How do you solve # |2x-3|=x+5# and find any extraneous solutions?

Question

Savannah Anderson · Accepted Answer

#x=8# or #x=-2/3#
(neither solution is extraneous)

Consider the two possibilities for #2x-3#: #{: ("if "2x-3 >= 0,color(white)("xxx"),"if "2x-3 < 0), (abs(2x-3)=x+5,,abs(2x-3)=x+5), (rarr 2x-3=x+5,,rarr 3-2x=x+5), (rarr x=8,,rarr-2=3x), (,,rarr x=-2/3) :}# Evaluating both results to check for extraneous solutions: #{: ("if " x=8,,color(white)("xxxxx"),"if "x=-2/3,), (LS,=abs(2x-3),,LS,=abs(2x-3)), (,=abs(2 * (8) -3),,,=abs( 2* (-2/3)-3)), (,=abs(15),,,=abs(-13/3)), (,=13,,,=13/3), (RS,=x+5,,RS,=x+5), (,=8+5,,,=-2/3+5), (,=13,,,=13/3), (LS,=RS,,LS,=RS), ("valid",,,"valid",) :}# Both solutions are valid (that is, neither is extraneous)

Isaiah Anthony · Answer

undefined To solve the equation $ |2x - 3| = x + 5 $ and identify any extraneous solutions, you need to consider two cases: when $ 2x - 3 $ is positive or zero, and when $ 2x - 3 $ is negative.

Case 1: $ 2x - 3 \geq 0 $
In this case, $ |2x - 3| = 2x - 3 $.
Solve $ 2x - 3 = x + 5 $ for $ x $ to find the solutions.

Case 2: $ 2x - 3 < 0 $
In this case, $ |2x - 3| = -(2x - 3) $.
Solve $ -(2x - 3) = x + 5 $ for $ x $ to find the solutions.

After finding the solutions for both cases, you need to check if any of the solutions make the original equation $ |2x - 3| = x + 5 $ undefined. If any solutions result in a division by zero in the original equation, they are extraneous solutions and should be discarded.

Genesis Allen · Answer

undefined To solve the equation $ |2x-3| = x + 5 $ and find any extraneous solutions, first, consider the cases where $ 2x - 3 \geq 0 $ and $ 2x - 3 < 0 $. Solve each case separately, then check for extraneous solutions by substituting the solutions back into the original equation and ensuring they satisfy it. If any solutions make the absolute value expression negative, they are extraneous.