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3 years ago

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WILL GIVE BRAINLEIST PLEASE HURRY DUE AT 9 PM SUPER EASY The algebraic expression shown below is missing two whole-number consta

nts. Determine the constants so that the expression simplifies to 14x+11.
4x+8(x+_)+_+2x

1 answer:

PilotLPTM [1.2K]3 years ago

5 0

Answer:1 and 3

Step-by-step explanation: 4x+8(x+1)+3+2x

4x+8x+8+3+2x=14x+11.

multiply and add the like terms. you get 14x+11.

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$\displaystyle \frac{4q^2-q+3}{q^2+6q+5}-\frac{3q^2-q-6}{q^2+6q+5}=\frac{q^2+9}{q^2+6q+5}$

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<u>Simplifying Rational Expressions</u>

If two or more rational expressions have the same denominator, the add and subtract operations are done only with the numerator. The final denominator will be the common of both.

The expression is:

$\displaystyle \frac{4q^2-q+3}{q^2+6q+5}-\frac{3q^2-q-6}{q^2+6q+5}$

Operating on the numerators:

$\displaystyle \frac{4q^2-q+3}{q^2+6q+5}-\frac{3q^2-q-6}{q^2+6q+5}=\frac{4q^2-q+3-(3q^2-q-6)}{q^2+6q+5}$

Removing parentheses:

$\displaystyle \frac{4q^2-q+3}{q^2+6q+5}-\frac{3q^2-q-6}{q^2+6q+5}=\frac{4q^2-q+3-3q^2+q+6}{q^2+6q+5}$

Simplifying:

$\boxed{\displaystyle \frac{4q^2-q+3}{q^2+6q+5}-\frac{3q^2-q-6}{q^2+6q+5}=\frac{q^2+9}{q^2+6q+5}}$

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