Question

В)1/?+4/?=?/6. 2/?+5/?=?/11

Answer 1

Відповідь:

8 . 9 . 3 . 6

Покрокове пояснення:

Answer 2

To solve these equations, let's assign variables to the unknowns. Let's say the first unknown is a, and the second unknown is b.

1/a + 4/b = 1/6 ...(Equation 1) 2/a + 5/b = 1/11 ...(Equation 2)

We have a system of two equations with two unknowns. To solve this system, we can use the method of substitution.

From Equation 1, we can rewrite it as: 4/b = 1/6 - 1/a

Multiplying both sides by 6ab to clear the denominators, we get: 24a = ab - 6b

Rearranging the terms, we have: ab - 24a - 6b = 0 ...(Equation 3)

Now, let's solve Equation 3 for b in terms of a: b(a - 6) = 24a b = 24a / (a - 6)

Substitute this value of b in Equation 2: 2/a + 5/(24a / (a - 6)) = 1/11

To simplify this equation, we can multiply both sides by 11a(a - 6) to clear the denominators: 22(a - 6) + 55a = a(a - 6)

Expanding and simplifying: 22a - 132 + 55a = a^2 - 6a

Rearranging the terms: a^2 - 83a + 132 = 0

Now we can solve this quadratic equation for a. Factoring it, we have: (a - 4)(a - 33) = 0

So, a = 4 or a = 33.

If a = 4, substitute it back into Equation 3 to find b: (4)b - 24(4) - 6b = 0 4b - 96 - 6b = 0 -2b = 96 b = -48

If a = 33, substitute it back into Equation 3 to find b: (33)b - 24(33) - 6b = 0 33b - 792 - 6b = 0 27b = 792 b = 792/27 b = 88/3

Therefore, the solutions to the system of equations are: a = 4, b = -48 a = 33, b = 88/3