There is no equation (or inequality) in the question and so there cannot be any solutions.
5x-3=2-8x 5x-3+3=2-8x+3 5x=5-8x 5x+8x=5-8x+8x 13x=5 x=5/13
7+8x-5 = 8x+7-5x Subtract 7 from both sides: 8x-5 = 8x-5x Subtract 8x from both sides: -5 = -5x Divide both sides by -5: x = 1 So the equation implies x = 1
28=8x+12-7 28=8x+5 23=8x x=23/8
8x + 20 = 4 (2x + 5)
8x^2 + 6x - 5 = 8x^2 + 10x - 4x - 5 = 2x * (4x + 5) - (4x + 5) = (2x -1) * (4x + 5)
x2 - 8x + 15
5x2 + 8x = 7 5x2 + 8x - 7 = 0 This cannot be factorised so the solutions need to be determined using the quadratic formula The solutions are {-8 ± sqrt[82 - 4*5*(-7)]}/(2*5) = {-8 ± sqrt[64 + 140]}/10 = {-8 ± sqrt[204]}/10 = -2.22829 and 0.62829 (to 5 dp)
Assuming you mean:(4x+36)(8x-40) = 0 then you can use the property that a product can only be zero if one of its factors is zero. In other words, you can change this to: 4x + 36 = 0 OR 8x - 40 = 0 Solve each of the individual solutions; their solutions are also solutions to the original equation.
x = 5 or x = 6
Each of the two equations has an infinite number. Simultaneously, they have only one.
8x - 9 = 9x - 5 8x - 8x - 9 = 9x - 8x - 5 -9 = x - 5 -9 + 5 = x - 5 + 5 -4 = x
None because without an equality sign the given terms can't be considered to be an equation
One.
5x-3=2-8x 5x-3+3=2-8x+3 5x=5-8x 5x+8x=5-8x+8x 13x=5 x=5/13
8x plus 4y equals 5 is 8x + 4y = 5.
7+8x-5 = 8x+7-5x Subtract 7 from both sides: 8x-5 = 8x-5x Subtract 8x from both sides: -5 = -5x Divide both sides by -5: x = 1 So the equation implies x = 1
5x + 5 - 8x - 8x + 4 = 9 - 11x