How do you solve #5( x + 1) = 7- 2( 5x + 8)#?

Answer 1

By the usual algebraic manipulations, we get #x = -14/15.#

#5(x+1)=7 -2(5x+8)#

First let's distribute the constant factors:

#5 x + 5 = 7 - (10 x + 16) #

Now we subtract:

#5x + 5 = - 9 - 10 x#

Now we move all the x terms to one side and the constant terms to the other. Terms negate when they move to the other side because what we're really doing is adding the negative to both sides:

#5x + 10 x = -9 - 5 #

Now we add and subtract:

#15 x = -14 #

Now we divide both sides by 15 for our solution:

#x = -14/15#

Check:

#5(-14/15 + 1) = 5/15 = 1/3 #
#7 - 2(5(-14/15) +8) =7 - 2(-14/3 + 24/3) ##= 7 - 2(10/3) = 1/3 quad sqrt#
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Answer 2

To solve the equation 5(x + 1) = 7 - 2(5x + 8), follow these steps:

  1. Distribute the terms inside the parentheses: 5x + 5 = 7 - 10x - 16

  2. Combine like terms on both sides of the equation: 5x + 5 = -10x - 9

  3. Add 10x to both sides to move all x terms to one side: 5x + 10x + 5 = -9

  4. Combine like terms again: 15x + 5 = -9

  5. Subtract 5 from both sides: 15x = -14

  6. Divide both sides by 15 to isolate x: x = -14/15

So, the solution to the equation is x = -14/15.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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