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Given, (2x + y - 3z)^{2} We know

(a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2ab + 2bc + 2ca

=> (2x + y - 3z)^{2 }= (2x)^{2} + (y)^{2} + (-3z)^{2} + 2 * 2x * y + 2 * y * (- 3z) + 2 * (- 3z) * 2x

= 4x^{2} + y^{2} + 9z^{ 2} + 4xy - 6yz - 6zx

=> (2x + y - 3z)^{2} = 4x^{2} + y^{2} + 9z^{ 2} + 4xy - 6yz - 6zx

(a + b + c)

=> (2x + y - 3z)

= 4x

=> (2x + y - 3z)

12x + 3y = 16

Given, 12x + 3y = 16

Step 1:

Subtract 12x from both sides

=> 12x + 3y -12x = 16 - 12x

=> 3y = 16 - 12x

Step 2:

Divide each side by 3

=> $\frac{3y}{3} = \frac{16 - 12x}{3}$

=> y = $ \frac{16 - 12x}{3}$

Step 3:

Put x = 1

=> y = $ \frac{16 - 12 * 1}{3}$

=> y = $ \frac{16 - 12}{3}$

=> y = $ \frac{4}{3}$, is the answer.

Step 1:

Subtract 12x from both sides

=> 12x + 3y -12x = 16 - 12x

=> 3y = 16 - 12x

Step 2:

Divide each side by 3

=> $\frac{3y}{3} = \frac{16 - 12x}{3}$

=> y = $ \frac{16 - 12x}{3}$

Step 3:

Put x = 1

=> y = $ \frac{16 - 12 * 1}{3}$

=> y = $ \frac{16 - 12}{3}$

=> y = $ \frac{4}{3}$, is the answer.

Two numbers are 36 and 24

GCD(24, 36) = 12

LCM = ?

We know that

Product of two numbers = Product of their LCM and GCD

=> 24 * 36 = LCM * 12

=> LCM = $\frac{24 * 36}{12}$

=>** LCM = 72**

GCD(24, 36) = 12

LCM = ?

We know that

Product of two numbers = Product of their LCM and GCD

=> 24 * 36 = LCM * 12

=> LCM = $\frac{24 * 36}{12}$

=>

Length of a rectangular garden = 25

width of a rectangular garden = 20

Perimeter of rectangle = 2(Length + width)

=> Perimeter of rectangular garden = 2(25 + 20)

= 2(45)

= 90

Hence,**Perimeter of rectangular garden is 90 feet**

width of a rectangular garden = 20

Perimeter of rectangle = 2(Length + width)

=> Perimeter of rectangular garden = 2(25 + 20)

= 2(45)

= 90

Hence,

Let first integer = x

Second integer = x + 1

According to statement:

=> 10(x + 1) + 3x = 36

Solve for x,

=> 10x + 10 + 3x = 36

=> 13x + 10 = 36

=> 13x = 36 - 10

=> 13x = 26

=> x = $\frac{26}{13}$

=> x = 2

The first integer = 2

Second integer = 2 + 1 = 3

Second integer = x + 1

According to statement:

=> 10(x + 1) + 3x = 36

Solve for x,

=> 10x + 10 + 3x = 36

=> 13x + 10 = 36

=> 13x = 36 - 10

=> 13x = 26

=> x = $\frac{26}{13}$

=> x = 2

The first integer = 2

Second integer = 2 + 1 = 3

Let first number = x

and second number = y

The problem states:

x + y = 10 ..........................(1)

x - y = 4 ............................(2)

Adding equation(1) and (2)

=> 2x = 14

=> x = 7, put in (1)

=> 7 + y = 10

Subtract 7 from both sides

=> 7 + y - 7 = 10 - 7

=> y = 3

Hence,**First number = 7**

Second number = 3

and second number = y

The problem states:

x + y = 10 ..........................(1)

x - y = 4 ............................(2)

Adding equation(1) and (2)

=> 2x = 14

=> x = 7, put in (1)

=> 7 + y = 10

Subtract 7 from both sides

=> 7 + y - 7 = 10 - 7

=> y = 3

Hence,

Second number = 3