Algebra 2 help (make denominators zero, and then solve equation)?
Yes, 8 would make it equal to zero. Here's what I would do to solve this equation: 1) On the left side of the equal sign, you want to make everything one fraction. To do this, multiply the +2 term by (x-8)/(x-8) so that it has the same denominator as the fraction, and then add them together. On the left side, this simplifies to (2x - 2)/(x-8). 2) Now, since we have two fractions equal to each other, you can cross multiply to solve for x. Take the numerator of the left fraction and multiply it with the denominator of the right fraction. Set this equal to the denominator of the left fraction multiplied by the numerator of the right fraction. 3) Expand, and solve for x! Hope that made sense! Good luck! (Note that, after step 1, both fraction on either side of the equal sign have the same denominator. Since this is true, you can essentially forget about the denominator and simplify to 2x-2 = 10. Solving, this would give x = 6. However, it's good practice to do it the long way, since cross-multiplication is a good skill to know!)
How do I solve an equation with constants in the denominators?
ok, for the 1st problem: so what you want to do first, is make a common denominator for the two fractions. do this by multiplying the left by 2. 2x/4=3x/4+5. move the 3x/4 to the left, and subract. 2x/4-3x/4=5 -x/4=5 cross multiply -x=20 x=-20. to check if this is right, plug it back into the equation. it works! try the 2nd prob, and email me if u cant get the rest! tUNZ
How do you solve for x when x is the denominator of a fraction?
Basically, we want to move the x out of the denominator by multiplying by the denominator. I have included some example equations that range in difficulty.Here’s an example problem:Solve for x2/x = 4We want to isolate the x, so first, we want to move it out of the denominator. We can do this by multiplying both sides by x.(2/x) * x = (4) * xThe x’s on the left side cancel out (dividing by x and multiplying are inverse operations, so we can take both out of the left side).We end up with:2 = 4xI personally like to have the x on the left side:4x = 2Now all we need to do is divide by 4 on both sides(4x) / 4 = 2 / 4Once again, we can cancel out. The two 4’s on the left side cancel each other out.We are left with:x = 2/4Or, simplified:x = 1/2To recap, we multiplied both sides by the denominator to get x out.Here’s another problem:2/x = 3/xA problem like this is impossible, because once you multiply both sides by x, x is cancelled out on both sides and2 = 3is left over. This is not possible, so the problem has no solution.Another problem:3/x = x/3(3/x) * x = (x/3) * x3 = (x/3) * (x/1)3 = (x^2)/3(3) * 3 = [(x^2)/3] * 39 = x^2square root {9} = square root {x^2}+/- 3 = xx = +/- 3Okay, last problem:-4/(x-1) = x-5(-4/(x-1)) * (x-1) = (x-5) * (x-1)-4 = (x^2) - 6x + 5(x^2) - 6x + 5 = -4 (flipped equation to put x on left side, {just my preference})([x^2] - 6x + 5) + 4 = -4 + 4 (setting equal to 0)x^2 - 6x + 9 = 0(x - 3) (x - 3) = 0x - 3 = 0x = +3Hope this helped!Thanks for reading through my answer. Here’s a little reward :)
Need help solving a rational equation?
The following rational equation has denominators that contain variables. for this equation. a. write the value or values of the variable that make a denominator zero. These are the restrictions on the variables. b. keeping the restrictions in mind, solve the equation. 2x/x+3 = 9 - 6/x+3 a. what are the value or values of the variable that makes the denominators zero? x= b. solve the equation. Select the correct choice below -A the solution set is: -B there is no solution Please need help helping how to remember this, big test coming up.
Solving a limit that can't remove the denominator zero?
evaluate lim as x approaches 2 from the right numerator is absolute value of (-x^2+6x-8) denomenator is (x^2-4x+4) So i can get the fraction to look like (abs((x-2)(-x+4))) over (x-2)(x-2) but that means the denomentaro still equals zero so whats up?? EDIT**** ok ya you can cancel out one of the sets on the denomenator with one of the top, butt hat still leaves you with one (x-2) on bottom and nothing to cancel it out with.
I need help in Alg. Its solving rational equations.?
I need help on like 5 questions ? My teacher taught me the lesson but it doesn't make sense . She taught us how to solve equations like 5/2s +3/4- 9/4s. And the answers -1/3 But i dont know hwo to do this problem . 1. y- 6/y=5 I got the same thing afterwards. Also. 2. 1/t-2=t/8 3. 2/c-2 = 2-4/c 4. 5/3p + 2/3= 5+p 5. 8/x+3 + 1/x +1 6. v+2/v +4/3v=11 Some one help me please . I been stuck on these problems forever. And explain how to do it . Thanks .
How do I solve rational equations that have exponents in the denominator?
This example has the same power in all denominators, therefore it is elementary. All you have to do is strike x² from all three denominators, PROVIDED that x≠0.Your equation becomes then: 1/6+(3x–2)/6 = 3, whence 1+3x–2 = 18, then 3x = 19 and, finally, x = 19/3, which is fine, since x turns out non-zero. It would be problematic otherwise, if e.g. the equation in question were:20/6x²+(3x-2)/6x² = 3/x²It gets decidedly more complicated if the powers of x in the denominators have different exponents.
What are the steps for solving equations with fractions?
When dealing with fractions, you need to be able to find the LCM - lowest common multiple of the denominator.e.g t+1/3 = 11/12how many 1/12s is 1/3?1/3 = 4/12so rewrite the equation:t + 4/12 = 11/12t= 11/12 - 4/12with common denominator:t= (11-4)/12t=7/12Next:2/5y = 1/8multiply both sides by 5yi.e. 5y*2/5y = 5y/85y*2/5y = 5y/8 ::::> 5y on the left cancelled2=5y/8multiply by 816=5yy=5/1618x-3(4 +6x)=-12 This actually does have no solution for x.look only at this chunk3(4 +6x)this is 3*4 + 3*6xi.e. 12+18xso the equation is 18x-(12+18x)=-12i.e. 18x -12 -18x = -12where 18x-18x=0x, there's no solution for x.