How do you solve #abs(4/3x+7)+5=14#?

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Answer 1 · 2015-03-30T14:20:50

Remembering that:

#|f(x)|=n# (#n# positive number), than #f(x)=+-n#, ...

We can say:

#|4/3x+7|=9rArr4/3x+7=+-9rArr4/3x=-7+-9rArr#

#x=3/4(-7+-9)rArr#

#x_1=3/4(-7-9)=3/4*(-16)=-12#

#x_2=3/4(-7+9)=3/4*2=3/2.#