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How do you solve the equation #abs(2x+6)=12#?

Answer 1

Genesis Allen

#x=3 orx=-9#

Here, #|2x+6|=12#

#=>2x+6=12 or 2x+6=-12#

When (-6) is added to each equation's two sides,

#=>2x+cancel(6)+(cancel((-)6)=12+(-6)=12-6#

#or2x+cancel(6)+(cancel(-6))=-12+(-6)=-12-6#

#=>2x=6 or 2x=-18#

When we divide each equation's two sides by 2, we obtain

#(cancel(2)x)/cancel(2)=cancel(6)^3/cancel(2) or(cancel(2)x)/cancel(2)=-cancel(18)^9/cancel(2)#

#x=3 orx=-9#

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Answer 2

Jameson Allen

To solve the equation ( |2x + 6| = 12 ), we first isolate the absolute value expression and then solve for ( x ).

Set up two equations: ( 2x + 6 = 12 ) and ( 2x + 6 = -12 )
Solve each equation separately: For ( 2x + 6 = 12 ), subtract 6 from both sides, then divide by 2. For ( 2x + 6 = -12 ), subtract 6 from both sides, then divide by 2.
Solve for ( x ) in each equation: For ( 2x + 6 = 12 ), ( 2x = 6 ), so ( x = 3 ). For ( 2x + 6 = -12 ), ( 2x = -18 ), so ( x = -9 ).

Therefore, the solutions to the equation ( |2x + 6| = 12 ) are ( x = 3 ) and ( x = -9 ).

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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.