How do you solve #5y+3y+16=48#?

Answer 1

Julia Adams

#y=4#

#5y+3y+16=48#

assemble similar terms on LHS

#8y+16=48#

subtract #16# from both sides

#8y+cancel(16-16)=48-16#

#8y=32#

#:.y=32/8#

#y=4#

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Answer 2

Kennedy Adams

#y=4#

#5y+3y+16=48#

#rArr8y+16=48#

#"subtract 16 from both sides"#

#8ycancel(+16)cancel(-16)=48-16#

#rArr8y=32#

#"divide both sides by 8"#

#(cancel(8) y)/cancel(8)=32/8#

#rArry=4" is the solution"#

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Answer 3

Avery Alexander

To solve the equation 5y + 3y + 16 = 48, you would first combine like terms on the left side of the equation. This gives you 8y + 16 = 48. Next, you would subtract 16 from both sides of the equation to isolate the term with y. This gives you 8y = 32. Finally, you would divide both sides by 8 to solve for y. So, y = 4.

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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.