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

2 Answers
May 31, 2018

Answer:

#x = 2a#

Explanation:

#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#

May 31, 2018

Answer:

#x=2a#

Explanation:

#"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#