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How do you solve #(1+2y)/(y-4 )= (4y^2+5y)/(2y^2-7y-4)#?

Answer 1

Ethan Adkins

Have a look:

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Answer 2

Ethan Adkins

#(1+2y)/(y−4)=(4y^2+5y)/(2y^2−7y−4)#

#(4y^2+5y)/color(red)(2y^2-8y+y-4)#

#(4y^2+5y)/[color(red)(2y(y-4)+1(y-4)]#

#=(4y^2+5y)/[(1+2y)(y-4)]#

#(1+2y)/color(red)cancel(y−4)=(4y^2+5y)/[(1+2y)color(red)cancel((y-4)]#

#color(red)[(1+2y)^2]=4y^2+5y#

By expansion: #->(a+b)^2=a^2+2ab+b^2#

#color(red)[1+4y+cancel(4y^2)]=cancel(4y^2)+5y#

ANSWER #y=1#

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Answer 3

Jace Anderson

To solve the equation (1+2y)/(y-4) = (4y^2+5y)/(2y^2-7y-4), we can start by cross-multiplying to eliminate the fractions. This gives us (1+2y)(2y^2-7y-4) = (y-4)(4y^2+5y). Expanding both sides of the equation, we get 2y^3 - 7y^2 - 4y + 4y^2 - 14y - 8 = 4y^3 + 5y^2 - 16y^2 - 20y. Combining like terms, we have 2y^3 - 3y^2 - 18y - 8 = 4y^3 - 11y^2 - 20y. Rearranging the equation, we get 0 = 2y^3 - 8y^2 + 2y - 8. Simplifying further, we have 0 = y^3 - 4y^2 + y - 4. To solve this cubic equation, we can use various methods such as factoring, synthetic division, or numerical methods like Newton's method.

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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.