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.
None because without an equality sign the given terms can't be considered to be an equation
8x - 9 = 9x - 5 8x - 8x - 9 = 9x - 8x - 5 -9 = x - 5 -9 + 5 = x - 5 + 5 -4 = x
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