this as two equations, and solve them separately. The two equations are:x - 19 = -3and:-x - 19 = -3
This question has an infinite number of solutions !
+ (plus) - (minus) / (divide) * (multiply) ^ (power) = (equals)
(x, y) = (√19/2, 9), (√19/2, -9), (-√19/2, 9), (-√19/2, -9).
The solutions to the quadratic equation are: x = -1 and x = 6
x = 4 and y = 7 which will satisfy both equations
The equation has infinitely many solutions.
There's an equals missing...