I think this might read "4x+5 = 4x-1". Subtracting 4x from both sides cannot upset the equality, and then you have +5 = -1, which is nonsense. One moral of this is "make sure your equation makes sense, and actually includes an = sign."
There are an indeterminate number of invisible solutions.
A quadratic equation is defined as an equation in which one or more of the terms ... In Geometry, we will concentrate on the graphical solutions to these systems. ... You can use the same table of values and simply find the y values for the straight line. ... Check (5,3) y = x2 - 4x - 2 3 = 52 - 4(5) - 2 3 = 3 check, y = x - 2 3 = 5 - 2
The equation is (x - 1)*(x - 3) = 0 or x2- 4x + 3 = 0The equation is (x - 1)*(x - 3) = 0 or x2- 4x + 3 = 0The equation is (x - 1)*(x - 3) = 0 or x2- 4x + 3 = 0The equation is (x - 1)*(x - 3) = 0 or x2- 4x + 3 = 0
y = 4x-3 is already a linear equation. The slope is 4 and the y-intercept is -3
x2 + 4x = 41
4x + 7 is an expression, not an equation. Only an equation (or an inequality) can have solutions, an expression cannot have a solution.
Use the quadratic equation formula to find the solutions to this equation.
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.
how many solutions does the equation have? 4x+1=5+2(2-4) a. one solution b. infinite solutions c. no solution
the equation has exactly 1 real solution. 8-4x=0 -> add 4x to both sides. 8=4 -> divide both sides by 4. 2=x.
using the t-table determine 3 solutions to this equation: y equals 2x
There are an indeterminate number of invisible solutions.
There are two solutions for x: x=11 and x=-7
x can be anything, if you don't have it in an equation form, can an be an infinite amount of solutions
None because without an equality sign the terms of the given expression is not an equation.
If you mean: x^2 -4x +20 = 0 then it has no real solutions because its discriminant is less than zero.
This is a quadratic equation which will have two solutions: X2 = 4x+5 Rearrange the equation: x2-4x-5 = 0 Factor the equation: (x+1)(x-5) = 0 So the solutions are: x = -1 or x = 5