How do you solve for x in #8ax-7a^2=19a^2-5ax#?

Answer 1

#x = 2a#

#8ax - 7a^2 = 19a^2 - 5ax#

Collecting like terms;

Note: When crossing over a Negative value or unknown in a algebraic expression, the sign changes the the Positive value, also when crossing over a Positive value or uknown the sign changes to a Negative value, (Vice-Versa)

#8ax + 5ax = 19a^2 + 7a^2#

Simplifying;

#13ax = 26a^2#
Dividing both sides by the coefficient of #x#;
#(13ax)/(13a) = (26a^2)/(13a)#
#(cancel(13a)x)/cancel(13a) = (26a^2)/(13a)#
#x = (26a^2)/(13a)#
#x = (2a xx 13a)/(13a)#
#x = (2a xx cancel(13a))/cancel(13a)#
#x = (2a)/1#
#x = 2a#
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Answer 2

#x=2a#

#"collect terms in x together on the left side and other"# #"terms on the right side"#
#"add "5ax" to both sides"#
#8ax+5ax-7a^2=19a^2cancel(-5x)cancel(+5x)#
#13ax-7a^2=19a^2#
#"add "7a^2" to both sides"#
#13ax=19a^2+7a^2#
#13ax=26a^2#
#"divide both sides by "13a#
#(cancel(13a) x)/cancel(13a)=(26a^2)/(13a)rArrx=2a#
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Answer 3

To solve the equation 8ax - 7a^2 = 19a^2 - 5ax, follow these steps:

  1. Group like terms: 8ax + 5ax = 19a^2 + 7a^2

  2. Combine like terms: 13ax = 26a^2

  3. Divide both sides by 13a: x = 2a

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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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