Question

If x − 2y = 5 and xy = 2, find:
i. x^3 − 8y^3.
ii. x^2 + 4y^2

Answer 1

Answer:

First, use the simultaneous equation method to find x and y by letting the first equation be 3x - 2y = 5 and the second equation be xy = 6.

3x - 2y = 5 (1st equation)

xy = 6 (2nd equation)

In the 2nd equation, move y over to 6 by dividing.

x = \frac{6}{y}y6

Then, substitute x into the 1st equation.

3(\frac{6}{y}y6 ) - 2y = 5

Move 2y over to 5.

3(\frac{6}{y}y6 ) = 5 + 2y

Multiply 3 with the numerator of the fraction, 6.

\frac{18}{y}y18 = 5 + 2y

Move the y on the left side of the equation to the right and multiply.

18 = y(5 + 2y)

18 = 5y + 2y²

Leave the equation in general form by letting the values equal to 0.

2y² + 5y - 18 = 0

Factorise the equation.

(y - 2)(2y + 9) = 0

Solve for the values of y.

y - 2 = 0

y = 2

2y + 9 = 0

2y = -9

y = -9/2

Substitute the values of y into the 2nd equation and solve for x.

x(2) = 6

2x = 6

x = 3

x(-9/2) = 6

x = 6 ÷ -9/2

x = 6 × -2/9

x = -4/3

Now, we will find the values of 27x³ - 8y³.

First, substitute the known values of x and y into the equation.

27(3)³ - 8(2)³

27(-4/3)³ - 8(-9/2)³

Find each of the values to the power of 3.

27(-27) - 8(-8)

27(-64/27) - 8(-729/8)

Multiply 27 and 8 with the numbers in the brackets beside them.

-729 + 64

-64 + 729

Add the numbers.

-665

665

Therefore, the values are -665 and 665.