2/x+3 - 3/4-x = 2x-2 / x^2-x-12?
[05] 2/(x+3) -3/(4-x)=(2x-2)/(x^2-x-12) 2/(x+3) -3/-(x-4)=2(x-1)/(x-4)(x+3) 2/(x+3) +3/(x-4)=2(x-1)/(x-4)(x+3) {2(x-4)+3(x+3)}/(x+3)(x-4) =2(x-1)/(x-4)(x+3) (2x-8+3x+9)/(x+3)(x-4)=2(x-1)/(x+3)(x-... (5x+1)=2(x-1)[cancelling (x+3)(x-40 from the denominators of both sides] 5x+1=2x-2 5x-2x= -2-1 3x= -3 x= -1
What is the solution set of 2(x+3) <5x-6?
First, remember that in order to solve this inequality we need to isolate x on one side of the equation. Then, when we prove it, we solve and prove it the same way we do an equation: after we find the value of x we prove it by substituting for x. Both sides of the inequality will be equal to each other.2(x+3) < 5x-6Distribute the 2:2x+6 < 5x-6Subtract 2x from both sides:6 < 3x-6Add 6 to both sides:12 < 3xDivide by 3:4 < xNow substitute for x in the original inequality:2(x+3) < 5x-6x > 42(4+3) < 5*4–62(7) < 20–614 < 14So x > 4.
What is the next term in the arithmetic sequence x+1, 2x-3, x+5?
In an arithmetic sequence, difference between successive terms is the same:(2x-3)-(x+1) = (x+5)-(2x-3) x - 4 = -x + 8 2x = 12 x = 6x + 1 = 7 2x - 3 = 9 x + 5 = 11Next term = 13If you want to write this in terms of x, then we notice a pattern in 1st and 3rd terms: x+1, x+5. These differ by 4. So then 2nd and 4th terms should also differ by 4: 2x-3, 2x+1Next term: 2x+1Sequence: x+1, 2x-3, x+5, 2x+1, x+9, 2x+5, x+13, 2x+9, …
How can one solve the following by using the method of factorization: [math]\frac{4}{x-3}=\frac{5}{2x+3}[/math]where [math]x[/math] is not equal to [math]0[/math] or [math]-\frac{3}{2}?[/math]
4/x-3=5/2x+3Multiply both sides by (x-3)(2x+3)4(2x+3)=5(x-3)=8x+12=5x-15Subtract 12 from both sides8x+12 -12 =5x-15 -12= 8x=5x-27Subtract 5x from both sides.8x -5x =5x-27 -5x= 3x=-27x=-9
If x=2+2*2/3+2*1/3 what is the value of x^3-6x^2+6x?
Given that x = 2 + 2^(2/3) + 2^(1/3)so x - 2 = 2^(2/3) + 2^(1/3) =(x-2)^3 = [2^(2/3)+2^(1/3)]^3use (a-b)^3= a^3 - b^3 - 3ab(a-b) and (a+b)^3= a^3 + b^3 + 3ab(a+b) formulae.or x^3 - 8 - 6x(x-2) = 2^2 + 2^1 + 3*[2^{(2/3)+(1/3)}[2^(2/3)+2^(1/3)or x^3 - 8 - 6x^2 + 12x = 4 + 2 + 6(x-2)or x^3 - 8 -6x^2 + 12x = 6 + 6x - 12or x^3 - 6x^2 +6x = 2
What should be the value of x so that x+2, 3x-2, and 7x-12, will form an arithmetic sequence? Justify the answer.
The terms x+2 (T1), 3x-2 (T2) and 7x-12 (T3) are in AP. To justify the answer, the common difference between T1 and T2 and between T2 and T3 must be the same. Hence (3x-2) -(x+2) = (7x-12)-(3x-2),3x-2 + 3x-2 = 7x-12 + x+2, or6x-4 = 8x- 10, or8x-6x = 10–4, or2x = 6, orx = 3.To justify whether x =3,T1: x+2 = 3+2 = 5T2: 3x-2 = 3*3–2 = 9–2 = 7T3: 7x-12 = 21–12 = 9.Thus T1, T2 and T3 are 5, 7 and 9 which is an AP. Hence x = 3 is correct.
If x+1, 3x, and 4x+2 are in A.P., then what is x?
since this is an arithmetic sequence then:3x - (x+1) = (4x + 2) - 3x3x - x - 1 = 4x + 2 - 3x2x -1 = x + 22x - x = 2 + 1x = 3x + 1, 3x, 4x + 24, 9, 14
How do I solve x-1/x-2 + x-3/x-4=10/3?
(x-1)/(x-2) + (x-3)/(x-4) =10/3or (x-2+1)/(x-2) + (x-4+1)/(x-4) = 10/3or 1 + 1/(x-2) + 1 + 1/(x-4) =10/3or 1/(x-2) + 1/(x-4) =10/3 - 2or (x-4+x-2)/(x-2)(x-4) =4/3or 2(x-3)/(x^2–6x+8) =4/3or (x-3) /(x^2–6x+8) =2/3or 2x^2-12x+16 = 3x-9or 2x^2 -15x +25=0or 2x^2–10x-5x +25 =0or 2x(x-5)-5(x-5)=0or (x-5)(2x-5)=0x = 5 , 5/2 . Answer., Answer.