9-2x

Krissy · by **Krissy** » 15 May 2010, 20:23

How many solutions does the following system have: 6x - 3y = 9 2x - y = 3?

Explain how to determine the number of solutions without solving the system. Then apply elimination, and interpret the resulting equation.

Calvin · by **Calvin** » 15 May 2010, 20:23

6x - 3y = 9
2x - y = 3

multiply the second equation by 3
6x - 3y = 9
3(2x - y) = 3*3

6x - 3y = 9
6x - 3y = 9

The equations are the same, than the system is indeterminate, it has infinite solutions.

Bye
gilvi

Lore · by **Lore** » 15 May 2010, 20:23

9(2x+8)=20-(x+5) I am trying to solve this problem but not sure where to start solving linear equations and in?

I am trying to solve this problem, but not sure where to start solving linear equations and inequalities

Mathilde · by **Mathilde** » 15 May 2010, 20:23

First you have to simplify everything. So distribute the 9 to the 2x+8 and distribute the negative to x+5
You get: 18x+72= 20 - x - 5
Then you want to get all like terms on one side (x's on the left)
add x to each side and add like terms (20-5): 19x+72 = 15
Then subtract 72 from each side
19x=-57
Divide each side by 19
x= -57/19

Dell · by **Dell** » 15 May 2010, 20:23

What does k equal when ((kx-2)(x-2))/((7x+9)(2x+5)) so that there is an horizontal assympote at y=2?

Please help and thankyou very much

Soo · by **Soo** » 15 May 2010, 20:23

Hi,

In order to get a horizontal asymptote, the degree of the numerator and denominator must be the same. In this case, they are the same because both would have an x² term as their largest exponents.

When that is true, the value of horizontal asymptote is found by dividing the coefficients of those leading terms. The numerator would have a leading term of kx² over the denominator's leading coefficient of 14x². That means that:

kx²
------- = 2
14x²

Multiplying both sides by 14x² to eliminate the fraction, you get:

kx² = 2(14x²)
kx² = 28x²
k = 28

I hope that helps!! :-)

Debbi · by **Debbi** » 15 May 2010, 20:23

How do I simplify this expression: y=((x^(2)-9)^(2)(-2x)-(-x^(2)-9)(s(x^(2)-9)(2x)))/(x^(2)-9)^(4))?

I'm trying to simplify this equation to then figure out my second derivative.

Rolland · by **Rolland** » 15 May 2010, 20:23

If you look really carefully, in the numerator you can factor out a (x^2 - 9)^2 - there's two in the first term of the numerator and there is a (x^2 - 9) listed twice in the second term. Factoring those and cancelling some in the denominator leaves you with:

(-2x -2xs) / (x^2 - 9)^2

Dina · by **Dina** » 15 May 2010, 20:23

Perform the operations and express the answer in simplest form:((2x^2)/(x^2+3x+9))/((2x^3-18x)/(x^3-27)) ? Perform the operations and express the answer in simplest form:((2x^2)/(x^2+3x+9))/((2x^3-18x)/(x^3-27)) ?

Lashonda · by **Lashonda** » 15 May 2010, 20:23

(2x²/(x² + 3x + 9)) / ((2x³ - 18x)/(x³ - 27))
= 2x²(x³ - 27) / [(x² + 3x + 9)(2x³ - 18x)]
= 2x²(x - 3)(x² + 3x + 9) / [2x(x² + 3x + 9)(x - 3)(x + 3)]
= x/(x + 3)

9-2x

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