-2
There are none because without an equality sign it can not be considered to be an equation and the + or - values of 3x and 10 are not given
An algebraic equation with an infinite number of solutions
Answer: no solution3x - 2 = 3x -53x - 3x - 2 = 3x - 3x - 5 (Subtract 3x from each side.)-2 = -5 Huh??When you are solving an equation, and the variable disappears, there are two possibilities:If you are left with a true statement (such as 2 = 2), then the equation is an identity, and any value of x will satisfy the equation. The number of solutions is infinite.If you are left with a false statement, as in this case, no value of x will satisfy the equation. There is no solution.
One.
None because without an equal sign it's not an equation
That's not an equation - it doesn't have an equal sign. Assuming you mean 2x2 - 3x - 90 = 0, you can find the solution, or usually the two solutions, of such equations with the quadratic formula. In this case, replace a = 2, b = -3, c = -90.
There are none because without an equality sign it can not be considered to be an equation and the + or - values of 3x and 10 are not given
If you mean: 3x^2+12x+12 = 0 then it has two equal solutions of -2
An algebraic equation with an infinite number of solutions
How many solutions are there to the equation below? 3x-10(x+2) = 13-7x 0
Answer: no solution3x - 2 = 3x -53x - 3x - 2 = 3x - 3x - 5 (Subtract 3x from each side.)-2 = -5 Huh??When you are solving an equation, and the variable disappears, there are two possibilities:If you are left with a true statement (such as 2 = 2), then the equation is an identity, and any value of x will satisfy the equation. The number of solutions is infinite.If you are left with a false statement, as in this case, no value of x will satisfy the equation. There is no solution.
There are two terms: 3x, -2b. Yeah, two terms. But where is the equation?
One.
None because without an equal sign it's not an equation
The answer has been right under your nose all this time.The equation of y equals -3x isy = -3x
If: 3x+2y = 5x+2y = 14 Then: 3x+2y = 14 and 5x+2y =14 Subtract the 1st equation from the 2nd equation: 2x = 0 Therefore by substitution the solutions are: x = 0 and y = 7
There are no real solutions because the discriminant of the quadratic equation is less than zero.