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2 years ago

Solve the system of linear equation. 2x-5= 14 3x+2y = 40

2 answers:

8 0

X=9.5, Y=5.75. Step by Step Explanation is-
First take 2x-5=14
= 2x=14+5 ; 2x=19 ; x=19/2=9.5
Now take 3x+2y=40; We substitute the value of x with 9.5. So the equation is-
=3*9.5+2y=40 ; 2y+28.5=40 ; 2y=40-28.5=11.5 ; y=11.5/2=5.75.
Thank you and hope it helped

n200080 [17]2 years ago

5 0

Answer:

x=9.5, y=5.75

Step-by-step explanation:

2x-5=14, 3x+2y=40

focus on 2x-5=14 first:

2x-5=14 (add 5 to both sides) 2x=19(divide both sides by 2) x=9.5

Now subtitute x into the second equation and find y:

3(9.5)+2y=40(simplify) 28.5+2y=40(subtract 28.5 on both sides) 2y=11.5(divide both sides by 2) y=5.75

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Similarly, the third footballer can be born on any day, except $d_1$ and $d_2$ :

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2 years ago

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Step-by-step explanation:

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Read 2 more answers

Prove that if x is an positive real number such that x + x^-1 is an integer, then x^3 + x^-3 is an integer as well.

Shkiper50 [21]

Answer:

By closure property of multiplication and addition of integers,

If $x + \dfrac{1}{x}$ is an integer

∴ $\left ( x + \dfrac{1}{x} \right) ^3 = x^3 + \dfrac{1}{x^3} +3\cdot \left (x + \dfrac{1}{x} \right )$ is an integer

From which we have;

$x^3 + \dfrac{1}{x^3}$ is an integer

Step-by-step explanation:

The given expression for the positive integer is x + x⁻¹

The given expression can be written as follows;

$x + \dfrac{1}{x}$

By finding the given expression raised to the power 3, sing Wolfram Alpha online, we we have;

$\left ( x + \dfrac{1}{x} \right) ^3 = x^3 + \dfrac{1}{x^3} +3\cdot x + \dfrac{3}{x}$

By simplification of the cube of the given integer expressions, we have;

$\left ( x + \dfrac{1}{x} \right) ^3 = x^3 + \dfrac{1}{x^3} +3\cdot \left (x + \dfrac{1}{x} \right )$

Therefore, we have;

$\left ( x + \dfrac{1}{x} \right) ^3 - 3\cdot \left (x + \dfrac{1}{x} \right )= x^3 + \dfrac{1}{x^3}$

By rearranging, we get;

$x^3 + \dfrac{1}{x^3} = \left ( x + \dfrac{1}{x} \right) ^3 - 3\cdot \left (x + \dfrac{1}{x} \right )$

Given that $x + \dfrac{1}{x}$ is an integer, from the closure property, the product of two integers is always an integer, we have;

$\left ( x + \dfrac{1}{x} \right) ^3$ is an integer and $3\cdot \left (x + \dfrac{1}{x} \right )$ is also an integer

Similarly the sum of two integers is always an integer, we have;

$\left ( x + \dfrac{1}{x} \right) ^3 + \left(- 3\cdot \left (x + \dfrac{1}{x} \right ) \right )$ is an integer

$\therefore x^3 + \dfrac{1}{x^3} = \left ( x + \dfrac{1}{x} \right) ^3 - 3\cdot \left (x + \dfrac{1}{x} \right )= \left ( x + \dfrac{1}{x} \right) ^3 + \left(- 3\cdot \left (x + \dfrac{1}{x} \right ) \right )$ is an integer

From which we have;

$x^3 + \dfrac{1}{x^3}$ is an integer.

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