Let a and b be real numbers satisfying a^3 - 3ab^2 = 47 and b^3 - 3a^2 b = 52. find a^2 + b^2.

a²+b² = -99/2

Step-by-step explanation:

Since we are given two equations, this equations will be solved simultaneously to get a² and b²

a³ - 3ab² = 47 ... 1

b³ - 3a² b = 52... 2

From 1, a(a² - 3b²) = 47...3

From 2, b(b² - 3a²) = 52... 4

Adding 3 and 4, we have;

a²+b²-3b²-3a² = 99 (note that a and b will no longer be part of the equations as they have been factored out)

a²+b²-(3b²+3a²) = 99

(a²+b²) -3(b²+a²)= 99

Taking the difference we have

- 2(a²+b²) = 99

a²+b² = -99/2

+25 = x

step-by-step explanation:

here x is the hypotenuse of a right triangle whose legs are 7 and 24.   by the pythagorean theorem, 7² + 24² must equal x²:

x² = 49 + 576 = 625.   thus, x = +√625 = +25 = x

37/3

step-by-step explanation:

step 1: simplify both sides of the equation.

7x+8(x+

1

4

)=3(6x−9)−8

7x+(8)(x)+(8)(

1

4

)=(3)(6x)+(3)(−9)+−8(distribute)

7x+8x+2=18x+−27+−8

(7x+8x)+(2)=(18x)+(−27+−8)(combine like terms)

15x+2=18x+−35

15x+2=18x−35

tep 2: subtract 18x from both sides.

15x+2−18x=18x−35−18x

−3x+2=−35

step 3: subtract 2 from both sides.

−3x+2−2=−35−2

−3x=−37

step 4: divide both sides by -3.

−3x

−3

=

−37

−3

x=

37

3