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3 years ago

Identifying Additive Inverses

Mathematics

1 answer:

mestny [16]3 years ago

4 0

Answer:

he additive inverse of:

a) $-6x^2-x-2$ is : $6x^2+x+2$

b) $6x^2-x+2$ is : $-6x^2+x-2$

c) $6x^2+x-2$ is : $-6x^2-x+2$

d) $6x^2+x+2$ is : $-6x^2-x-2$

Step-by-step explanation:

You need to consider that the additive inverse of a polynomial is that polynomial that consists of the opposite of each term of the polynomial given.

Then, the additive inverse of:

a) $-6x^2-x-2$ is : $6x^2+x+2$

b) $6x^2-x+2$ is : $-6x^2+x-2$

c) $6x^2+x-2$ is : $-6x^2-x+2$

d) $6x^2+x+2$ is : $-6x^2-x-2$

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Solve the system of equations by substitution. 3/8 x + 1/3 y =17/24 and <br> x + 7y = 8

goblinko [34]

$\left \{ {{ \frac{3}{8}x+ \frac{1}{3}y = \frac{17}{14} } \atop {x+7y=8}} \right.$

To solve this system by substitution, first isolate x in the second equation.

$x+7y=8$
$x+7y-7y=8-7y$
$x=8-7y$

Now, plug this expression ( $8-7y$ ) for x in the top equation to solve for y.

$\frac{3}{8} (8-7y)+ \frac{1}{3} y= \frac{17}{14}$
$3- \frac{21}{8} y+ \frac{1}{3} y= \frac{17}{24}$
$72-63y+8y=17$
$72-55y=17$
$-55y=17-72$
$-55y=-55$
$y=1$

Now that you have y, plug it into the second equation and solve for x.

$x+7y=8$
$x+7(1)=8$
$x+7=8$
$x=1$

Last step is to plug your x- and y-values in to both equations to check your work.

$\frac{3}{8} (1)+ \frac{1}{3} y= \frac{17}{24}$

$\frac{3}{8} * \frac{3}{3} = \frac{9}{24} ; \frac{1}{3} * \frac{8}{8} = \frac{8}{24}$

$\frac{9}{24} + \frac{8}{24} = \frac{17}{24}$ <--True

$1+7(1)=8$
$1+7=8$ <--True

Answer:
$x=1 \\ y=1$

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Katen [24]

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Elodia [21]

Answer:

Step-by-step explanation:

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Nastasia [14]

Hey there! I'm happy to help!

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tino4ka555 [31]

Step-by-step explanation

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