Question

Case. 4x + 5y = 3, ax-7y=11.

Answer 1

4x + 5y = 3, у = (3 – 4х)/5 = (3/5) – (4/5)х.

ax-7y=11.

ax-7((3/5) – (4/5)х) = 11,

ах + (28/5)х = 11 + (21/5),

((5а + 28)/5)х = (76/5),

(5а + 28)х = 76,

x = 76/(5a + 28).

Есть точка разрыва полученного выражения при знаменателе, равном нулю:

5a + 28 = 0

а = -28/5 = -5,6.

Answer 2

To solve the system of equations:

1. 4x + 5y = 3
2. ax - 7y = 11

We'll use the method of substitution to eliminate one variable and solve for the other.

Let's solve equation 2) for x:

ax - 7y = 11

Rearranging the equation, we have:

ax = 7y + 11

Dividing through by 'a', we get:

x = (7y + 11) / a

Now substitute this value of x into equation 1):

4((7y + 11) / a) + 5y = 3

Multiply both sides of the equation by 'a' to eliminate the fraction:

4(7y + 11) + 5ay = 3a

Expand and simplify:

28y + 44 + 5ay = 3a

Rearrange the terms:

28y + 5ay = 3a - 44

Factor out 'y':

y(28 + 5a) = 3a - 44

Now, there are two cases to consider:

Case 1: When (28 + 5a) ≠ 0

In this case, we can divide both sides of the equation by (28 + 5a) to solve for y:

y = (3a - 44) / (28 + 5a)

Substitute this value of y back into equation 2) to solve for x:

x = (7((3a - 44) / (28 + 5a))) / a

Simplifying further, we have:

x = (21a - 308) / (28 + 5a)

So the solutions for x and y in terms of 'a' are:

x = (21a - 308) / (28 + 5a) y = (3a - 44) / (28 + 5a)

Case 2: When (28 + 5a) = 0

In this case, (28 + 5a) = 0 implies a = -28/5.

If a = -28/5, substitute this value into equation 2) to solve for y:

(-28/5)x - 7y = 11 -28x - 35y = 55

Since the equation -28x - 35y = 55 is a linear equation, it can't be satisfied simultaneously with the equation 4x + 5y = 3. Therefore, there are no solutions in this case.

In summary, the solutions to the system of equations 4x + 5y = 3 and ax - 7y = 11 are:

When (28 + 5a) ≠ 0: x = (21a - 308) / (28 + 5a) y = (3a - 44) / (28 + 5a)

When (28 + 5a) = 0: There are no solutions.