5y-4-40y2-16=1

Rewrite 16 as 42.

5y-4-40y2-42=1

Since both terms are perfect squares, factor using the difference of squares formula, a2-b2=(a+b)(a-b) where a=y and b=4.

5y-4-40(y+4)(y-4)=1

5y-4-40(y+4)(y-4)=1

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

y-4,(y+4)(y-4),1

The LCM is the smallest positive number that all of the numbers divide into evenly.

1. List the prime factors of each number.

2. Multiply each factor the greatest number of times it occurs in either number.

The number 1 is not a prime number because it only has one positive factor, which is itself.

Not prime

The LCM of 1,1,1 is the result of multiplying all prime factors the greatest number of times they occur in either number.

1

The factor for y-4 is y-4 itself.

(y-4)=y-4

(y-4) occurs 1 time.

The factor for y+4 is y+4 itself.

(y+4)=y+4

(y+4) occurs 1 time.

The factor for y-4 is y-4 itself.

(y-4)=y-4

(y-4) occurs 1 time.

The LCM of y-4,y+4,y-4 is the result of multiplying all factors the greatest number of times they occur in either term.

(y-4)(y+4)

(y-4)(y+4)

Multiply each term in 5y-4-40(y+4)(y-4)=1 by (y-4)(y+4) in order to remove all the denominators from the equation.

5y-4⋅((y-4)(y+4))-40(y+4)(y-4)⋅((y-4)(y+4))=1⋅((y-4)(y+4))

Simplify 5y-4⋅((y-4)(y+4))-40(y+4)(y-4)⋅((y-4)(y+4)).

Simplify each term.

Cancel the common factor of y-4.

Cancel the common factor.

5y-4⋅((y-4)(y+4))-40(y+4)(y-4)⋅((y-4)(y+4))=1⋅((y-4)(y+4))

Rewrite the expression.

5⋅(y+4)-40(y+4)(y-4)⋅((y-4)(y+4))=1⋅((y-4)(y+4))

5⋅(y+4)-40(y+4)(y-4)⋅((y-4)(y+4))=1⋅((y-4)(y+4))

Apply the distributive property.

5y+5⋅4-40(y+4)(y-4)⋅((y-4)(y+4))=1⋅((y-4)(y+4))

Multiply 5 by 4.

5y+20-40(y+4)(y-4)⋅((y-4)(y+4))=1⋅((y-4)(y+4))

Cancel the common factor of (y-4)(y+4).

Move the leading negative in -40(y+4)(y-4) into the numerator.

5y+20+-40(y+4)(y-4)⋅((y-4)(y+4))=1⋅((y-4)(y+4))

Factor (y-4)(y+4) out of (y+4)(y-4).

5y+20+-40(y-4)(y+4)(1)⋅((y-4)(y+4))=1⋅((y-4)(y+4))

Cancel the common factor.

5y+20+-40(y-4)(y+4)⋅1⋅((y-4)(y+4))=1⋅((y-4)(y+4))

Rewrite the expression.

5y+20-40=1⋅((y-4)(y+4))

5y+20-40=1⋅((y-4)(y+4))

5y+20-40=1⋅((y-4)(y+4))

Subtract 40 from 20.

5y-20=1⋅((y-4)(y+4))

5y-20=1⋅((y-4)(y+4))

Simplify 1⋅((y-4)(y+4)).

Multiply (y-4)(y+4) by 1.

5y-20=(y-4)(y+4)

Expand (y-4)(y+4) using the FOIL Method.

Apply the distributive property.

5y-20=y(y+4)-4(y+4)

Apply the distributive property.

5y-20=y⋅y+y⋅4-4(y+4)

Apply the distributive property.

5y-20=y⋅y+y⋅4-4y-4⋅4

5y-20=y⋅y+y⋅4-4y-4⋅4

Simplify terms.

Combine the opposite terms in y⋅y+y⋅4-4y-4⋅4.

Reorder the factors in the terms y⋅4 and -4y.

5y-20=y⋅y+4y-4y-4⋅4

Subtract 4y from 4y.

5y-20=y⋅y+0-4⋅4

Add y⋅y and 0.

5y-20=y⋅y-4⋅4

5y-20=y⋅y-4⋅4

Simplify each term.

Multiply y by y.

5y-20=y2-4⋅4

Multiply -4 by 4.

5y-20=y2-16

5y-20=y2-16

5y-20=y2-16

5y-20=y2-16

5y-20=y2-16

Subtract y2 from both sides of the equation.

5y-20-y2=-16

Move 16 to the left side of the equation by adding it to both sides.

5y-20-y2+16=0

Add -20 and 16.

5y-y2-4=0

Factor the left side of the equation.

Factor -1 out of 5y-y2-4.

Reorder 5y and -y2.

-y2+5y-4=0

Factor -1 out of -y2.

-(y2)+5y-4=0

Factor -1 out of 5y.

-(y2)-(-5y)-4=0

Rewrite -4 as -1(4).

-(y2)-(-5y)-1⋅4=0

Factor -1 out of -(y2)-(-5y).

-(y2-5y)-1⋅4=0

Factor -1 out of -(y2-5y)-1(4).

-(y2-5y+4)=0

-(y2-5y+4)=0

Factor.

Factor y2-5y+4 using the AC method.

Consider the form x2+bx+c. Find a pair of integers whose product is c and whose sum is b. In this case, whose product is 4 and whose sum is -5.

-4,-1

Write the factored form using these integers.

-((y-4)(y-1))=0

-((y-4)(y-1))=0

Remove unnecessary parentheses.

-(y-4)(y-1)=0

-(y-4)(y-1)=0

-(y-4)(y-1)=0

Multiply each term in -(y-4)(y-1)=0 by -1

Multiply each term in -(y-4)(y-1)=0 by -1.

(-(y-4)(y-1))⋅-1=0⋅-1

Simplify (-(y-4)(y-1))⋅-1.

Simplify by multiplying through.

Apply the distributive property.

(-y–4)(y-1)⋅-1=0⋅-1

Multiply -1 by -4.

(-y+4)(y-1)⋅-1=0⋅-1

(-y+4)(y-1)⋅-1=0⋅-1

Expand (-y+4)(y-1) using the FOIL Method.

Apply the distributive property.

(-y(y-1)+4(y-1))⋅-1=0⋅-1

Apply the distributive property.

(-y⋅y-y⋅-1+4(y-1))⋅-1=0⋅-1

Apply the distributive property.

(-y⋅y-y⋅-1+4y+4⋅-1)⋅-1=0⋅-1

(-y⋅y-y⋅-1+4y+4⋅-1)⋅-1=0⋅-1

Simplify and combine like terms.

Simplify each term.

Multiply y by y by adding the exponents.

Move y.

(-(y⋅y)-y⋅-1+4y+4⋅-1)⋅-1=0⋅-1

Multiply y by y.

(-y2-y⋅-1+4y+4⋅-1)⋅-1=0⋅-1

(-y2-y⋅-1+4y+4⋅-1)⋅-1=0⋅-1

Multiply -y⋅-1.

Multiply -1 by -1.

(-y2+1y+4y+4⋅-1)⋅-1=0⋅-1

Multiply y by 1.

(-y2+y+4y+4⋅-1)⋅-1=0⋅-1

(-y2+y+4y+4⋅-1)⋅-1=0⋅-1

Multiply 4 by -1.

(-y2+y+4y-4)⋅-1=0⋅-1

(-y2+y+4y-4)⋅-1=0⋅-1

Add y and 4y.

(-y2+5y-4)⋅-1=0⋅-1

(-y2+5y-4)⋅-1=0⋅-1

Apply the distributive property.

-y2⋅-1+5y⋅-1-4⋅-1=0⋅-1

Simplify.

Multiply -y2⋅-1.

Multiply -1 by -1.

1y2+5y⋅-1-4⋅-1=0⋅-1

Multiply y2 by 1.

y2+5y⋅-1-4⋅-1=0⋅-1

y2+5y⋅-1-4⋅-1=0⋅-1

Multiply -1 by 5.

y2-5y-4⋅-1=0⋅-1

Multiply -4 by -1.

y2-5y+4=0⋅-1

y2-5y+4=0⋅-1

y2-5y+4=0⋅-1

Multiply 0 by -1.

y2-5y+4=0

y2-5y+4=0

Factor y2-5y+4 using the AC method.

Consider the form x2+bx+c. Find a pair of integers whose product is c and whose sum is b. In this case, whose product is 4 and whose sum is -5.

-4,-1

Write the factored form using these integers.

(y-4)(y-1)=0

(y-4)(y-1)=0

If any individual factor on the left side of the equation is equal to 0, the entire expression will be equal to 0.

y-4=0

y-1=0

Set the first factor equal to 0 and solve.

Set the first factor equal to 0.

y-4=0

Add 4 to both sides of the equation.

y=4

y=4

Set the next factor equal to 0 and solve.

Set the next factor equal to 0.

y-1=0

Add 1 to both sides of the equation.

y=1

y=1

The final solution is all the values that make (y-4)(y-1)=0 true.

y=4,1

y=4,1

Exclude the solutions that do not make 5y-4-40y2-16=1 true.

y=1

Solve for y 5/(y-4)-40/(y^2-16)=1