2t=t4t-6

Multiply the numerator of the first fraction by the denominator of the second fraction. Set this equal to the product of the denominator of the first fraction and the numerator of the second fraction.

2⋅(4t-6)=t⋅t

Since t is on the right side of the equation, switch the sides so it is on the left side of the equation.

t⋅t=2⋅(4t-6)

Multiply t by t.

t2=2⋅(4t-6)

Simplify 2⋅(4t-6).

Apply the distributive property.

t2=2(4t)+2⋅-6

Multiply.

Multiply 4 by 2.

t2=8t+2⋅-6

Multiply 2 by -6.

t2=8t-12

t2=8t-12

t2=8t-12

Subtract 8t from both sides of the equation.

t2-8t=-12

Move 12 to the left side of the equation by adding it to both sides.

t2-8t+12=0

Factor t2-8t+12 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 12 and whose sum is -8.

-6,-2

Write the factored form using these integers.

(t-6)(t-2)=0

(t-6)(t-2)=0

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

t-6=0

t-2=0

Set the first factor equal to 0 and solve.

Set the first factor equal to 0.

t-6=0

Add 6 to both sides of the equation.

t=6

t=6

Set the next factor equal to 0 and solve.

Set the next factor equal to 0.

t-2=0

Add 2 to both sides of the equation.

t=2

t=2

The final solution is all the values that make (t-6)(t-2)=0 true.

t=6,2

t=6,2

Solve for t 2/t=t/(4t-6)