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$\sqrt{2x + 3a}$ = 5

Step 1:

Given equation

$\sqrt{2x + 3a}$ = 5

Step 2:

Squaring both side to remove the radical

=> 2x + 3a = 25

=> 3a = 25 - 2x ......................(1)

Step 3:

Substituting x = 2 in equation (1),

=> 3a = 25 - 2 * 2

=> 3a = 25 - 4 = 21

=>** a = 7**

Given equation

$\sqrt{2x + 3a}$ = 5

Step 2:

Squaring both side to remove the radical

=> 2x + 3a = 25

=> 3a = 25 - 2x ......................(1)

Step 3:

Substituting x = 2 in equation (1),

=> 3a = 25 - 2 * 2

=> 3a = 25 - 4 = 21

=>

Given quadratic equation 6x^{2} + 25x - 25 = 0

=> 6x^{2} + 25x - 25 = 0

Factors the left hand side of the equation,

=> 6x^{2} + 30x - 5x - 25 = 0

=> 6x(x + 5) - 5(x + 5) = 0

=> (6x - 5)(x + 5) = 0

=> either 6x - 5 = 0 or x + 5 = 0

=> 6x = 5

=> x = $\frac{5}{6}$

or x = - 5

=> Solutions of equation are** x = 0.83, - 5.**

=> 6x

Factors the left hand side of the equation,

=> 6x

=> 6x(x + 5) - 5(x + 5) = 0

=> (6x - 5)(x + 5) = 0

=> either 6x - 5 = 0 or x + 5 = 0

=> 6x = 5

=> x = $\frac{5}{6}$

or x = - 5

=> Solutions of equation are

- 3 + 5x = y

2x + 5y = 12

Given

- 3 + 5x = y

2x + 5y = 12

or

5x - y = 3 ................(1)

2x + 5y = 12 ...............(2)

Step 1:

No equation given solved for a variable, Equation (1) can be solved literal for y,

y = 5x - 3

Substituting the expression in equation (2)

=> 2x + 5(5x - 3) = 12

=> 2x + 25x - 15 = 12

=> 27x - 15 = 12

=> 27x = 12 + 15 = 27

=> x = 1

Step 2:

Substituting x = 1 in y = 5x - 3,

y = 5 x 1 - 3 = 5 - 3 = 2

=> y = 2

Hence the solution to the given system,** x = 1 and y = 2.**

- 3 + 5x = y

2x + 5y = 12

or

5x - y = 3 ................(1)

2x + 5y = 12 ...............(2)

Step 1:

No equation given solved for a variable, Equation (1) can be solved literal for y,

y = 5x - 3

Substituting the expression in equation (2)

=> 2x + 5(5x - 3) = 12

=> 2x + 25x - 15 = 12

=> 27x - 15 = 12

=> 27x = 12 + 15 = 27

=> x = 1

Step 2:

Substituting x = 1 in y = 5x - 3,

y = 5 x 1 - 3 = 5 - 3 = 2

=> y = 2

Hence the solution to the given system,

Let width of the rectangle = x

then length = 2x + 20

(Length be 20 more than twice its width)

Perimeter = 280

We know, Perimeter of rectangle = 2(length + width)

=> 2(length + width) = 280

=> 2(2x + 20 + x) = 280

=> 2(3x + 20) = 280

=> 6x + 40 = 280

=> 6x = 240

=> x = 40 feet

=> Width of the rectangle = 40 feet.

then length = 2x + 20

(Length be 20 more than twice its width)

Perimeter = 280

We know, Perimeter of rectangle = 2(length + width)

=> 2(length + width) = 280

=> 2(2x + 20 + x) = 280

=> 2(3x + 20) = 280

=> 6x + 40 = 280

=> 6x = 240

=> x = 40 feet

=> Width of the rectangle = 40 feet.

Let the present age of Stefy = x

Macshon is twice as old as Stefy

=> Age of Macshon = 2x

**Five years ago:**

Age of Stefy = x - 5

and age of Macshon = 2x - 5

The problem states:

Macshon three times as old as Stefy was

=> 2x - 5 = 3(x - 5 )

=> 2x - 5 = 3x - 15

=> x = 10, age of Stefy

So age of Macshon = 2 * 10 = 20

Hence Macshon is of 20.

Macshon is twice as old as Stefy

=> Age of Macshon = 2x

Age of Stefy = x - 5

and age of Macshon = 2x - 5

The problem states:

Macshon three times as old as Stefy was

=> 2x - 5 = 3(x - 5 )

=> 2x - 5 = 3x - 15

=> x = 10, age of Stefy

So age of Macshon = 2 * 10 = 20

Hence Macshon is of 20.