moalnamer
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PLEASE HELP


If a = 1, find the values of b, c, and d that make the given expression equivalent to the expression below.

PLEASE HELP If a 1 find the values of b c and d that make the given expression equivalent to the expression below class=

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susanwiederspan
To do this, you need to multiply out the expressions. This is a bit tedious, but remember like FOIL for binomials, for these trinomials you must multiply each term. If you need a step-by-step, I'd be happy to provide it. Let me know.

Once you have simplified the expression, you get
[tex] \dfrac{-x-9}{2x-4} [/tex].

But, the problem stipulates that a must equal 1. We can equivalently factor out the negative sign and put it on the denominator with no change to write 
[tex] \dfrac{x+9}{-(2x-4)} = \dfrac{x+9}{-2x+4}[/tex].

So, seeing where each coefficient corresponds between the two expressions, you get a = 1, b = 9, c = –2, and d = 4.
MrRoyal

Equivalent expressions are expressions with the same value.

The values of the variables are:

[tex]\mathbf{a = 1}[/tex]      [tex]\mathbf{b = 9}[/tex]      [tex]\mathbf{c = -2}[/tex]       [tex]\mathbf{d = 4}[/tex]

The expression is given as:

[tex]\mathbf{\frac{3x^2 + 2x - 21}{-2x^2 -2x + 12} \cdot \frac{2x^2 + 25x + 63}{6x^2 + 7x - 49}}[/tex]

Expand

[tex]\mathbf{\frac{3x^2 + 2x - 21}{-2x^2 -2x + 12} \cdot \frac{2x^2 + 25x + 63}{6x^2 + 7x - 49 } = \frac{3x^2 + 9x - 7x - 21}{-2x^2 +4x -6x + 12} \cdot \frac{2x^2 + 14x + 9x + 63}{6x^2 + 21x - 14x - 49 } }[/tex]

Factorize

[tex]\mathbf{\frac{3x^2 + 2x - 21}{-2x^2 -2x + 12} \cdot \frac{2x^2 + 25x + 63}{6x^2 + 7x - 49 } = \frac{3x(x + 3) - 7(x + 3)}{-2x(x -2) -6(x - 2)} \cdot \frac{2x(x + 7) + 9(x + 7)}{3x(2x + 7) - 7(2x - 7) } }[/tex]

Factor out the terms

[tex]\mathbf{\frac{3x^2 + 2x - 21}{-2x^2 -2x + 12} \cdot \frac{2x^2 + 25x + 63}{6x^2 + 7x - 49 } = \frac{(3x - 7) (x + 3)}{(-2x -6)(x - 2)} \cdot \frac{(2x + 9)(x + 7)}{(3x - 7) (2x - 7) } }[/tex]

Cancel out 3x - 7

[tex]\mathbf{\frac{3x^2 + 2x - 21}{-2x^2 -2x + 12} \cdot \frac{2x^2 + 25x + 63}{6x^2 + 7x - 49 } = \frac{(x + 3)}{(-2x -6)(x - 2)} \cdot \frac{(2x + 9)(x + 7)}{ (2x - 7) } }[/tex]

Factor out -2

[tex]\mathbf{\frac{3x^2 + 2x - 21}{-2x^2 -2x + 12} \cdot \frac{2x^2 + 25x + 63}{6x^2 + 7x - 49 } = \frac{(x + 3)}{-2(x +3)(x - 2)} \cdot \frac{(2x + 9)(x + 7)}{ (2x - 7) } }[/tex]

Cancel out x + 3

[tex]\mathbf{\frac{3x^2 + 2x - 21}{-2x^2 -2x + 12} \cdot \frac{2x^2 + 25x + 63}{6x^2 + 7x - 49 } = \frac{1}{-2(x - 2)} \cdot \frac{(2x + 9)(x + 7)}{ (2x - 7) }}[/tex]

Rewrite as:

[tex]\mathbf{\frac{3x^2 + 2x - 21}{-2x^2 -2x + 12} \cdot \frac{2x^2 + 25x + 63}{6x^2 + 7x - 49 } = \frac{(2x + 9)(x + 7)}{ -2(x - 2)(2x - 7) } }[/tex]

Expand

[tex]\mathbf{\frac{3x^2 + 2x - 21}{-2x^2 -2x + 12} \cdot \frac{2x^2 + 25x + 63}{6x^2 + 7x - 49 } = \frac{2x^2 + 25x + 63}{ -4x^2 + 22x - 28}}[/tex]

Factorize again

[tex]\mathbf{\frac{3x^2 + 2x - 21}{-2x^2 -2x + 12} \cdot \frac{2x^2 + 25x + 63}{6x^2 + 7x - 49 } = \frac{(2x+ 7)(x + 9)}{(2x + 7)(-2x + 4)}}[/tex]

Cancel out common factors

[tex]\mathbf{\frac{3x^2 + 2x - 21}{-2x^2 -2x + 12} \cdot \frac{2x^2 + 25x + 63}{6x^2 + 7x - 49 } = \frac{x + 9}{-2x + 4}}[/tex]

From the question, we have:

[tex]\mathbf{\frac{ax + b}{cx + d}}[/tex]

So, we have:

[tex]\mathbf{\frac{ax + b}{cx + d} = \frac{x + 9}{-2x + 4}}[/tex]

By comparison, we have:

[tex]\mathbf{a = 1}[/tex]

[tex]\mathbf{b = 9}[/tex]

[tex]\mathbf{c = -2}[/tex]

[tex]\mathbf{d = 4}[/tex]

Read more about equivalent expressions at:

https://brainly.com/question/24242989

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