Solve the absolute value equation. Check Solutions.?
If p=5 the left side is 0 but the right side is 10, so 5 is not a solution. If p=-15 the right side is negative, but the left side is positive, so 15 is not a solution either. Here is how I would do it: case 1: if p-5 > 0 then |p-5|= p-5, so the equation is: 3(p-5) = 2p, so multiply out and get 3p - 15 = 2p, subtract 2p from both sides and add 15 to both sides: p = 15, and this satisfies the case hypothesis that p-5 > 0, so p = 15 is a solution. case 2: if p-5<0 the |p-5| = -(p-5), so the equation is: -3(p-5) = 2p, multiply out and get: -3p + 15 = 2p, add 3p to both sides 15 = 5p, divide by 5 3 = p. Since 3-5 is less than zero this satisfies the case hypothesis, so p=3 is a solution.
Solve the absolute value equation or indicate that the equation has no solution?
| 4x - 6 | -7 = -12 1) | 4x - 6 | = 4x -6 when x >6/4 4x -6 -7 = -12 4x = -12 +13 =1 x =1/4 , this solution is not in the domain, therefore is rejected. 2) | 4x - 6 | = -4x +6 when x < 6/4 -4x +6 -7 = -12 -4x = -12 +1 = -11 x = 11/4 ( not in the domain), solution is rejected Summary: No Solutions exist for this equation
Solve the absolute value equation or indicate the equation has no solution.?
Simplifying x2 + 6x + 1 = 8 Reorder the terms: 1 + 6x + x2 = 8 Solving 1 + 6x + x2 = 8 Solving for variable 'x'. Reorder the terms: 1 + -8 + 6x + x2 = 8 + -8 Combine like terms: 1 + -8 = -7 -7 + 6x + x2 = 8 + -8 Combine like terms: 8 + -8 = 0 -7 + 6x + x2 = 0 Factor a trinomial. (-7 + -1x)(1 + -1x) = 0 Subproblem 1Set the factor '(-7 + -1x)' equal to zero and attempt to solve: Simplifying -7 + -1x = 0 Solving -7 + -1x = 0 Move all terms containing x to the left, all other terms to the right. Add '7' to each side of the equation. -7 + 7 + -1x = 0 + 7 Combine like terms: -7 + 7 = 0 0 + -1x = 0 + 7 -1x = 0 + 7 Combine like terms: 0 + 7 = 7 -1x = 7 Divide each side by '-1'. x = -7 Simplifying x = -7 Subproblem 2Set the factor '(1 + -1x)' equal to zero and attempt to solve: Simplifying 1 + -1x = 0 Solving 1 + -1x = 0 Move all terms containing x to the left, all other terms to the right. Add '-1' to each side of the equation. 1 + -1 + -1x = 0 + -1 Combine like terms: 1 + -1 = 0 0 + -1x = 0 + -1 -1x = 0 + -1 Combine like terms: 0 + -1 = -1 -1x = -1 Divide each side by '-1'. x = 1 Simplifying x = 1Solutionx = {-7, 1}
Solve the absolute value equation or indicate that the equation has no solution:?
| 4x - 6 | -7 = -12 a million) | 4x - 6 | = 4x -6 while x >6/4 4x -6 -7 = -12 4x = -12 +13 =a million x =a million/4 , this answer isn't interior the area, as a result's rejected. 2) | 4x - 6 | = -4x +6 while x < 6/4 -4x +6 -7 = -12 -4x = -12 +a million = -11 x = 11/4 ( no longer interior the area), answer is rejected precis: No strategies exist for this equation
Solving Absolute Value Equations?
isolate the absolute 27. 5 |q + 6| =20 |q+6|=4 q+6=+/-20 30. 8 |5w - 1 |= 0 0/8=0 |5w-1|=0 5w-1=+/-0 LOL 5w=1 31. 2 |3x - 4| + 8=6 2|3x-4|= -2 |3x-4|= -1 3x-4= +/-1 34. -5|3z + 8| - 5= -20 -5|3z+8|= -15 |3z+8|=3 3z+8=+/-3 42. -3y-2 = |6y + 25| |6y+25|=+/-(-3y-2) 6y+25= -3y-2 6y+25= 3y+2 9y= -27 3y= -23
Solve the absolute value equation?
5x+3=2x+3 or 5x+3=-(2x+3) 5x+3=2x-5x+3 5x+3=-2x-3, -2x-5x=3+3 -5x -7x=6 x= -6/7 3=-3x+3 3-3=3x 0/3 x=0, -6/7
Solve the equation for the indicated value?
Given a - 2[b - 3(c - x)] = 6; solve for x 1) Clear the parentheses by distributing the - 3 a - 2 [ b - 3c + 3x ] = 6 2) Clear the brackets by distributing the - 2 a - 2b + 6c - 6x = 6 3) Subtract 6 from both sides a - 2b + 6c - 6 - 6x = 0 4) Add 6x to both sides a - 2b + 6c - 6 = 6x 5) Divide both sides by 6 ( a - 2b + 6c - 6 ) / 6 = x
Absolute Value Equation question?
Remember that the absolute value of anything will either be 0 or a positive number. The absolute value could never be a negative number. That means that the absolute value of anything could never be -12. I agree with your statement: |x+2| = -12 But this says "the absolute value of something is a NEGATIVE 12"...Can't be. Therefore, the equation you gave has no solution.
How do you know when an absolute value equation has no solution?
Well, one sufficient condition is when the absolute value is, apparently, equal to a negative value. This contradicts the definition of absolute value (which is always >= 0). Also, another sufficient condition is when the sets end up mismatching, eg you have an absolute Real, but the other side of the equality is not a subset of Reals (eg it's Complex with a non-zero complex component (aka imaginary part)), or in general if the absolute value ends up being equated to anything which is not a subset of it's parent set
I really like this question. It’s one I wish more of my students would ask, and one which I wish more differential equations classes carefully addressed.The answer is that you can’t ignore it, but if you do make a mistake and ignore it, and then make a pair of subsequent mistakes, then you’ll end up with the correct general solution because your later mistakes serendipitously correct your first. Basically, you get lucky that you eventually get the right answer because your solution is fundamentally not correct.Here’s the correct solution using separability. (I’ll skip the change of variables that takes place behind the scenes, but if you don’t know what I’m referring to, you should ask in the comments.)[math]y'=y\ \implies\ \int \frac {dy}y = \int dt\ \implies \log|y| = t+c[/math]Now, we need to be careful and point out that [math]c[/math] can be any real number.Next, we exponentiate both sides of the equation (which is legal because the exponential function is invertible) to get:[math]|y|=e^{t+c}= e^c\cdot e^t= ke^t[/math]Since [math]c[/math] could be an arbitrary real number, [math]e^c=k[/math] can be an arbitrary POSITIVE number. (Recall that the image of the set of reals under the exponential function is the set of positive reals.)Then[math]y=\pm ke^t[/math]But since [math]k[/math] can be an arbitrary positive number, [math]\pm k[/math] can take on any non-zero value. So we can write[math]y=\hat k e^t[/math]with [math]\hat k\ne 0[/math].But as happens often, this method of solving separable equations fails to find the equilibrium solution which, in this case, is [math]y=0[/math]. (Equilibrium solutions are missed because we divide both sides of the original separable equation [math]y'=f(y)g(t)[/math] by [math]f(y)[/math] to get the equation into a form we can integrate. But when [math]f(y)=0[/math] this amounts to division by zero which is not allowed. And notice that when [math]f(y)=0[/math], we have an equilibrium solution.)So when we combine the equilibrium solution with the solution we previously found, we get the general solution. Either [math]y=0[/math] or [math]y=\hat k e^t[/math] with [math]\hat k\ne 0[/math]. But we can combine these parts to just write[math]y=\tilde k e^t[/math]with [math]\tilde k\in\mathbb R[/math].And here, you see the same solution that you would have gotten if you ignored the absolute value, ignored the limitations on the various constants, and ignored the equilibrium solution.