Verified answer

Thanks for adding the parentheses!

1) No real solution

Put everything under a common denominator:

x / (x − 6) + 8 /( x − 4) = x2 / (x2 − 10x + 24)

[x(x - 4) + 8(x - 6)]/[(x - 6)(x - 4)] = x^2/(x^2 − 10x + 24)

(x^2 - 4x + 8x - 24)/(x^2 - 10x + 24) = x^2/(x^2 − 10x + 24)

x^2 + 4x - 24 = x^2

4x = 24

x = 6

Check by substituting back into the original equation:

6/(6 - 6) + 8/(6 - 4) = 6^2/(6^2 - 10*6 + 24)

The first term has a denominator of 0, therefore x ≠ 6. Therefore, the equation has no real solution.

2)

Same thing as 1):

2 / (y − 8) + y / (y − 5) = −3 / (y2 − 13y + 40)

[2(y - 8) + y(y - 8)]/[(y - 8)(y - 5)] = −3/(y^2 − 13y + 40)

(2y - 16 + y^2 - 8y)/(y^2 - 13 y + 40) = −3/(y^2 − 13y + 40)

y^2 - 6y - 16 = -3

y^2 - 6y - 13 = 0

y = [-b ± √(b^2 - 4ac)] / (2a)

= [-(-6) ± √((-6)^2 - 4*1*-13)] / (2*1)

= [6 ± √(36 + 52)] / 2

= [6 ± √(88)] / 2

= [6 ± 2√(11)] / 2

= 3 ± √(11)

Substitute back into the original equation to check for extraneous solutions.

3) x = {5/3, 4}

4x = −80 / (3x − 17)

x = -20/(3x - 17)

x(3x - 17) = -20

3x^2 - 17x + 20 = 0

(3x - 5)(x - 4) = 0

x = {5/3, 4}

Substitute back into the original equation to check for extraneous solutions:

For x = 5/3:

4(5/3) = -80/(3(5/3) - 17)

20/3 = -80/(5 - 17)

20/3 = -80/-12

20/3 = 20/3

Therefore, x = 5/3 is a solution to the equation.

For x = 4:

4(4) = -80/(3*4 - 17)

16 = -80/-5

16 = 16

Therefore, x = {5/3, 4}

4)

12 + 6/t = 16 - 3/t

6/t + 3/t = 16 - 12

9/t = 4

9 = 4t

t = 9/4

Check by substituting.

Hope I helped!

Could you include parentheses for denominators in the equation? For instance, is x/x-6

x/x -6

or

x/(x-6)

and so on.