Can you explain how to solve 4(y+1)=2(y-2)

24n = 120. What is n?

25c^2+10c+1

2x+5=x^2

Solve (D2-3D+2)y =x(x+4) and show that its general solution is given by y=Aex + Be2x +(x2/2) + (7x/2) + (19/4)

My work :

auxiliary eqn , m2 - 3m + 2= 0 m=1 , m=2

Reduced eqn : yn = Aex + Be2x

P.I. = 1/f(D)F(x) = (1/(D2-3D+2))(x(x+4))

P.I. = [(x2+4x) + ( (1/2)( 2- 3(2x) - 3(4) ) ) + (1/4)(92) )

P.I. = x2 + (4x/2) - (6x/4) - (5/2) + (18/4)

P.I. =(x2/2) + x + 8/4

General solution : y= Aex + Be2x + (x2/2) + x + 8/4

Could anyone point out my mistakes?

differentiate f(t)=2-(2/3)t

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find x = 2(3x×x)+(2x×)=14400

9x+5=7x+11

f(x)=3x^4+9e^x^6

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Find what y will be when x=40 degrees 2) Sketch the curve for one positive and one negative cycle and then on your graph determine the x value when y is 9

dimmsdale discount golf supply sells titleist golf balls for 60c each, the nike golf balls for 94c each and the Calloway golf balls for 140.50, how many golf balls of each kind were bought?

-4=8x+y

5x+3/4x-7≥2, can anyone solve this inequality and show the step by step method of doing it, Many thanks!!

Solve:

3x(x+4)+3(x+4)=0

what is 1cmx1cmx1cm im mm2?

3x⁴-2x³-5x²-4/x²-4

HOW CAN I SOVLE QUADRATIC SIMTENEOUS EQUATION

how do i figure out x2+7x+10=0

please can you give me a step by step guide

2(3x+8)=-23+9

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3x+y=7 2x+y=6

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Eg. 4x2 + 3x - 8?

like i dont get how to do this problem:3y-4=2

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f(x)=-3(x+5)(x-11)

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Solve the simultaneous equations: x2 +2 y2 = 8 and x + 2y =8

I need help with this please help?

N+N=30 what is the number for N

nth tems

2x - 3y = 21 5x + 2y = 5

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How would you solve x^ + y^ = 25 y - 3x = 13

x2 + 2xy = 5 x + y = 3

Hi, I need a bit of help on how to solve the following. y = -4x + 4 y = x2 -6x - 8 I have tried to substitute equation one into equation two ( RHS must be equal) to give -4x + 4 = x2 -6x -8 I have then added all like terms to give but im then stuck and I think i have got the method wrong. Bit confused due to x2 and y = as I thought i did understand how to solve these so im now annoyed. x2 -10x -4 Any help on how to solve this would be helpful. Thanks

I really need help with topic. I don't get it at ALL. this is one of the questions used in class:- y=2x-1. y=x2 (Both x and y equal to 1 I just don't understand how or why) please explain

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how do i solve x for (2/x+3)=9/4=5 and none is in parenthesis.

(X-2) over 3 + 1 = x over 7

3x-2=-29 6+1/2x=4 X/10+4=5 5x+3+x=75

4(8-x)=40

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3x+2 -------- + 3 = 4 thank you ,:) x-1 the ---- is meant to be divided or over (:

7-(x+1)=9-(2x-1)

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x + y = 8 x - y = 2

equation: 23-4x=-9 ——- 2

23-4x=-18 is the next step then -4x=18-23 -4x=-18-23 -4x=-41 can i get a step by step answer on how to conver this into x=41/4

please can i have help im really stuck , thanks.

find the value of x

9x+5=7x+11

93-3x=9x+21 what is x ?

Hey! Could anyone help me with this question? It would be much appreciated, thank you!

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X+ (1/(x+5)^2) how do I show this is one to one and onto?

solve using elimination

expand and simplify the expression

x²+8x-9

http://tinyurl.com/7dvcbtb

I looked at the steps but I don't understand how that chain rule worked.

Shouldn't f(u) = usin (x) ?

Solving Quadratic Simultaneous Equations

Try to figure out how to solve this equation: x2 + y2 = 29 y - x = 3  

How to Find the Answer

Lets begin by rewriting our two equations, x²+y²=29…..iy-x=3……….ii Next we take our equation ii and equate y in terms of x as follows, y-x=3y=x+3… Now we substitute y term in equation i:. x²+y²=29 where there’s the value of y we replace it with x+3 x²+(x+3) (x+3) =29 we rewrite the equation in quadratic formatx²+x²+3x+3x+9=292x²+6x+9=29 2x²+6x+9-29=02x²+6x-20=0……now we have formed our quadratic equation. To solve the equation we may use the factorization method or completing square method. So in the case here,  I prefer to use the completing square method because our equation doesn’t have perfect squares. Which is [- b±(b²_4ab)¹/²]/2a Next, we substitute the coefficients of x in our equation above, where a=2, b=6, and c=-20[-(6)±(6²-{4×2×-20})¹/²]/2×2{-6±(36+160)¹/²}/4 {-6±(196)¹/²}/4 where the root of 196=14(-6±14)/4x=(-6+14)/4=2x=(-6-14)/4=-5 Now we have the possible values of x 2&-5, then we substitute these values on our y equation y=x+3y=2+3y=5y=-5+3y=-2 Therefore our values for both x and y will be x=2 or – 5y=5 or – 2 This is because our simultaneous equation had a quadratic equation.  

  (X-2) over 3 + 1 = x over 7