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3 years ago

10

Solve by elimination -5x+3y=20 5x-6y=-10

1 answer:

allochka39001 [22]3 years ago

6 0

Y= -10/3
X= -6

Steps:
-3y=10
Y=-10/3

-5x+3(-10/3)=20
X=-6

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$\displaystyle \mathbf{\frac{5}{(c+4)}}$

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<u>Division of Rational Expressions</u>

When dividing two rational expressions like f(c) / g(c), it's usually easier to multiply the numerator by the reciprocal of the denominator: f(c) * (1/g(c)). The reciprocal is just flipping the numerator and the denominator.

Let's find the following division:

$\displaystyle \frac{\frac{c^2-64}{3c^2+26c+16}} {\frac{c^2-4c-32}{15c+10}}$

First, we factor the polynomials where possible.

$c^2-64 = (c-8)(c+8)$

$3c^2+26c+16=(3c+2)(c+8)$

$c^2-4c-32=(c-8)(c+4)$

$15c+10=5(3c+2)$

Substituting:

$\displaystyle \frac{\frac{c^2-64}{3c^2+26c+16}} {\frac{c^2-4c-32}{15c+10}}=\frac{\frac{(c-8)(c+8)}{(3c+2)(c+8)}} {\frac{(c-8)(c+4)}{5(3c+2)}}$

Multiplying by the reciprocal of the denominator:

$\displaystyle \frac{\frac{c^2-64}{3c^2+26c+16}} {\frac{c^2-4c-32}{15c+10}}=\frac{(c-8)(c+8)}{(3c+2)(c+8)}\cdot \frac{5(3c+2)}{(c-8)(c+4)}$

Simplifying by (c+8)(3c+2)(c-8):

$\displaystyle \frac{\frac{c^2-64}{3c^2+26c+16}} {\frac{c^2-4c-32}{15c+10}}=\frac{5}{(c+4)}$

Answer:

$\displaystyle \mathbf{\mathbf{\frac{5}{(c+4)}}}$

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