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

Answer 1

#x = 0#

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

Simplify the left side by combining like terms:

#3x + 5 = x + 5#
Subtract #color(blue)x# from both sides: #3x + 5 quadcolor(blue)(-quadx) = x + 5 quadcolor(blue)(-quadx)#
#2x + 5 = 5#
Subtract #color(blue)5# from both sides: #2x + 5 quadcolor(blue)(-quad5) = 5 quadcolor(blue)(-quad5)#
#2x = 0#
Divide both sides by #color(blue)2#: #(2x)/color(blue)2 = 0/color(blue)2#
#x = 0#

Hope this helps!

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Answer 2

To solve the equation 8x - 2 - 5x + 7 = 1x + 5, you would first combine like terms on both sides of the equation. This gives you 3x + 5 = x + 5. Then, you would isolate the variable by subtracting x from both sides of the equation. This yields 2x + 5 = 5. Next, you would subtract 5 from both sides to isolate the variable term, resulting in 2x = 0. Finally, divide both sides by 2 to solve for x. This gives you x = 0.

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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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