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3 years ago

1x+11y=55 substitution

Mathematics

1 answer:

STatiana [176]3 years ago

4 0

Answer:

y = -1/11 + 5

Step-by-step explanation:

1x + 11y = 55

-1x -1x

11y = -1x + 55

/11 /11

y = -1/11 + 5

Hope this helps!

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4 years ago

What is the difference?

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Answer:

The option $\frac{(x+5)(x+2)}{x^3-9x}$ is correct

The difference of the given expression is

$\frac{2x+5}{x^2-3x}-(\frac{3x+5}{x^3-9x})-({\frac{x+1}{x^2-9})=\frac{(x+5)(x+2)}{x^3-9x}$

Step-by-step explanation:

Given expression is $\frac{2x+5}{x^2-3x}-(\frac{3x+5}{x^3-9x})-({\frac{x+1}{x^2-9})$

To find the difference of the given expression as below :

$\frac{2x+5}{x^2-3x}-(\frac{3x+5}{x^3-9x})-({\frac{x+1}{x^2-9})$

$=\frac{2x+5}{x(x-3)}-(\frac{3x+5}{x(x^2-9)})-({\frac{x+1}{x^2-9})$

$=\frac{2x+5}{x(x-3)}-(\frac{3x+5}{x(x^2-3^2)})-({\frac{x+1}{x^2-3^2})$

$=\frac{2x+5}{x(x-3)}-(\frac{3x+5}{x(x-3)(x+3)})-({\frac{x+1}{(x-3)(x+3)})$

( using the formula $a^2-b^2=(a+b)(a-b)$ )

$=\frac{2x+5(x+3)-(3x+5)-x(x+1)}{x(x-3)(x+3)}$

$=\frac{2x^2+6x+5x+15-3x-5-x^2-x}{x(x-3)(x+3)}$ (adding the like terms)

$=\frac{x^2+7x+10}{x(x^2-9)}$ ( by factoring the quadratic polynomial )

$=\frac{(x+5)(x+2)}{x^3-9x}$

Therefore $\frac{2x+5}{x^2-3x}-(\frac{3x+5}{x^3-9x})-({\frac{x+1}{x^2-9})=\frac{(x+5)(x+2)}{x^3-9x}$

Therefore the difference of the given expression is

$\frac{(x+5)(x+2)}{x^3-9x}$

Therefore option $\frac{(x+5)(x+2)}{x^3-9x}$ is correct

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